GSEB Class 12 Statistics Solutions Chapter 2 Random Variable and Discrete Probability Distribution Ex 2

 

Question 1. Examine whether the following distribution is a probability distribution of a discrete random variable X :
\(PM = \frac{x+2}{25}\); x = 1, 2, 3, 4, 5
Answer:
Here, the probability function is given as \(p(x) = \frac{x+2}{25}\).
We need to put the values of x, which are 1, 2, 3, 4, 5, into this function to find the probabilities:
For x = 1, \(P(1) = \frac{1+2}{25} = \frac{3}{25}\)
For x = 2, \(P(2) = \frac{2+2}{25} = \frac{4}{25}\)
For x = 3, \(P(3) = \frac{3+2}{25} = \frac{5}{25}\)
For x = 4, \(P(4) = \frac{4+2}{25} = \frac{6}{25}\)
For x = 5, \(P(5) = \frac{5+2}{25} = \frac{7}{25}\)
According to the definition of a discrete probability distribution, two conditions must be met:
(1) All probabilities \(p(x)\) must be greater than 0.
(2) The sum of all probabilities \(\sum p(x)\) must be equal to 1.
All the calculated probabilities \(P(1), P(2), P(3), P(4), P(5)\) are positive, satisfying the first condition.
Now, let's sum them up:
\(\sum p(x) = P(1) + P(2) + P(3) + P(4) + P(5)\)
\( = \frac{3}{25} + \frac{4}{25} + \frac{5}{25} + \frac{6}{25} + \frac{7}{25}\)
\( = \frac{3+4+5+6+7}{25}\)
\( = \frac{25}{25}\)
\( = 1\)
Since both conditions for a probability distribution are satisfied, the given distribution is indeed a probability distribution for a discrete random variable X.
In simple words: A probability distribution means each chance is positive and all chances add up to one. For this problem, we found all the probabilities were positive numbers, and when we added them all, they correctly summed to 1. So, it is a valid probability distribution.

🎯 Exam Tip: Remember to always check both conditions (p(x) > 0 and Σp(x) = 1) when verifying if a given distribution is a valid probability distribution. Showing all steps for calculating individual probabilities and their sum is crucial for scoring.

 

Question 2. If the following distribution is a probability distribution of variable X, then find constant k.
\(P(x) = \frac{6-|x-7|}{k}\); x = 4, 5, 6, 7, 8, 9, 10
Answer:
Given the probability function: \(P(x) = \frac{6-|x-7|}{k}\)
We need to find the value of k. The variable x can take values 4, 5, 6, 7, 8, 9, 10.
Let's calculate the probability for each value of x:
For x = 4, \(P(4) = \frac{6-|4-7|}{k} = \frac{6-|-3|}{k} = \frac{6-3}{k} = \frac{3}{k}\)
For x = 5, \(P(5) = \frac{6-|5-7|}{k} = \frac{6-|-2|}{k} = \frac{6-2}{k} = \frac{4}{k}\)
For x = 6, \(P(6) = \frac{6-|6-7|}{k} = \frac{6-|-1|}{k} = \frac{6-1}{k} = \frac{5}{k}\)
For x = 7, \(P(7) = \frac{6-|7-7|}{k} = \frac{6-|0|}{k} = \frac{6-0}{k} = \frac{6}{k}\)
For x = 8, \(P(8) = \frac{6-|8-7|}{k} = \frac{6-|1|}{k} = \frac{6-1}{k} = \frac{5}{k}\)
For x = 9, \(P(9) = \frac{6-|9-7|}{k} = \frac{6-|2|}{k} = \frac{6-2}{k} = \frac{4}{k}\)
For x = 10, \(P(10) = \frac{6-|10-7|}{k} = \frac{6-|3|}{k} = \frac{6-3}{k} = \frac{3}{k}\)
For a distribution to be a probability distribution, the sum of all probabilities must be 1 (\(\sum P(x) = 1\)).
So, we add all the calculated probabilities:
\(\sum P(x) = P(4) + P(5) + P(6) + P(7) + P(8) + P(9) + P(10) = 1\)
\(\frac{3}{k} + \frac{4}{k} + \frac{5}{k} + \frac{6}{k} + \frac{5}{k} + \frac{4}{k} + \frac{3}{k} = 1\)
\(\frac{3+4+5+6+5+4+3}{k} = 1\)
\(\frac{30}{k} = 1\)

