Class 12 Mathematics Chapter 20 Theoretical Probability Distribution

Key Concept: Binomial Distribution: Properties of Bernoulli's Trials

d) All trials must be independent and have a constant probability of success.
[Solution Description]
Bernoulli's trials must satisfy four conditions:
1. The trials must be finite.
2. The trials must be independent.
3. Each trial must have exactly two outcomes: success or failure.
4. The probability of success ($p$) must remain constant for each trial.
Among the options, only one correctly matches all these conditions.

Key Concept: Types of Random Variables, Continuous Random Variable

c) $\frac{3}{8}$
[Solution Description]
For $f(z)$ to be a valid probability density function, the integral over all possible values must equal 1:
$\int_{-\infty}^{\infty} f(z) \, dz = 1$
Since $f(z) = 0$ outside $[0, 2]$, we have:
$\int_{0}^{2} k z^2 \, dz = 1$
Evaluate the integral:
$k \left[ \frac{z^3}{3} \right]_0^2 = k \left( \frac{8}{3} - 0 \right) = \frac{8k}{3}$
Set this equal to 1 and solve for $k$:
$\frac{8k}{3} = 1 \implies k = \frac{3}{8}$

Key Concept: Bernoulli’s Trials Conditions

c) Probability of defect remains 0.05 for each item
[Solution Description]
For the production of items to be Bernoulli's trials, the following conditions must hold:
1. The trials must be independent.
2. The probability of success (defective item) must remain constant ($p = 0.05$) for each trial.
3. Each trial must have only two possible outcomes: defective or non-defective.
Among these, the most critical condition that defines Bernoulli's trials is the constant probability of success ($p$) across all trials.

Key Concept: Mean of Probability Distribution

c) The expected value of the random variable
[Solution Description]
The mean $\mu$ of a probability distribution represents the expected value or average value of the random variable $X$. It is calculated as the weighted sum of all possible values $x_i$ multiplied by their respective probabilities $p_i$.

Key Concept: Probability Density Function, Continuous Random Variable

c) $\frac{3}{8}$
[Solution Description]
To make $f(x)$ a valid PDF, the integral over its entire range must equal 1:
$\int_{0}^{2} kx^2 \, dx = 1$
Integrate $kx^2$ with respect to $x$:
$k \left[ \frac{x^3}{3} \right]_0^2 = 1 \implies k \left( \frac{8}{3} - 0 \right) = 1 \implies k = \frac{3}{8}$

Key Concept: Probability Distribution, Variance

b) $\frac{5}{9}$
[Solution Description]
First, verify that probabilities sum to 1:
$\sum_{y=1}^{3} \frac{y}{6} = \frac{1}{6} + \frac{2}{6} + \frac{3}{6} = 1$
Calculate the expectation $E(Y)$:
$E(Y) = \sum y P(y) = 1 \times \frac{1}{6} + 2 \times \frac{2}{6} + 3 \times \frac{3}{6} = \frac{1 + 4 + 9}{6} = \frac{14}{6} = \frac{7}{3}$
Next, find $E(Y^2)$:
$E(Y^2) = \sum y^2 P(y) = 1^2 \times \frac{1}{6} + 2^2 \times \frac{2}{6} + 3^2 \times \frac{3}{6} = \frac{1 + 8 + 27}{6} = \frac{36}{6} = 6$
Compute the variance:
$\sigma^2 = E(Y^2) - [E(Y)]^2 = 6 - \left( \frac{7}{3} \right)^2 = 6 - \frac{49}{9} = \frac{54}{9} - \frac{49}{9} = \frac{5}{9}$

Key Concept: Standard Normal Distribution

c) 0.8664
[Solution Description]
For a standard normal distribution, we can find $P(-1.5 < X < 1.5)$ using the standard normal table or calculator.
Step 1: Find $P(X < 1.5)$. From standard normal tables, this is approximately $0.9332$.
Step 2: Find $P(X < -1.5)$. Due to symmetry, this is equal to $1 - P(X < 1.5) = 1 - 0.9332 = 0.0668$.
Step 3: Calculate the required probability:
$P(-1.5 < X < 1.5) = P(X < 1.5) - P(X < -1.5) = 0.9332 - 0.0668 = 0.8664.$
So, the probability is approximately $0.8664$.

Key Concept: Binomial Distribution

b) $0.016$
[Solution Description]
For each question, the probability of guessing correctly ($p$) is $\frac{1}{4}$ and incorrectly ($q$) is $\frac{3}{4}$. Using the binomial formula for $n = 10$ and $r = 6$:
$P(X = 6) = \binom{10}{6} \left(\frac{1}{4}\right)^6 \left(\frac{3}{4}\right)^{4}$
Calculating the combination:
$\binom{10}{6} = 210$
Now compute the probability:
$P(X = 6) = 210 \times \left(\frac{1}{4}\right)^6 \times \left(\frac{3}{4}\right)^{4} \approx 210 \times 0.000244 \times 0.3164 \approx 0.01622$
The probability is approximately $0.01622$.

Key Concept: Mean and Variance

b) 0.2
[Solution Description]
The mean of a binomial distribution is $\mu = np$ and the variance is $\sigma^2 = np(1 - p)$. Given $\mu = 6$ and $\sigma^2 = 4.8$, we have:
$np = 6 \quad \text{and} \quad np(1 - p) = 4.8$
Substituting the first equation into the second:
$6(1 - p) = 4.8$
Solving for $p$:
$1 - p = \frac{4.8}{6} = 0.8$
$p = 1 - 0.8 = 0.2$
Therefore, the probability of success is $0.2$.

Key Concept: Variance and Standard Deviation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the standard deviation of a dataset is always non-negative. This is true because standard deviation is derived from the square root of variance, and variance is always non-negative since it is a sum of squared deviations. Thus, the reason correctly explains why the assertion is true, as both involve the fundamental property of variance being non-negative.

Key Concept: Probability, Empirical Rule

c) 0.9544
[Solution Description]
By the empirical rule, approximately 95\% of data falls within $\mu \pm 2\sigma$.
Here, $\mu \pm 2\sigma = 100 \pm 30 = (70, 130)$.
The probability is approximately 0.9544 (or 95.44\%).

Key Concept: Probability Density Function

b) $\sum_{i} p(x_i) = 1$ and $p(x_i) \geq 0$ for all $i$.
[Solution Description]
For a valid probability density function of a discrete random variable, two conditions must hold:
1) All probabilities $p(x_i)$ must be non-negative, i.e., $p(x_i) \geq 0$ for all $i$.
2) The sum of all probabilities must equal 1, i.e., $\sum_{i} p(x_i) = 1$.
Option b) states both conditions correctly.

Key Concept: Probability under Normal Curve

b) 0.1587
[Solution Description]
First, calculate the z-score for $X = 45$:
$z = \frac{45 - 50}{5} = \frac{-5}{5} = -1.$
Now, find $P(X < 45) = P(Z < -1)$ using the standard normal table.
From the table, $P(Z < -1) \approx 0.1587$.
So, the probability is approximately $0.1587$.

Key Concept: Non-Bernoulli Identification, Dependency Analysis

c) Both (a) and (b)
[Solution Description]
Bernoulli trials require constant probability of success and independence. Here:
1) Probability changes as items are not replaced (dependent trials).
2) The stopping condition (3 non-defective items) violates finite, fixed trials.
Thus, the scenario fails both independence and constant probability conditions.

Key Concept: Probability Distribution, Discrete and Continuous Random Variables

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
**Assertion Analysis:**
The assertion states that for a discrete random variable $X$, the sum of probabilities over all possible values must equal 1. This is correct because it aligns with the fundamental property of probability distributions: the total probability must be 1.
**Reason Analysis:**
The reason describes how the cumulative distribution function (CDF) of a continuous random variable is derived from its probability density function (PDF). This is also correct because, by definition, the CDF $F(x)$ is given by:
$F(x) = \int_{-\infty}^{x} f(t) \, dt,$
where $f(t)$ is the PDF.
**Connection Between Assertion and Reason:**
While both statements are true individually, they address different aspects of probability distributions (discrete vs. continuous). The reason does not explain why the sum of probabilities for a discrete random variable equals 1. Hence, the reason is not the correct explanation of the assertion.
Both the assertion and reason are true, but the reason does not explain the assertion.

Key Concept: Notation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that $P(X = x)$ represents the probability that the random variable $X$ takes the value $x$. This is correct as per the standard notation used in probability theory. The reason explains the convention of using capital letters for random variables and small letters for their observed values, which is also correct and provides the basis for understanding the assertion. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Bernoulli's Trials - Probability of Success

b) 0.1
[Solution Description]
For a Bernoulli trial, the probability of success (defective item here) is given as $p = 0.1$. The probability of failure (non-defective) is $q = 1 - p = 0.9$.
Since the question asks for the probability of exactly one defective item (success), we use the Bernoulli probability formula:
$P(X = x) = p^x q^{1-x}$.
Here, $x = 1$ (for defective item).
Substituting values:
$P(X = 1) = (0.1)^1 (0.9)^{0} = 0.1$.

Key Concept: Poisson Distribution Mean and Variance

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Poisson distribution has a property that its mean and variance are equal. The mean of a Poisson distribution is given by $\mu = \lambda$ and the variance is also $\sigma^2 = \lambda$. Therefore, the Assertion (A) stating "For a Poisson distribution, the mean is equal to the variance" is true. The Reason (R) states "In the Poisson distribution, both mean and variance are given by $\lambda$," which is also true and directly explains why the mean equals the variance in this distribution. Hence, the correct answer is option a).

Key Concept: Discrete Probability Distribution

c) 0.6
[Solution Description]
To find the probability that X is greater than 2, we sum the probabilities of X taking values 3 and 4.
$P(X > 2) = P(X = 3) + P(X = 4) = 0.4 + 0.2 = 0.6$

Key Concept: Probability in Bernoulli’s Trials

c) 0.6
[Solution Description]
In Bernoulli's trials, the sum of the probability of success ($p$) and failure ($q$) is always 1, i.e., $p + q = 1$. Given $p = 0.4$, we can find $q$ as follows:
$q = 1 - p = 1 - 0.4 = 0.6$
Therefore, the probability of failure is $0.6$.

Key Concept: Mean and Variance of Random Variable

b) 0.61
[Solution Description]
First, calculate the mean ($\mu$):
$\mu = E(X) = \sum x_i P(X = x_i) = 1 \times 0.2 + 2 \times 0.3 + 3 \times 0.5 = 0.2 + 0.6 + 1.5 = 2.3$
Now, calculate $E(X^2)$:
$E(X^2) = \sum x_i^2 P(X = x_i) = 1 \times 0.2 + 4 \times 0.3 + 9 \times 0.5 = 0.2 + 1.2 + 4.5 = 5.9$
Finally, the variance is:
$\sigma^2 = E(X^2) - \mu^2 = 5.9 - (2.3)^2 = 5.9 - 5.29 = 0.61$

Key Concept: Bernoulli's Trials - Conditions

d) All of the above
[Solution Description]
For trials to be Bernoulli trials, the following conditions must hold:
1. Finite number of trials.
2. Independence between trials.
3. Exactly two outcomes: success or failure.
4. Constant probability of success ($p$) for each trial.
Therefore, all four conditions must be met.

Key Concept: Probability Density Function

a) $P(X=1)=0.5$, $P(X=2)=0.3$, $P(X=3)=0.2$
[Solution Description]
A valid probability distribution must satisfy two conditions: 1) The sum of all probabilities must be equal to 1, and 2) Each probability must lie between 0 and 1. Now let's check each option:
Option 1: $P(X=1)=0.5$, $P(X=2)=0.3$, $P(X=3)=0.2$. Here, $0.5 + 0.3 + 0.2 = 1$ and all probabilities are between 0 and 1. This is a valid probability distribution.
Option 2: $P(X=1)=0.8$, $P(X=2)=0.4$, $P(X=3)=-0.2$. Here, one probability is negative, so it is invalid.
Option 3: $P(X=1)=0.6$, $P(X=2)=0.6$, $P(X=3)=0.6$. Here, $0.6 + 0.6 + 0.6 = 1.8$, which is greater than 1, so it is invalid.
Option 4: $P(X=1)=0.9$, $P(X=2)=0.05$, $P(X=3)=0.05$. Here, $0.9 + 0.05 + 0.05 = 1$ and all probabilities are between 0 and 1. This is also a valid probability distribution.
However, only one option is completely correct. Thus, the correct answer is a).

Key Concept: Notation for observed values

b) Lowercase letters like x, y, z
[Solution Description]
The syllabus states that observed values are denoted as $x, y, z$, while random variables are written as $X, Y, Z$.

Key Concept: Probability distribution, Bar chart visualization

c) 0.30
[Solution Description]
For a valid probability distribution, the sum of all probabilities must equal 1. Given:
$P(X=1) + P(X=2) + P(X=3) + P(X=4) = 1$
Substituting the given values:
$0.2 + k + 0.4 + 0.1 = 1$
$0.7 + k = 1$
$k = 1 - 0.7 = 0.3$

Key Concept: Binomial Distribution

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that for $n = 10$ and $p = 0.3$, the probability of exactly 5 successes ($X=5$) is calculated using the formula $P(X=5) = \,^{10}C_5 \times (0.3)^5 \times (0.7)^5$. This is correct because the general formula for the binomial probability mass function is indeed $P(X=r) = \,^nC_r \, p^r (1-p)^{n-r}$, which matches the form given in the reason.
Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Conditions for Bernoulli's Trials

c) The trials must be independent of each other
[Solution Description]
Bernoulli's trials must satisfy these conditions:
1. The number of trials is finite.
2. Each trial has two outcomes: success or failure.
3. The trials are independent.
4. The probability of success ($p$) remains constant across trials.
Hence, all the given options except option c) are necessary conditions.

Key Concept: Mean and Expectation of a Discrete Random Variable

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Let us evaluate both statements step by step.
1. **Assertion Analysis**:
- Given: $p_i = \frac{1}{n}$ for all $i$ (uniform distribution).
- Variance $\sigma^2 = \sum p_i (x_i - \mu)^2 = \frac{1}{n} \sum (x_i - \mu)^2$.
- If all $x_i$ are equal to some constant $c$, then $\mu = c$, so $(x_i - \mu)^2 = 0$ for all $i$, leading to $\sigma^2 = 0$.
- Conversely, if $\sigma^2 = 0$, then $\sum (x_i - \mu)^2 = 0$, which implies each $(x_i - \mu)^2 = 0$ and thus all $x_i = \mu$.
2. **Reason Analysis**:
- The reason correctly states that if all $x_i$ are equal to $c$, then $\mu = c$ because $\mu = \sum p_i x_i = c \sum p_i = c$.
- It further explains why each term $(x_i - \mu)^2$ becomes zero, directly justifying the assertion.
Conclusion: Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Binomial Distribution

a) $\frac{5}{16}$
[Solution Description]
The number of trials $n = 5$. Probability of success (head) $p = 0.5$, and failure $q = 1 - p = 0.5$. Using the binomial formula:
$P(X = r) = ^nC_r p^r q^{n-r}$
For $r = 3$:
$P(X = 3) = ^5C_3 (0.5)^3 (0.5)^{2} = 10 \times 0.125 \times 0.25 = 0.3125$

Key Concept: Bernoulli’s Trials, Binomial Distribution

b) 0.3456
[Solution Description]
We use the binomial probability formula:
$P(X = r) = \binom{n}{r} p^r q^{n-r}$, where:
- $n = 5$ (number of trials),
- $r = 3$ (desired successes),
- $p = 0.6$ (probability of success),
- $q = 1 - p = 0.4$ (probability of failure).
Step 1: Calculate $\binom{5}{3}$:
$\binom{5}{3} = \frac{5!}{3!2!} = 10$.
Step 2: Compute $p^r \cdot q^{n-r}$:
$0.6^3 \cdot 0.4^{2} = 0.216 \cdot 0.16 = 0.03456$.
Step 3: Multiply results from Step 1 and Step 2:
$10 \times 0.03456 = 0.3456$.
Thus, $P(X = 3) = 0.3456$.

Key Concept: Probability Distribution Representation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a discrete random variable, the probability distribution assigns probabilities to each possible outcome. The total probability of all possible outcomes must sum to 1 because one of the outcomes must occur. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Variance of a Discrete Probability Distribution

b) 0.50
[Solution Description]
First, compute the mean $\mu$:
$\mu = 0 \times 0.25 + 1 \times 0.50 + 2 \times 0.25 = 0 + 0.50 + 0.50 = 1.0$
Next, compute $E(X^2)$:
$E(X^2) = 0^2 \times 0.25 + 1^2 \times 0.50 + 2^2 \times 0.25 = 0 + 0.50 + 1.0 = 1.5$
Now, compute variance $\sigma^2$:
$\sigma^2 = E(X^2) - \mu^2 = 1.5 - 1.0 = 0.5$

Key Concept: Binomial Distribution, Card Draw Example

d) Assertion is false, but Reason is true.
[Solution Description]
To determine whether the given scenario follows a binomial distribution, we analyze the conditions:
- **Fixed number of trials ($n = 2$)**: This condition is satisfied as there are exactly 2 draws.
- **Independent trials**: This is only true if the draws are with replacement. Here, they are without replacement, so the trials are not independent.
- **Constant probability of success**: The probability changes because the deck composition changes after the first draw (e.g., $P(\text{King on first draw}) = \frac{4}{52}$, but $P(\text{King on second draw})$ depends on the first draw).
Since the trials are not independent and the probability of success is not constant, the distribution is not strictly binomial. Thus, Assertion is false. However, the Reason incorrectly assumes independence and constant probability, making it false as well.