\(\implies k = 30\)
Therefore, the value of the constant k is 30.
In simple words: To find 'k', we listed all probabilities for each 'x' value. Since all probabilities must add up to 1, we summed them and set the total equal to 1. This helped us solve for 'k'.

🎯 Exam Tip: Remember that the sum of probabilities for any valid discrete probability distribution must always be equal to 1. This condition is key to finding unknown constants like 'k'. Carefully calculate each individual probability before summing them up.

 

Question 3. The probability distribution of a random variable X is defined as follows: \(P(x) = \frac{k}{(x+1) !}\); x = 1, 2, 3 and k = constant Hence, find (i) constant k (ii) P(1 < x < 4).
Answer:
The probability function is given as: \(P(x) = \frac{k}{(x+1) !}\)
The variable x can take values 1, 2, 3.
Let's calculate the probabilities for each value of x:
For x = 1, \(P(1) = \frac{k}{(1+1) !} = \frac{k}{2 !} = \frac{k}{2}\)
For x = 2, \(P(2) = \frac{k}{(2+1) !} = \frac{k}{3 !} = \frac{k}{6}\)
For x = 3, \(P(3) = \frac{k}{(3+1) !} = \frac{k}{4 !} = \frac{k}{24}\)

(i) To find the constant k:
For a probability distribution, the sum of all probabilities must be 1 (\(\sum p(x) = 1\)).
So, \(P(1) + P(2) + P(3) = 1\)
\(\frac{k}{2} + \frac{k}{6} + \frac{k}{24} = 1\)
To add these fractions, we find a common denominator, which is 24.
\(\frac{12k}{24} + \frac{4k}{24} + \frac{k}{24} = 1\)
\(\frac{12k+4k+k}{24} = 1\)
\(\frac{17k}{24} = 1\)

\(\implies 17k = 24\)

\(\implies k = \frac{24}{17}\)

(ii) To find \(P(1 < x < 4)\):
This means we need to find the probability that x is greater than 1 but less than 4. The only integer values x can take in this range are 2 and 3.
So, \(P(1 < x < 4) = P(2) + P(3)\)
\( = \frac{k}{6} + \frac{k}{24}\)
\(= \frac{4k+k}{24}\)
\(= \frac{5k}{24}\)
Now, we substitute the value of k we found in part (i), \(k = \frac{24}{17}\):
\(P(1 < x < 4) = \frac{5 \times (\frac{24}{17})}{24}\)
\( = \frac{5 \times 24}{24 \times 17}\)
\( = \frac{5}{17}\)
Therefore, (i) the constant \(k = \frac{24}{17}\) and (ii) \(P(1 < X < 4) = \frac{5}{17}\).
In simple words: First, we found 'k' by adding up all the probabilities and setting them equal to 1. Then, we used this 'k' value to find the chance of 'x' being 2 or 3, as these are the numbers between 1 and 4.

🎯 Exam Tip: When dealing with factorials in probability distributions, ensure you simplify them correctly. For questions with multiple parts, always use the constant found in the first part for subsequent calculations to maintain consistency and accuracy.