Key Concept: Mean and Variance of Binomial Distribution

c) 3.464
[Solution Description]
For a binomial distribution:
- Mean ($\mu$) = $np$
- Variance ($\sigma^2$) = $npq$
- Standard deviation ($\sigma$) = $\sqrt{npq}$
Given:
- $n$ = 50
- $p$ = 0.4
- $q$ = 1 - $p$ = 0.6
Calculating variance:
$\sigma^2 = npq = 50 \times 0.4 \times 0.6 = 12$
Standard deviation = $\sqrt{12} \approx 3.464$.

Key Concept: Variance Properties, Linear Transformation

c) 8
[Solution Description]
Variance of a linear transformation $V = aW + b$ is given by $\sigma_V^2 = a^2 \sigma_W^2$. Here, $a = 2$, $b = -3$, and $\sigma_W^2 = 16$. Thus,
$\sigma_V^2 = 2^2 \times 16 = 4 \times 16 = 64$
The standard deviation is the square root of the variance:
$\sigma_V = \sqrt{64} = 8$

Key Concept: Probability Density Function

b) 1.0
[Solution Description] The sum of all probabilities for a probability distribution must be 1. Here, $0.2 + 0.3 + 0.5 = 1$.

Key Concept: Bernoulli's Trials, Probability of Independent Events

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that for a fair coin ($p = 0.5$) tossed 10 times, the probability of getting exactly 5 heads and 5 tails is $\binom{10}{5}(0.5)^{10}$. This is correct because in Bernoulli trials, the probability of any specific sequence with $k$ successes (here, heads) in $n$ trials is given by the binomial probability formula $\binom{n}{k}p^{k}(1-p)^{n-k}$. Substituting $n=10$, $k=5$, and $p=0.5$ yields $\binom{10}{5}(0.5)^{5}(0.5)^{5} = \binom{10}{5}(0.5)^{10}$.
The reason states two facts: First, that the probability of any sequence of outcomes in Bernoulli trials is the product of individual probabilities, which is true due to independence. Second, it mentions that the number of ways to arrange $k$ successes in $n$ trials is $\binom{n}{k}$, which is also correct since this counts distinct sequences. However, while both statements are true, the reason does not directly explain why the assertion's calculation is correct; it merely provides general properties of Bernoulli trials. Thus, the reason is not the correct explanation of the assertion.

Key Concept: Mixed Random Variables, Validity Conditions

b) 1.0
[Solution Description]
The total probability must equal 1. Calculate the discrete part:
$P(X = 0) + P(X = 1) = 0.2 + 0.3 = 0.5$
The continuous part should contribute the remaining $0.5$:
$\int_{1}^{2} c(2 - x) \, dx = c \left[2x - \frac{x^2}{2}\right]_{1}^{2} = c \left((4 - 2) - \left(2 - \frac{1}{2}\right)\right) = c \times 0.5$
Solve for $c$:
$0.5c = 0.5 \Rightarrow c = 1$
Check non-negativity: $c(2 - x) \geq 0$ for $x \in (1, 2]$ implies $c \geq 0$.

Key Concept: Probability, Sample Space

b) $\frac{1}{2}$
[Solution Description]
The sample space for three coin tosses has 8 outcomes: $\{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$. To find $P(X \geq 2)$, count the favorable outcomes where the number of heads is 2 or 3. These are: $HHH, HHT, HTH, THH$. There are 4 such outcomes. Therefore, $P(X \geq 2) = \frac{4}{8} = \frac{1}{2}$.

Key Concept: Mean, Variance, Combined Distributions

c) 23
[Solution Description]
For two independent binomial random variables $X_1$ and $X_2$:
$\text{Var}(Y) = \text{Var}(X_1) + \text{Var}(X_2)$
Calculate the variances of $X_1$ and $X_2$:
$\text{Var}(X_1) = n_1 p_1 (1 - p_1) = 50 \times 0.3 \times 0.7 = 10.5$
$\text{Var}(X_2) = n_2 p_2 (1 - p_2) = 50 \times 0.5 \times 0.5 = 12.5$
Thus, the variance of $Y$ is:
$\text{Var}(Y) = 10.5 + 12.5 = 23$

Key Concept: Probability Density Function

c) The probability that $X$ is between $a$ and $b$
[Solution Description]
For a continuous random variable $X$, the integral of the probability density function (PDF) over an interval $[a, b]$ gives the probability that $X$ lies within that interval. Mathematically, this is represented as:
$P(a < X < b) = \int_{a}^{b} f(x) \, dx$
The solution involves recalling the definition of the PDF for a continuous random variable.

Key Concept: Binomial Distribution Parameters

c) $\frac{4}{3}$
[Solution Description]
The variance ($\sigma^2$) of a binomial distribution is calculated as:
$\sigma^2 = npq$
Here, $n = 6$, $p = \frac{1}{3}$, and $q = 1 - p = \frac{2}{3}$.
Substituting these values:
$\sigma^2 = 6 \times \frac{1}{3} \times \frac{2}{3} = \frac{12}{9} = \frac{4}{3}$

Key Concept: Probability Distribution of a Discrete Random Variable

c) $\frac{1}{2}$
[Solution Description]
The possible values of $X$ when rolling a fair die are $1, 2, 3, 4, 5, 6$, each with probability $\frac{1}{6}$. To find $P(X \leq 3)$, we sum the probabilities for $X = 1, 2, 3$:
$P(X \leq 3) = P(X=1) + P(X=2) + P(X=3) = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}.$

Key Concept: Binomial Probability

b) $\frac{5}{16}$
[Solution Description]
We use the Binomial Probability formula:
$P(X = r) = \binom{n}{r} p^r (1 - p)^{n - r}$
Here, $n = 5$, $r = 3$, and $p = 0.5$ (since it's a fair coin). Plugging in the values:
$P(X = 3) = \binom{5}{3} (0.5)^3 (0.5)^{2} = 10 \times 0.125 \times 0.25 = 0.3125$
The probability is 0.3125.

Key Concept: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the total area under the normal distribution curve is equal to 1, which is true because it represents the total probability of all outcomes. The reason given is also true as the integral of the probability density function from $-\infty$ to $+\infty$ equals 1. Moreover, the reason correctly explains the assertion since the area under the curve corresponds to the integral of the PDF.

Key Concept: Variance Formula

c) $\sigma^2 = \sum p_i (x_i - \mu)^2$
[Solution Description]
The variance $\sigma^2$ is calculated using the formula:
$\sigma^2 = \sum p_i (x_i - \mu)^2$
where $p_i$ is the probability of $x_i$.

Key Concept: Probability distribution, Expected value

c) 2.4
[Solution Description]
First, find the value of $k$ such that the sum of all probabilities equals 1.
$0.2 + 0.3 + 0.4 + k = 1$
Solving for $k$:
$k = 1 - (0.2 + 0.3 + 0.4) = 0.1$
Now, calculate the expected value $\mu$ using the formula:
$\mu = \sum_{i=1}^4 x_i P(X=x_i) = 1 \times 0.2 + 2 \times 0.3 + 3 \times 0.4 + 4 \times 0.1$
$\mu = 0.2 + 0.6 + 1.2 + 0.4 = 2.4$

Key Concept: Bernoulli’s Trials, Variance, Probability Density Function

c) 0.96
[Solution Description]
For Bernoulli’s Trials, the variance for $n$ trials is given by $\sigma^2 = n p q$, where $q = 1 - p$.
Here, $n = 4$, $p = 0.6$, $q = 0.4$.
Variance = $4 \times 0.6 \times 0.4 = 0.96$.

Key Concept: Binomial Distribution, Probability of Success

c) 0.3456
[Solution Description]
To find the probability of getting exactly 3 heads in 5 flips of a biased coin, we use the binomial distribution formula:
$P(X = r) = \, ^nC_r \, p^r q^{n-r}$
Here, $n = 5$, $r = 3$, $p = 0.6$, and $q = 1 - p = 0.4$.
$P(X = 3) = \, ^5C_3 \, (0.6)^3 (0.4)^{2}$
Calculate $_5C_3$:
$^5C_3 = \frac{5!}{3!2!} = 10$
Now compute $(0.6)^3$ and $(0.4)^2$:
$(0.6)^3 = 0.216$
$(0.4)^2 = 0.16$
Multiply all parts together:
$10 \times 0.216 \times 0.16 = 0.3456$
So, the probability is 0.3456.

Key Concept: Binomial Distribution

a) $\frac{5}{16}$
[Solution Description]
To find the probability of getting exactly 3 heads in 5 tosses of a fair coin, we use the binomial probability formula:
$P(X = r) = \binom{n}{r} p^r q^{n-r}$
Here, $n = 5$ (number of trials), $r = 3$ (desired successes), $p = 0.5$ (probability of success, i.e., getting a head), and $q = 0.5$ (probability of failure, i.e., getting a tail). Plugging in the values:
$P(X = 3) = \binom{5}{3} (0.5)^3 (0.5)^{2} = 10 \times 0.125 \times 0.25 = 0.3125$
So, the probability is $0.3125$.

Key Concept: Probability Density Function Properties

b) $\int_{-\infty}^{\infty} f(x) \, dx = 1$ and $f(x) \geq 0$ for all $x$
[Solution Description]
A probability density function (PDF) must satisfy two key properties: 1) The integral over its entire range must be equal to 1, i.e., $\int_{-\infty}^{\infty} f(x) \, dx = 1$. 2) The PDF must be non-negative for all $x$, i.e., $f(x) \geq 0$ for all $x$.

Key Concept: Bernoulli’s Trials

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Tossing a fair coin multiple times satisfies all conditions for Bernoulli's trials:
1. Finite trials (here, 10 tosses).
2. Each toss is independent.
3. Two outcomes per trial (head or tail).
4. Constant success probability ($p = 0.5$ for heads). Thus, both Assertion (A) and Reason (R) are true, and R correctly explains A.

Key Concept: Probability Calculation

a) $\frac{24}{169}$
[Solution Description]
To find the probability of drawing exactly one king in two successive draws:
Case 1: King first, non-king second = $\frac{4}{52} \times \frac{48}{52}$
Case 2: Non-king first, king second = $\frac{48}{52} \times \frac{4}{52}$
Total probability = $\frac{192}{2704} + \frac{192}{2704} = \frac{384}{2704} = \frac{24}{169}$

Key Concept: Mean of Binomial Distribution

b) 3.0
[Solution Description]
The mean ($\mu$) of a binomial distribution is calculated using:
$\mu = n \cdot p$
Substituting $n = 10$ and $p = 0.3$:
$\mu = 10 \times 0.3 = 3$
The mean is 3.

Key Concept: Properties of Probability Distribution

c) Equal to 1
[Solution Description]
For any probability distribution, the sum of probabilities for all possible outcomes must equal 1. This ensures that the distribution accounts for all possible events.

Key Concept: Binomial Distribution, Bernoulli Trials

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that for a binomial distribution with $n = 20$ and $p = 0.4$, the mean ($μ$) is 8 and variance ($\sigma^2$) is 4.8.
Step 1: Calculate the mean using the formula $\mu = n p$:
$\mu = 20 \times 0.4 = 8$
Step 2: Calculate the variance using the formula $\sigma^2 = n p (1 - p)$:
$\sigma^2 = 20 \times 0.4 \times (1 - 0.4) = 20 \times 0.4 \times 0.6 = 4.8$
Both the assertion and reason are true. Moreover, the reason correctly explains why the assertion holds true because the formulas provided in the reason are indeed the correct formulas to compute the mean and variance of a binomial distribution.

Key Concept: Probability Distribution Conditions

c) $P(X = 1) = 0.4, \quad P(X = 2) = 0.3, \quad P(X = 3) = 0.2, \quad P(X = 4) = 0.1$
[Solution Description]
To determine if a distribution is valid, check:
1) Each probability satisfies $0 \leq p(x_i) \leq 1$.
2) The sum of all probabilities equals 1.
For option c):
$P(X = 1) = 0.4, \quad P(X = 2) = 0.3, \quad P(X = 3) = 0.2, \quad P(X = 4) = 0.1$
Sum: $0.4 + 0.3 + 0.2 + 0.1 = 1$. All probabilities are between 0 and 1. Hence, it is valid.

Key Concept: Concept: Binomial Probability

c) 0.24609375
[Solution Description]
For a binomial distribution, the probability of exactly $k$ successes in $n$ trials is given by:
$P(X = k) = ^nC_k \cdot p^k \cdot (1-p)^{n-k}$
Substituting $n = 10$, $p = 0.5$, and $k = 5$:
$P(X = 5) = ^{10}C_5 \cdot (0.5)^5 \cdot (0.5)^{5} = 252 \times 0.03125 \times 0.03125 = 0.24609375$

Key Concept: Variance and Standard Deviation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the variance of a dataset is always non-negative. This is true because variance is defined as the average of squared deviations from the mean. Since squaring any real number (including deviations from the mean) results in a non-negative value, the variance must also be non-negative. The reason correctly explains the assertion by stating that variance is calculated using squared deviations, which are always non-negative. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Z-score Calculation

c) 1.5
[Solution Description]
The z-score formula is given by:
$z = \frac{X - \mu}{\sigma},$
where $X$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation.
Given: $X = 175$, $\mu = 160$, $\sigma = 10$.
Substitute the values into the formula:
$z = \frac{175 - 160}{10} = \frac{15}{10} = 1.5.$
So, the z-score is $1.5$.

Key Concept: Properties of Standard Deviation

b) Standard deviation is zero if all values are identical
[Solution Description]
The standard deviation $\sigma$ is always non-negative, i.e., $\sigma \geq 0$. It becomes zero only when all values in the dataset are equal.

Key Concept: Variance calculation, Alternative formula

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To verify this, let us expand the second formula:
$\sigma^2 = \sum_{i=1}^n p_i (x_i - \mu)^2 = \sum_{i=1}^n p_i (x_i^2 - 2\mu x_i + \mu^2)$
$= \sum_{i=1}^n p_i x_i^2 - 2\mu \sum_{i=1}^n p_i x_i + \mu^2 \sum_{i=1}^n p_i$
Since $\mu = \sum_{i=1}^n p_i x_i$ and $\sum_{i=1}^n p_i = 1$, substituting these gives:
$\sigma^2 = \sum_{i=1}^n p_i x_i^2 - 2\mu \cdot \mu + \mu^2 \cdot 1 = \sum_{i=1}^n p_i x_i^2 - \mu^2$
This matches the first formula exactly. Thus, both Assertion and Reason are true, and Reason correctly explains why the two formulas are equivalent.

Key Concept: Normal Distribution Properties

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that for a normal distribution, the mean, median, and mode are equal, which is true because the normal distribution is perfectly symmetric. The reason states that the normal distribution is symmetric about its mean, which is also true. Moreover, the symmetry of the normal distribution about its mean is the correct explanation for why the mean, median, and mode are equal in this distribution.

Key Concept: Probability, Random Variables

b) 1
[Solution Description]
First, list all possible outcomes and their probabilities. When three fair coins are tossed, the possible values of $X$ are 0, 1, 2, and 3. The probability mass function for $X$ is:
$P(X = 0) = \frac{1}{8}, \quad P(X = 1) = \frac{3}{8}, \quad P(X = 2) = \frac{3}{8}, \quad P(X = 3) = \frac{1}{8}.$
Given $P(X = x) = \frac{3}{8}$, the possible values of $x$ are 1 and 2.

Key Concept: Probability Density Function

b) $f(x) \geq 0$ for all \textit{x}, $\int_{-\infty}^{\infty} f(x) \, dx = 1$
[Solution Description]
The probability density function (pdf) of a continuous random variable must be non-negative for all \textit{x}, and the total area under its curve must equal 1.

Key Concept: Identifying Bernoulli’s Trials

b) No, because the probability of success changes.
[Solution Description]
Bernoulli's trials require:
1. Finite trials → satisfied.
2. Two outcomes → satisfied (success: green, failure: red).
3. Constant probability → not satisfied (probability changes as balls are not replaced).
Hence, these are not Bernoulli's trials.