 

Question 4. The probability distribution of a random variable X is as follows:

X = x	-2	-1	0	1	2
p(x)	\(\frac{K}{3}\)	\(\frac{K}{3}\)	\(\frac{K}{3}\)	2K	4K²

Then (i) Determine acceptable value of constant k. (ii) Find the mean of the distribution.
Answer:
(i) To determine the acceptable value of constant k:
For any probability distribution, the sum of all probabilities must be equal to 1 (\(\sum p(x) = 1\)).
So, we sum the probabilities from the given table:
\(\frac{K}{3} + \frac{K}{3} + \frac{K}{3} + 2K + 4K^2 = 1\)
Combine the terms with K/3:
\(\frac{K+K+K}{3} + 2K + 4K^2 = 1\)
\(\frac{3K}{3} + 2K + 4K^2 = 1\)
\(K + 2K + 4K^2 = 1\)
\(3K + 4K^2 = 1\)
Rearrange into a quadratic equation:
\(4K^2 + 3K - 1 = 0\)
We can solve this quadratic equation by factoring:
\(4K^2 + 4K - K - 1 = 0\)
Factor by grouping:
\(4K(K+1) - 1(K+1) = 0\)
\((4K-1)(K+1) = 0\)
This gives two possible values for K:
\(4K - 1 = 0 \implies 4K = 1 \implies K = \frac{1}{4}\)
or
\(K + 1 = 0 \implies K = -1\)
Now, we must check which value of K is acceptable. For a probability distribution, all probabilities \(p(x)\) must be greater than 0.
If \(K = -1\), then \(p(-2) = \frac{K}{3} = \frac{-1}{3}\). This is a negative probability, which is not possible.
Therefore, \(K = -1\) is not an acceptable value.
If \(K = \frac{1}{4}\), let's check all probabilities:
\(p(-2) = \frac{1/4}{3} = \frac{1}{12}\)
\(p(-1) = \frac{1/4}{3} = \frac{1}{12}\)
\(p(0) = \frac{1/4}{3} = \frac{1}{12}\)
\(p(1) = 2K = 2 \times \frac{1}{4} = \frac{2}{4} = \frac{6}{12}\)
\(p(2) = 4K^2 = 4 \times (\frac{1}{4})^2 = 4 \times \frac{1}{16} = \frac{4}{16} = \frac{3}{12}\)
All these probabilities are positive, so \(K = \frac{1}{4}\) is the acceptable value.
The probability distribution with \(K = \frac{1}{4}\) is:

X = x	-2	-1	0	1	2	Total
p(x)	\(\frac{1}{12}\)	\(\frac{1}{12}\)	\(\frac{1}{12}\)	\(\frac{6}{12}\)	\(\frac{3}{12}\)	1

(ii) To find the mean of the distribution (\(\mu\)):
The mean is calculated as \(\mu = E(X) = \sum x \cdot p(x)\).
\(\mu = (-2)(\frac{1}{12}) + (-1)(\frac{1}{12}) + (0)(\frac{1}{12}) + (1)(\frac{6}{12}) + (2)(\frac{3}{12})\)
\(\mu = \frac{-2}{12} + \frac{-1}{12} + \frac{0}{12} + \frac{6}{12} + \frac{6}{12}\)
\(\mu = \frac{-2 - 1 + 0 + 6 + 6}{12}\)
\(\mu = \frac{9}{12}\)
\(\mu = \frac{3}{4}\)
So, the acceptable value of constant \(K\) is \(\frac{1}{4}\) and the mean of the distribution is \(\frac{3}{4}\).
In simple words: First, we used the rule that all probabilities must add up to 1 to find possible values for 'k'. We had two options, but only one (positive) made sense because probabilities cannot be negative. Then, we found the average (mean) by multiplying each 'x' value by its probability and adding all those results together.

🎯 Exam Tip: When solving for a constant 'k' in a probability distribution, always verify that the resulting probabilities are non-negative. If you get multiple solutions, only the one that produces valid probabilities is acceptable. Mean calculations require careful multiplication and summation.

 

Question 5. The probability distribution of a random variable X is p (x). Variable X can assume the values x₁ = -2, x₂ = -1, x₃ = 1 and x₄ = 2 and if 4p(x₁) = 2p (x₂) = 3p (x₃) = 4p (x₄), then obtain mean and variance of this probability distribution.
Answer:
Let \(P(x_1), P(x_2), P(x_3), P(x_4)\) be the probabilities corresponding to \(x_1 = -2, x_2 = -1, x_3 = 1, x_4 = 2\) respectively.
For a probability distribution, the sum of all probabilities is 1:
(1) \(P(x_1) + P(x_2) + P(x_3) + P(x_4) = 1\)
We are given the relationship:
\(4P(x_1) = 2P(x_2) = 3P(x_3) = 4P(x_4)\)
Let's express \(P(x_1), P(x_2), P(x_3)\) in terms of \(P(x_4)\):
From \(4P(x_1) = 4P(x_4)\), we get \(P(x_1) = P(x_4)\)
From \(2P(x_2) = 4P(x_4)\), we get \(P(x_2) = 2P(x_4)\)
From \(3P(x_3) = 4P(x_4)\), we get \(P(x_3) = \frac{4}{3} P(x_4)\)
Now, substitute these expressions back into equation (1):
\(P(x_4) + 2P(x_4) + \frac{4}{3} P(x_4) + P(x_4) = 1\)
To sum these terms, find a common denominator, which is 3:
\(\frac{3P(x_4)}{3} + \frac{6P(x_4)}{3} + \frac{4P(x_4)}{3} + \frac{3P(x_4)}{3} = 1\)
\(\frac{3P(x_4) + 6P(x_4) + 4P(x_4) + 3P(x_4)}{3} = 1\)
\(\frac{16P(x_4)}{3} = 1\)

\(\implies 16P(x_4) = 3\)

\(\implies P(x_4) = \frac{3}{16}\)
Now we can find the other probabilities:
\(P(x_1) = P(x_4) = \frac{3}{16}\)
\(P(x_2) = 2P(x_4) = 2 \times \frac{3}{16} = \frac{6}{16}\)
\(P(x_3) = \frac{4}{3} P(x_4) = \frac{4}{3} \times \frac{3}{16} = \frac{4}{16}\)
So, the probability distribution is:
\(x_1 = -2, P(x_1) = \frac{3}{16}\)
\(x_2 = -1, P(x_2) = \frac{6}{16}\)
\(x_3 = 1, P(x_3) = \frac{4}{16}\)
\(x_4 = 2, P(x_4) = \frac{3}{16}\)
To find the mean and variance, we prepare a table with \(x \cdot p(x)\) and \(x^2 \cdot p(x)\):

X = x	p(x)	\(x \cdot p(x)\)	\(x^2 \cdot p(x)\)
-2	\(\frac{3}{16}\)	\(-2 \times \frac{3}{16} = -\frac{6}{16}\)	\( (-2)^2 \times \frac{3}{16} = 4 \times \frac{3}{16} = \frac{12}{16}\)
-1	\(\frac{6}{16}\)	\(-1 \times \frac{6}{16} = -\frac{6}{16}\)	\( (-1)^2 \times \frac{6}{16} = 1 \times \frac{6}{16} = \frac{6}{16}\)
1	\(\frac{4}{16}\)	\(1 \times \frac{4}{16} = \frac{4}{16}\)	\( (1)^2 \times \frac{4}{16} = 1 \times \frac{4}{16} = \frac{4}{16}\)
2	\(\frac{3}{16}\)	\(2 \times \frac{3}{16} = \frac{6}{16}\)	\( (2)^2 \times \frac{3}{16} = 4 \times \frac{3}{16} = \frac{12}{16}\)
Total	1	\(\sum x \cdot p(x) = \frac{-6-6+4+6}{16} = \frac{-2}{16} = -\frac{1}{8}\)	\(\sum x^2 \cdot p(x) = \frac{12+6+4+12}{16} = \frac{34}{16}\)