Key Concept: Binomial Distribution, Bernoulli Trials

d) Assertion is false, but Reason is true.
[Solution Description]
To verify the assertion, we calculate:
1. Probability of at least 7 successes:
$P(X \geq 7) = \sum_{r=7}^{10} {}^{10}C_r (0.4)^r (0.6)^{10-r}$
Calculating individual terms:
$P(7) = {}^{10}C_7 (0.4)^7 (0.6)^3 = 120 \times 0.00016384 \times 0.216 \approx 0.002642$
$P(8) = {}^{10}C_8 (0.4)^8 (0.6)^2 = 45 \times 0.000065536 \times 0.36 \approx 0.001075$
$P(9) = {}^{10}C_9 (0.4)^9 (0.6)^1 = 10 \times 0.0000262144 \times 0.6 \approx 0.000157$
$P(10) = {}^{10}C_{10} (0.4)^{10} (0.6)^0 = 1 \times 0.0001048576 \times 1 \approx 0.000105$
Summing up: $P(X \geq 7) \approx 0.003979$.
2. Probability of exactly 5 successes:
$P(5) = {}^{10}C_5 (0.4)^5 (0.6)^5 = 252 \times 0.01024 \times 0.07776 \approx 0.200658$
Clearly, $P(X \geq 7) < P(X=5)$, so the Assertion is false.
The Reason states that probabilities decrease monotonically beyond the mean ($np = 4$). However, for smaller $n$, this may not hold strictly due to discrete nature. The general trend is correct but not universally true for all cases. Hence, the Reason is true but does not correctly explain the Assertion.

Key Concept: Binomial Distribution

c) 0.3125
[Solution Description]
For a binomial distribution with $n = 5$ trials and probability of success $p = 0.5$, the probability of exactly $k = 3$ successes (heads) is:
$P(X = 3) = \binom{5}{3} (0.5)^3 (0.5)^{5-3} = \binom{5}{3} (0.5)^5$
Calculating the binomial coefficient:
$\binom{5}{3} = \frac{5!}{3!2!} = 10$
Thus:
$P(X = 3) = 10 \times (0.5)^5 = 10 \times 0.03125 = 0.3125$

Key Concept: Bernoulli's Trials, Probability Consistency

d) Assertion is false, but Reason is true.
[Solution Description]
For the given scenario:
1. The number of trials is finite (as there are 15 balls total).
2. However, when balls are drawn **without replacement**, the probability of drawing a green ball changes in each trial. For example:
- First trial: Probability of green ball = $\frac{10}{15}$.
- If first draw is green, second trial probability becomes $\frac{9}{14}$.
- If first draw is red, second trial probability becomes $\frac{10}{14}$.
Thus, the probability of success is **not constant** across trials, violating the condition for Bernoulli trials.
Therefore, Assertion (A) is false because the trials are not Bernoulli trials, while Reason (R) is also false as the probability does not remain constant.

Key Concept: Probability Density Function

d) 2.7
[Solution Description]
The expected value (mean) of a discrete random variable is calculated as:
$\mu = \sum x_i P(x_i)$
Substituting the given probabilities:
$\mu = (1 \times 0.1) + (2 \times 0.3) + (3 \times 0.4) + (4 \times 0.2)$
Calculating each term:
$\mu = 0.1 + 0.6 + 1.2 + 0.8 = 2.7$

Key Concept: Mean of a Probability Distribution

c) 1.9
[Solution Description]
The expected value $E(X)$ is calculated using the formula:
$E(X) = \sum x_i \cdot P(x_i) = (1 \times 0.4) + (2 \times 0.3) + (3 \times 0.3).$
Substituting the values:
$E(X) = 0.4 + 0.6 + 0.9 = 1.9.$

Key Concept: Binomial Distribution

d) $0.642$
[Solution Description]
The probability that at least one bulb is defective is the complement of none being defective. First, find $P(X = 0)$ using the binomial formula:
$P(X = 0) = \binom{20}{0} (0.05)^0 (0.95)^{20} = 1 \times 1 \times (0.95)^{20}$
Calculating $(0.95)^{20}$:
$(0.95)^{20} \approx 0.358$
Therefore, the probability of at least one defective bulb is:
$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.358 = 0.642$
The probability is approximately $0.642$.

Key Concept: Binomial Distribution, Probability Calculation

a) 0.40951
[Solution Description]
To find the probability of at least one defective item in 5 trials, it's easier to calculate the complement (no defective items) and subtract from 1.
Step 1: Find $P(\text{no defective items})$:
$P(X = 0) = \binom{5}{0} (0.1)^0 (0.9)^5 = 1 \times 1 \times (0.9)^5 \approx 0.59049$
Step 2: Subtract from 1 to get $P(\text{at least one defective})$:
$1 - 0.59049 = 0.40951$

Key Concept: Bernoulli's Trials

c) 0.3125
[Solution Description]
This is a binomial probability problem where:
Number of trials $n = 5$, probability of success (getting a head) $p = 0.5$, and $q = 1 - 0.5 = 0.5$.
The probability of getting exactly $k = 3$ successes in $n$ trials is given by:
$P(X = 3) = ^5C_3 \cdot p^3 q^{2} = 10 \times (0.5)^3 \times (0.5)^2 = 10 \times 0.125 \times 0.25 = 0.3125.$
Therefore, the probability is 0.3125.

Key Concept: Random Variable Notation

b) $P(Y = y)$
[Solution Description]
According to standard notation, the probability that a random variable $Y$ takes the value $y$ is denoted as $P(Y = y)$.

Key Concept: Probability Density Function

c) $k = \frac{1}{2}$
[Solution Description]
For a valid PDF, the total integral over its range must equal 1.
$\int_{0}^{2} k(2 - y) \, dy = 1$
Solve the integral:
$k \int_{0}^{2} (2 - y) \, dy = k \left[ 2y - \frac{y^2}{2} \right]_0^2 = k \left( 4 - 2 \right) = 2k = 1$
Therefore, $k = \frac{1}{2}$.

Key Concept: Probability Mass Function, Cumulative Probability

b) 0.4967
[Solution Description]
This is a binomial distribution problem with:
- $n$ = 8
- $p$ = 0.2
- $q$ = 0.8
- We need $P(X \geq 2) = 1 - P(X < 2) = 1 - [P(X=0) + P(X=1)]$.
First, calculate $P(X=0)$:
$P(X=0) = \binom{8}{0} (0.2)^0 (0.8)^8 = 1 \times 1 \times 0.1678 = 0.1678$
Next, calculate $P(X=1)$:
$P(X=1) = \binom{8}{1} (0.2)^1 (0.8)^7 = 8 \times 0.2 \times 0.2097 \approx 0.3355$
Now, compute $P(X < 2)$:
$P(X < 2) = 0.1678 + 0.3355 = 0.5033$
Finally, $P(X \geq 2)$:
$P(X \geq 2) = 1 - 0.5033 = 0.4967$
The probability is approximately 0.4967.

Key Concept: Binomial Distribution

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that in a binomial distribution, the probability of success remains constant for each trial, which is true as per the definition of binomial distribution. The Reason states that binomial distribution is only applicable when the trials are independent and have exactly two outcomes, which is also true as these are the conditions for Bernoulli trials, which form the basis of binomial distribution. Moreover, the Reason correctly explains why the probability of success remains constant (as trials are independent). Hence, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Discrete Random Variable

c) $\frac{1}{2}$
[Solution Description]
A fair six-sided die has outcomes: 1, 2, 3, 4, 5, 6. The even numbers are 2, 4, 6. Total favorable outcomes = 3, total possible outcomes = 6. Probability = $\frac{3}{6} = \frac{1}{2}$.

Key Concept: Bernoulli’s Trials - Constant Probability

d) Constant probability of success
[Solution Description]
Bernoulli’s trials require the probability of success ($p$) to remain constant across all trials. If the coin becomes biased after the first toss, the probability of success changes, violating the condition of constant probability.

Key Concept: Probability of success and failure in Bernoulli's Trials

c) $\frac{1}{2}$
[Solution Description]
In a Bernoulli trial, the probability of success ($p$) plus the probability of failure ($q$) equals 1. Here, success is defined as getting heads ($p = 0.5$). Therefore, the probability of failure (tails) is $q = 1 - p = 1 - 0.5 = 0.5$.

Key Concept: Continuous Probability Density Function

b) $\frac{3}{4}$
[Solution Description]
For f(x) to be a valid probability density function, the total area under the curve must be 1:
$\int_{-1}^{1} k(1 - x^2) \, dx = 1$
Solving the integral:
$k \left[ \int_{-1}^{1} (1 - x^2) \, dx \right] = k \left[ x - \frac{x^3}{3} \right]_{-1}^1 = k \left( \left(1 - \frac{1}{3}\right) - \left(-1 + \frac{1}{3}\right) \right) = k \left( \frac{4}{3} \right) = 1$
Thus, $k = \frac{3}{4}$.

Key Concept: Probability Distribution Representation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a discrete random variable, the probability distribution assigns specific probabilities to distinct values. A bar chart is indeed suitable for such representations as it visually separates each value and its corresponding probability, making it easier to understand the distribution. The assertion correctly states that a bar chart can represent the probability distribution of a discrete random variable, and the reason provides the correct justification for using a bar chart due to its clarity in displaying individual probabilities. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Probability Notation: P(X=x

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Assertion states that for the given probabilities of $P(X = 1) = 0.6$ and $P(X = 2) = 0.4$, their sum is 1. Calculating this: $0.6 + 0.4 = 1$, which matches the property stated in Reason. Hence, both Assertion and Reason are true, and Reason correctly explains Assertion as it is a fundamental property of probability distributions for discrete random variables.

Key Concept: Probability Distribution Representation

d) 0.4
[Solution Description]
The height of each bar in a probability distribution bar chart corresponds to the probability of the respective value of the random variable. From the table, the probability $p(x = 2) = 0.4$. Hence, the height of the bar for $x = 2$ is 0.4.

Key Concept: Dice Rolling, Binomial Probability

c) $\frac{3}{8}$
[Solution Description]
This is a binomial distribution problem with $n=4$ trials and probability of success (odd number) $p=\frac{3}{6}=0.5$. The probability of exactly two successes is:
$P(X=2) = ^4C_2 (0.5)^2 (0.5)^2 = 6 \times \frac{1}{4} \times \frac{1}{4} = \frac{6}{16} = \frac{3}{8}$

Key Concept: Probability Distribution - Bar Chart Representation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a discrete random variable, a bar chart is indeed the correct graphical representation as each distinct value the random variable can take is represented by a separate bar whose height corresponds to its probability. The assertion is true. The reason states that the sum of the heights (probabilities) of all bars must equal 1 which is correct because the total probability for all possible outcomes of a random variable must sum to 1. Moreover, the reason logically explains why the bar chart is appropriate, as it emphasizes this fundamental property of probability distributions. Hence, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Binomial Distribution, Mean and Variance

b) 4.8
[Solution Description]
The variance $\sigma^2$ of a binomial distribution is given by:
$\sigma^2 = n p q$
where $q = 1 - p$. Here, $n = 20$, $p = 0.4$, so $q = 0.6$.
Step 1: Plug the values into the formula:
$\sigma^2 = 20 \times 0.4 \times 0.6 = 4.8$

Key Concept: Mean of Discrete Random Variable

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The mean of a discrete random variable is calculated as $\mu = \sum p_i x_i$. For a fair six-sided die, each outcome $x_i$ (1 to 6) has probability $p_i = \frac{1}{6}$. Thus, the mean is:
$\mu = \frac{1+2+3+4+5+6}{6} = 3.5$. The reason correctly states that all outcomes are equally likely. Further, the reason explains why the assertion is true because equal probabilities lead to this mean calculation.

Key Concept: Bar Chart Interpretation

c) 0.7
[Solution Description]
To find $P(1 \leq Y \leq 2)$, we add the probabilities of $Y = 1$ and $Y = 2$ since these are distinct events. From the bar chart:
- $P(Y = 1) = 0.2$
- $P(Y = 2) = 0.5$
Thus, $P(1 \leq Y \leq 2) = 0.2 + 0.5 = 0.7$.

Key Concept: Binomial Distribution, Mean and Variance

b) 2.4
[Solution Description]
For a binomial distribution, the variance $\sigma^2$ is given by:
$\sigma^2 = n p (1 - p)$
Substituting $n = 10$ and $p = 0.6$:
$\sigma^2 = 10 \times 0.6 \times 0.4 = 2.4$

Key Concept: Bernoulli’s Trials, Independent Events, Non-replacement

b) No, because the trials are not independent.
[Solution Description]
Bernoulli’s Trials require:
- Finite number of trials (condition met, since 2 trials).
- Independence (not met, as the first draw affects the second).
- Constant probability (not met, since $P(\text{black})$ changes after the first draw).
Therefore, these trials are **not** Bernoulli’s Trials.

Key Concept: Mean and Variance of Binomial Distribution

c) 21
[Solution Description]
The variance of a binomial distribution is given by $\sigma^2 = npq$, where $q = 1 - p$. Here, $n = 100$, $p = 0.3$, so $q = 1 - 0.3 = 0.7$. Thus, the variance is calculated as:
$\sigma^2 = 100 \times 0.3 \times 0.7 = 21$

Key Concept: Bernoulli’s Trials

d) Each trial has exactly two outcomes: success or failure.
[Solution Description]
According to the definition, Bernoulli’s trials must satisfy the following conditions:
1) Finite number of trials.
2) Independence of each trial.
3) Exactly two outcomes (success or failure).
4) Constant probability of success in each trial.

Key Concept: Bernoulli Trials

d) Each trial has exactly two outcomes: success or failure.
[Solution Description]
Bernoulli trials must satisfy the following conditions:
1. Finite number of trials.
2. Independence of trials.
3. Two possible outcomes (success or failure).
4. Constant probability of success ($p$) for each trial.

Key Concept: Bernoulli's Trials

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For the given scenario:
- Number of trials is finite (condition satisfied).
- Each trial is independent as the ball is replaced (condition satisfied).
- There are only two outcomes: drawing a black ball (success) or not (failure) (condition satisfied).
- Probability of success (drawing black ball) remains $p = \frac{3}{8}$ for all trials since the ball is replaced (condition satisfied).
Thus, Assertion (A) is true. Reason (R) correctly explains why these trials qualify as Bernoulli's trials because it mentions the constancy of probability due to replacement.

Key Concept: Combined distributions, Variance comparison

d) 180
[Solution Description]
Variance properties:
1) Var(aX + b) = $a^2$Var(X)
2) For independent variables, Var(X ± Y) = Var(X) + Var(Y)
Applying these:
Var(Z) = Var(2X - 3Y + 5) = Var(2X) + Var(-3Y) + Var(5)
Constants have zero variance: Var(5) = 0
So Var(Z) = $2^2$Var(X) + $(-3)^2$Var(Y) = 4×9 + 9×16 = 36 + 144 = 180

Key Concept: Standard Deviation Interpretation

b) 1.549
[Solution Description]
For a binomial distribution, the variance $\sigma^2$ is calculated as:
$\sigma^2 = n p q = 10 \times 0.4 \times (1 - 0.4) = 10 \times 0.4 \times 0.6 = 2.4$
The standard deviation $\sigma$ is the square root of the variance:
$\sigma = \sqrt{2.4} \approx 1.549$
The standard deviation is approximately $1.549$.

Key Concept: Bernoulli Trials, Probability Distribution

b) $\frac{125}{3888}$
[Solution Description]
This is a binomial distribution problem with $n = 5$ trials, success probability $p = \frac{1}{6}$ (rolling a six), and failure probability $q = \frac{5}{6}$. The probability of exactly 3 successes (sixes) is given by:
$P(X = 3) = \binom{5}{3} \left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^{2}$
Calculate $\binom{5}{3} = 10$.
Now compute:
$10 \times \left(\frac{1}{216}\right) \times \left(\frac{25}{36}\right) = \frac{250}{7776} = \frac{125}{3888}$
Simplifying further, $\frac{125}{3888} \approx 0.03215$.

Key Concept: Variance and Standard Deviation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion (A) states that the standard deviation cannot be negative. This is correct because standard deviation is defined as the positive square root of the variance, $\sigma = \sqrt{\sigma^2}$. Since the variance ($\sigma^2$) is always non-negative (as it is a sum of squared terms multiplied by probabilities), its square root must also be non-negative.
The reason (R) correctly explains why the standard deviation cannot be negative: it is derived from the variance, which is inherently non-negative due to the squaring of deviations from the mean. Thus, both (A) and (R) are true, (R) provides the correct explanation for (A).

Key Concept: Binomial Distribution: Probability Calculation

c) 0.3125
[Solution Description]
For a binomial distribution with $n = 5$ trials and probability of success (getting heads) $p = 0.5$, the probability of exactly 3 successes (heads) is:
$P(X=3) = ^{5}C_3 (0.5)^3 (0.5)^{2} = 10 \times 0.125 \times 0.25 = 0.3125$
Thus, the probability is 0.3125.