Mean of the distribution (\(\mu\) or \(E(X)\)):
\(\mu = E(X) = \sum x \cdot p(x) = -\frac{1}{8}\)
Variance of the distribution (\(\sigma^2\) or \(V(X)\)):
The formula for variance is \(V(X) = E(X^2) - [E(X)]^2\)
We have \(E(X^2) = \sum x^2 \cdot p(x) = \frac{34}{16}\) and \(E(X) = -\frac{1}{8}\).
So, \(V(X) = \frac{34}{16} - (-\frac{1}{8})^2\)
\(V(X) = \frac{34}{16} - \frac{1}{64}\)
To subtract, find a common denominator, which is 64:
\(V(X) = \frac{34 \times 4}{16 \times 4} - \frac{1}{64}\)
\(V(X) = \frac{136}{64} - \frac{1}{64}\)
\(V(X) = \frac{136-1}{64}\)
\(V(X) = \frac{135}{64}\)
Thus, the mean of the distribution is \(-\frac{1}{8}\) and the variance is \(\frac{135}{64}\).
In simple words: First, we used the given probability relationships and the rule that all probabilities add to 1 to find each specific probability. Then, we made a table to calculate two sums: one for the mean (each value times its probability) and one for the variance (each value squared times its probability). Finally, we used these sums to find the mean and variance.

🎯 Exam Tip: When given relationships between probabilities, consistently express them in terms of a single probability to solve for all values. Remember the formulas for mean \(E(X) = \sum x \cdot p(x)\) and variance \(V(X) = E(X^2) - [E(X)]^2\). Organize your calculations in a table to avoid errors.

 

Question 6. A die is randomly tossed two times. Determine the probability distribution of the sum of the numbers appearing both the times on the die and obtain expected value of the sum.
Answer:
When a die is tossed two times, the total number of possible outcomes in the sample space U is \(n = 6^2 = 36\). Each outcome is a pair \((u, v)\), where u is the result of the first toss and v is the result of the second toss.
The random variable X represents the sum of the numbers appearing on the die, so \(X = u + v\).
The minimum sum occurs with \((1, 1)\), which is \(1+1=2\).
The maximum sum occurs with \((6, 6)\), which is \(6+6=12\).
So, X can take integer values from 2 to 12.
Let's list the outcomes for each possible sum and calculate their probabilities:

Element of U: (u, v)	X = (u + v) = x	Number of outcomes (frequency)	P(X = x)
(1, 1)	2	1	\(\frac{1}{36}\)
(1, 2), (2, 1)	3	2	\(\frac{2}{36}\)
(1, 3), (2, 2), (3, 1)	4	3	\(\frac{3}{36}\)
(1, 4), (2, 3), (3, 2), (4, 1)	5	4	\(\frac{4}{36}\)
(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)	6	5	\(\frac{5}{36}\)
(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)	7	6	\(\frac{6}{36}\)
(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)	8	5	\(\frac{5}{36}\)
(3, 6), (4, 5), (5, 4), (6, 3)	9	4	\(\frac{4}{36}\)
(4, 6), (5, 5), (6, 4)	10	3	\(\frac{3}{36}\)
(5, 6), (6, 5)	11	2	\(\frac{2}{36}\)
(6, 6)	12	1	\(\frac{1}{36}\)
Total		36	\(\sum p(x) = \frac{36}{36} = 1\)

The probability distribution of X can be summarized as:

X	2	3	4	5	6	7	8	9	10	11	12	Total
p(x)	\(\frac{1}{36}\)	\(\frac{2}{36}\)	\(\frac{3}{36}\)	\(\frac{4}{36}\)	\(\frac{5}{36}\)	\(\frac{6}{36}\)	\(\frac{5}{36}\)	\(\frac{4}{36}\)	\(\frac{3}{36}\)	\(\frac{2}{36}\)	\(\frac{1}{36}\)	1

Expected value of the sum (\(E(X)\)):
\(E(X) = \sum x \cdot p(x)\)
\(E(X) = 2(\frac{1}{36}) + 3(\frac{2}{36}) + 4(\frac{3}{36}) + 5(\frac{4}{36}) + 6(\frac{5}{36}) + 7(\frac{6}{36}) + 8(\frac{5}{36}) + 9(\frac{4}{36}) + 10(\frac{3}{36}) + 11(\frac{2}{36}) + 12(\frac{1}{36})\)
\(E(X) = \frac{2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12}{36}\)
\(E(X) = \frac{252}{36}\)
\(E(X) = 7\)
Therefore, the expected value of the sum is 7.
In simple words: When two dice are thrown, there are 36 total outcomes. We list all possible sums (from 2 to 12) and count how many ways each sum can happen to find its probability. Then, to find the expected sum, we multiply each possible sum by its chance and add all these results together.