Key Concept: Probability Distribution of a Random Variable

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that the sum of probabilities of all possible outcomes in a probability distribution must be 1, which is true as per the definition of a probability distribution. The Reason explains that this is because probability measures the likelihood of events, and the total likelihood must account for all possibilities, which is also true and correctly explains the Assertion.
Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Discrete Probability Distribution

c) 0.5
[Solution Description]
For a valid discrete probability distribution, the sum of all probabilities must equal 1:
$P(X = 1) + P(X = 2) + P(X = 3) = 1$
Substituting the given values:
$0.2 + 0.3 + k = 1 \implies k = 1 - 0.5 = 0.5$
Therefore, the value of $k$ must be 0.5.

Key Concept: Probability Density Function

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Solution:
Step 1: Understand the assertion. A continuous random variable can take any value in an interval, and the probability of it taking any exact value is given by the integral from $a$ to $a$, which is $\int_{a}^{a} f(x) \, dx = 0$.
Step 2: Understand the reason. Probability for a continuous random variable is calculated as the area under the curve of the probability density function $f(x)$ over an interval. Since a single point has no width, the area (probability) at that exact point is zero.
Step 3: Both Assertion (A) and Reason (R) are true. Additionally, the Reason correctly explains why the Assertion is true because the definition of probability for continuous variables involves intervals, not points.
Conclusion: The correct answer is the option where both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Example of Bernoulli Trials

c) Yes, it satisfies all the required conditions.
[Solution Description]
Tossing a fair coin 10 times satisfies all the conditions for Bernoulli trials:
1. The number of trials (10) is finite.
2. Each toss is independent of the others.
3. Each toss has exactly two outcomes: Heads (success) or Tails (failure).
4. The probability of success ($p = 0.5$) remains constant for each toss.

Key Concept: Binomial Probability, Cumulative Probability

b) $\frac{7}{128}$
[Solution Description]
To find the probability of getting at least 8 heads, we calculate the sum of probabilities for exactly 8, 9, and 10 heads. For a fair coin, $p = 0.5$ and $n = 10$.
Using the binomial formula:
$P(X = r) = {}^{10}C_r \cdot (0.5)^r \cdot (0.5)^{10-r} = {}^{10}C_r \cdot (0.5)^{10}$.
Calculate for $r = 8, 9, 10$:
- $P(8) = {}^{10}C_8 \cdot (0.5)^{10} = 45 \cdot (0.0009765625) = 0.0439453125$
- $P(9) = {}^{10}C_9 \cdot (0.5)^{10} = 10 \cdot (0.0009765625) = 0.009765625$
- $P(10) = {}^{10}C_{10} \cdot (0.5)^{10} = 1 \cdot (0.0009765625) = 0.0009765625$
Summing these probabilities:
$P(\text{at least } 8) = 0.0439453125 + 0.009765625 + 0.0009765625 = 0.0546875$
Alternatively, in fraction form:
$\frac{56}{1024} = \frac{7}{128} \approx 0.0546875$.

Key Concept: Tree Diagram, Conditional Probability

b) $\frac{1}{4}$
[Solution Description]
Given the first toss is heads ($H$), focus on the subtree after $H$. The remaining two tosses can result in $\{HH, HT, TH, TT\}$. Only one outcome ($HT$) results in exactly one tail, meaning only one outcome ($HTT$) gives exactly two tails in total. However, since $HTT$ is not a valid outcome under the condition that the first toss is heads, re-evaluate: The condition restricts to $\{HHH, HHT, HTH, HTT\}$. Exactly two tails would require two tails in the last two tosses, i.e., $HTT$. So, there's only 1 favorable outcome out of 4 possible. Thus, the probability is $\frac{1}{4}$.

Key Concept: Probability notation: P(X=x

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the probability $P(X=3)$ for a fair die is $\frac{1}{6}$. This is true because a fair die has six equally likely outcomes (1 through 6), so the probability of any specific outcome, including 3, is $\frac{1}{6}$.
The reason states that for a fair die, each outcome from 1 to 6 has an equal probability of $\frac{1}{6}$. This is also true and correctly explains why the assertion is true, as the equality of probabilities for all outcomes justifies the specific probability in the assertion.

Key Concept: Bernoulli’s Trials, Independence of trials

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
When sampling is done with replacement, the composition of the bag remains unchanged after each draw. Here, the number of trials is finite, each trial has two outcomes (success: black ball, failure: red ball), and the probability of success ($p = \frac{5}{10} = 0.5$) remains constant for each trial. Since the trials are independent due to replacement, the Assertion is true. The Reason correctly explains why the Assertion is true, as the constant probability condition is met because of replacement.

Key Concept: Random Variables Notation

c) $P(X = x)$
[Solution Description]
The probability that a random variable $X$ takes the value $x$ is denoted as $P(X = x)$. This notation is standard in probability theory.

Key Concept: Continuous Probability Distribution, PDF properties

c) Assertion is true, but Reason is false.
[Solution Description]
The Assertion states that for a continuous random variable $X$ defined on $[0, 1]$, $P(X = 0.5) = 0$. This is true because, in continuous probability distributions, the probability at any single exact point is zero. Probabilities are only meaningful over intervals.
The Reason states that the probability density function $f(x)$ gives the exact probability at any point $x$. This is false because $f(x)$ does not give the exact probability at a point but rather the probability per unit length around that point. The actual probability at any single point $x$ is zero for continuous distributions.
Thus, Assertion is true, but Reason is false.

Key Concept: Bernoulli’s Trials, Probability

c) 0.3125
[Solution Description]
This is a binomial probability problem where:
Number of trials $n = 5$,
Probability of success (heads) $p = 0.5$,
Desired successes $k = 3$.
The probability is given by:
$P(X = 3) = \binom{5}{3} (0.5)^3 (0.5)^{2} = 10 \times 0.125 \times 0.25 = 0.3125$

Key Concept: Mean of Binomial Distribution for Coin Tosses

b) $5$
[Solution Description]
For a binomial distribution with $n$ trials and success probability $p$, the mean is $np$. Here, $n = 10$ and $p = 0.5$ (since the coin is fair).
$\mu = np = 10 \times 0.5 = 5$
The expected number of heads is $5$.

Key Concept: Mean and Expectation of a Discrete Random Variable

d) Assertion is false, but Reason is true.
[Solution Description]
The Assertion states that the expected value of a discrete random variable must be one of its possible values. However, this is not always true. For example, consider a fair six-sided die with outcomes 1 to 6. The expected value is $E(X) = \frac{1+2+3+4+5+6}{6} = 3.5$, but 3.5 is not one of the possible outcomes. The Reason correctly defines the expectation formula, but it does not support the Assertion. Hence, the Assertion is false, and the Reason is true.

Key Concept: Coin Tossing, Binomial Probability

a) $\frac{5}{16}$
[Solution Description]
This is a binomial distribution problem with $n=5$ trials and probability of success (head) $p=0.5$. The probability of getting exactly 3 heads is given by:
$P(X=3) = ^5C_3 (0.5)^3 (0.5)^{2} = 10 \times \frac{1}{8} \times \frac{1}{4} = \frac{10}{32} = \frac{5}{16}$

Key Concept: Mean and Variance of Discrete Random Variable

c) 2.3
[Solution Description]
The expected value (mean) of $X$ is calculated as:
$\mu = \sum x_i P(x_i) = 1 \times 0.2 + 2 \times 0.3 + 3 \times 0.5 = 0.2 + 0.6 + 1.5 = 2.3.$

Key Concept: Bernoulli's Trials and Binomial Distribution

b) $\frac{5}{16}$
[Solution Description]
This is a binomial distribution problem where:
- Number of trials ($n$) = 5
- Probability of success (head) ($p$) = $\frac{1}{2}$
- Probability of failure (tail) ($q$) = $1 - p = \frac{1}{2}$
- Number of successes ($r$) = 3
Using the binomial PMF formula:
$P(X = r) = \binom{n}{r} p^r q^{n-r}$
Substituting values:
$P(X = 3) = \binom{5}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^{2} = 10 \times \frac{1}{8} \times \frac{1}{4} = \frac{10}{32} = \frac{5}{16}$

Key Concept: Binomial Variance

b) $21$
[Solution Description]
The variance for a binomial distribution is given by:
$\sigma^2 = npq$
where $q = 1 - p = 1 - 0.3 = 0.7$.
$\sigma^2 = 100 \times 0.3 \times 0.7 = 21$

Key Concept: Bernoulli’s Trials - Probability Relation

b) $p + q = 1$
[Solution Description]
In Bernoulli's trials, the sum of the probabilities of success ($p$) and failure ($q$) must always equal 1 because these are the only two possible outcomes.
Therefore, $p + q = 1$ or equivalently $q = 1 - p$.

Key Concept: Transformation, Variance, Standard Deviation

b) 6
[Solution Description]
Step 1: Variance scales by the square of the coefficient.
$\text{Var}(Z) = 2^2 \cdot \text{Var}(Y) = 4 \cdot (3)^2 = 36$
Step 2: Take the square root to find the standard deviation.
$\sigma_Z = \sqrt{36} = 6$

Key Concept: Conditions for Bernoulli's Trials

b) Probability of success must remain constant across trials.
[Solution Description]
For trials to be Bernoulli trials, they must satisfy the following conditions:
1) They are finite.
2) They are independent.
3) Each trial has exactly two outcomes: success or failure.
4) The probability of success ($p$) remains constant for each trial.
Only option b) matches these conditions.

Key Concept: Conditions for Bernoulli's Trials

c) Exactly two possible outcomes.
[Solution Description]
For a trial to be a Bernoulli trial, it must satisfy the following conditions:
1) Finite number of trials.
2) Independence between trials.
3) Exactly two possible outcomes: success or failure.
4) Constant probability of success ($p$) for each trial.
Among the options, "Exactly two possible outcomes" is one of these conditions.

Key Concept: Graphical Representation of Probability Distribution

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
For a continuous random variable, it is standard practice to define its probability density function (PDF) as zero outside its actual range $[A, B]$ to ensure the total probability equals 1 when integrated over $(-\infty, \infty)$. Thus, the Assertion is true. The Reason states the fundamental property that the integral of the PDF over its range must be 1, which is also a correct statement. However, while both statements are true, the Reason does not explain why the PDF must be zero outside its range—it only describes a separate property. Hence, the Reason is not the correct explanation of the Assertion.

Key Concept: Continuous Probability Density Function

b) 0.5
[Solution Description]
To find $k$, we use the property that the integral of the probability density function over its range must equal 1:
$\int_{0}^{2} kx \, dx = 1$
Integrating:
$k \left[ \frac{x^2}{2} \right]_0^2 = 1 \implies k \left( \frac{4}{2} - 0 \right) = 1 \implies 2k = 1 \implies k = 0.5$
Hence, the value of $k$ is 0.5.

Key Concept: Mean and Variance of Poisson Distribution

c) 10
[Solution Description]
For a Poisson distribution, the mean ($\mu$) and variance ($\sigma^2$) are equal to the rate parameter $\lambda$. Here, $\lambda = 5$ per hour. For two hours, the rate becomes $2\lambda = 10$. Thus, the variance is also 10.

Key Concept: Bernoulli's Trials Example

c) Tossing a coin.
[Solution Description]
An example of Bernoulli's trials should satisfy all the conditions mentioned in the syllabus:
1. Finite trials.
2. Independent trials.
3. Two outcomes per trial ('success' and 'failure').
4. Constant probability of success ($p$).
Tossing a coin meets all these conditions as:
1. It has finite trials.
2. Each toss is independent.
3. Outcomes are 'head' (success) or 'tail' (failure).
4. Probability of heads remains constant (e.g., 0.5 for a fair coin).

Key Concept: Bernoulli Trials, Conditional Probability

d) 1.000
[Solution Description]
Let’s define:
- Total balls = 10 (6 red, 4 blue)
- Probability of drawing a red ball, $P(R) = \frac{6}{10} = 0.6$
- Probability of drawing a blue ball, $P(B) = \frac{4}{10} = 0.4$
We need to find $P(\text{at least 1 blue} | \text{not all red})$.
Since it's given that not all three balls are red, we can calculate this as:
$P(\text{at least 1 blue} | \text{not all red}) = \frac{P(\text{at least 1 blue})}{P(\text{not all red})}$
First, find $P(\text{all red})$:
$P(\text{all red}) = (0.6)^3 = 0.216$
Then, $P(\text{not all red}) = 1 - P(\text{all red}) = 1 - 0.216 = 0.784$
Next, find $P(\text{at least 1 blue})$:
$P(\text{at least 1 blue}) = 1 - P(\text{all red}) = 0.784$
Therefore:
$P(\text{at least 1 blue} | \text{not all red}) = \frac{0.784}{0.784} = 1.0$
However, this seems counterintuitive because if not all are red, at least one must be blue.
Thus, the answer is 1.0.

Key Concept: Continuous Random Variable, Probability Density Function

c) $\frac{7}{27}$
[Solution Description]
First find k by ensuring total probability is 1:
$\int_{0}^{3} kx^2 dx = 1$ ⇒ k[x^3/3]_0^3 = 1 ⇒ k(9) = 1 ⇒ k = 1/9
Now calculate P(1 < X < 2):
$\int_{1}^{2} \frac{1}{9}x^2 dx = \frac{1}{9}[\frac{x^3}{3}]_1^2 = \frac{1}{27}[8 - 1] = \frac{7}{27}$

Key Concept: Bernoulli Trials

a) 2.1
[Solution Description]
The variance of a binomial distribution is given by:
$\sigma^2 = npq$
Here, $n = 10$, $p = 0.3$, and $q = 1 - p = 0.7$.
$\sigma^2 = 10 \times 0.3 \times 0.7 = 2.1$

Key Concept: Mean and Variance of Binomial Distribution

c) Assertion is true, but Reason is false.
[Solution Description]
The mean of a binomial distribution is indeed given by $\mu = np$. However, the variance of a binomial distribution is $\sigma^2 = npq$, where $q = 1 - p$. Only when $p = q = 0.5$, the variance equals the mean ($np(0.5) = \frac{n}{2}$), but this is not always true for all values of $p$. Thus, the assertion is true, but the reason is false.

Key Concept: Random Variables, Binomial Distribution

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To verify the assertion, we compute the probability using the binomial probability formula:
$P(X = 5) = \binom{10}{5} (0.5)^5 (0.5)^{5} = \frac{252}{1024} = \frac{63}{256}$
This matches the assertion. The reason correctly states the formula for the binomial probability mass function, which is used to derive the assertion. Hence, both are true, and the reason correctly explains the assertion.

Key Concept: Binomial Distribution, Bernoulli Trials

b) $\frac{105}{512}$
[Solution Description]
To find the probability of getting exactly 6 heads in 10 flips of a fair coin, we use the binomial distribution formula:
$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$
Here, $n = 10$ (number of trials), $k = 6$ (number of successes), and $p = 0.5$ (probability of success).
Step 1: Calculate $\binom{10}{6}$:
$\binom{10}{6} = \frac{10!}{6!4!} = 210$
Step 2: Compute $(0.5)^6$:
$(0.5)^6 = \frac{1}{64}$
Step 3: Compute $(0.5)^4$:
$(0.5)^4 = \frac{1}{16}$
Step 4: Multiply all terms together:
$P(X = 6) = 210 \times \frac{1}{64} \times \frac{1}{16} = \frac{210}{1024} = \frac{105}{512}$

Key Concept: Probability Density Function - Properties

c) $f(x) \geq 0$ and $\int_{-\infty}^{\infty} f(x) \, dx = 1$
[Solution Description]
For a function $f(x)$ to be a valid probability density function, it must satisfy two conditions:
1) $f(x) \geq 0$ for all $x$ in its domain.
2) The integral of $f(x)$ over its entire range must equal 1: $\int_{-\infty}^{\infty} f(x) \, dx = 1$.

Key Concept: Standard Deviation of a Continuous Random Variable

b) $\frac{\sqrt{2}}{6}$
[Solution Description]
First, calculate the mean $\mu$:
$\mu = \int_{0}^{1} x f(x) \, dx = \int_{0}^{1} x \cdot 2x \, dx = 2 \int_{0}^{1} x^2 \, dx = 2 \left[ \frac{x^3}{3} \right]_0^1 = \frac{2}{3}$
Next, compute $E(X^2)$:
$E(X^2) = \int_{0}^{1} x^2 f(x) \, dx = \int_{0}^{1} x^2 \cdot 2x \, dx = 2 \int_{0}^{1} x^3 \, dx = 2 \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{2}$
Variance $\sigma^2$ is:
$\sigma^2 = E(X^2) - \mu^2 = \frac{1}{2} - \left( \frac{2}{3} \right)^2 = \frac{1}{2} - \frac{4}{9} = \frac{1}{18}$
Standard deviation $\sigma$ is:
$\sigma = \sqrt{\sigma^2} = \sqrt{\frac{1}{18}} = \frac{\sqrt{2}}{6}$

Key Concept: Binomial Distribution, Variance Calculation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a binomial distribution with $n = 2$ trials, the variance is $\text{Var}(X) = npq = 2pq$. Here, $q = 1 - p$. The assertion correctly states the variance for this case. The reason provides the general formula for the variance of a binomial distribution, which includes the specific case when $n = 2$. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.