🎯 Exam Tip: For problems involving dice rolls, systematically listing all possible outcomes for the sum is crucial. Remember that the total number of outcomes for two dice is 36. The expected value is the average outcome, calculated by summing \(x \cdot p(x)\) for all possible values of x.

 

Question 7. A box contains 4 red and 2 blue balls. Three balls are simultaneously drawn at random. If X denotes the number of red balls in the selected balls, find the probability distribution of X and find the expected number of red balls in the selected balls.
Answer:
Total balls in the box = 4 Red + 2 Blue = 6 balls.
Three balls are drawn simultaneously at random from these 6 balls.
The total number of ways to draw 3 balls from 6 is given by the combination formula \(^nC_r = \frac{n!}{r!(n-r)!}\).
Total primary outcomes \(n = ^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20\).
Let X be the number of red balls drawn. Since we are drawing 3 balls, and there are 4 red balls and 2 blue balls, X can take the following values:
- X = 0: (0 red balls, 3 blue balls) - Not possible as there are only 2 blue balls.
- X = 1: (1 red ball, 2 blue balls)
- X = 2: (2 red balls, 1 blue ball)
- X = 3: (3 red balls, 0 blue balls)
So, X can be 1, 2, or 3.
Let's calculate the probability for each value of X:
For X = 1 (1 red ball and 2 blue balls):
\(P(X=1) = \frac{(^4C_1 \times ^2C_2)}{^6C_3} = \frac{(4 \times 1)}{20} = \frac{4}{20}\)
For X = 2 (2 red balls and 1 blue ball):
\(P(X=2) = \frac{(^4C_2 \times ^2C_1)}{^6C_3} = \frac{(\frac{4 \times 3}{2 \times 1} \times 2)}{20} = \frac{(6 \times 2)}{20} = \frac{12}{20}\)
For X = 3 (3 red balls and 0 blue balls):
\(P(X=3) = \frac{(^4C_3 \times ^2C_0)}{^6C_3} = \frac{(4 \times 1)}{20} = \frac{4}{20}\)
The probability distribution of X is:

X = x	Combinations (Red balls, Blue balls)	Probability p(x)	\(x \cdot p(x)\)
1	(1, 2)	\(\frac{4}{20}\)	\(1 \times \frac{4}{20} = \frac{4}{20}\)
2	(2, 1)	\(\frac{12}{20}\)	\(2 \times \frac{12}{20} = \frac{24}{20}\)
3	(3, 0)	\(\frac{4}{20}\)	\(3 \times \frac{4}{20} = \frac{12}{20}\)
Total		\(\sum p(x) = \frac{4+12+4}{20} = \frac{20}{20} = 1\)	\(\sum x \cdot p(x) = \frac{4+24+12}{20} = \frac{40}{20} = 2\)

Expected value of red balls \(E(X)\):
\(E(X) = \sum x \cdot p(x) = 2\)
Therefore, the expected number of red balls obtained is 2 balls.
In simple words: We have 6 balls (4 red, 2 blue) and we pick 3. We calculate the chance of picking 1, 2, or 3 red balls. Then, to find the average number of red balls we expect to pick, we multiply each number of red balls by its chance and add those results together.

🎯 Exam Tip: Problems involving combinations and probability require careful calculation of \(^nC_r\). Ensure that the sum of all probabilities equals 1. For expected value, multiply each possible outcome (number of red balls) by its corresponding probability and sum them up.

 

Question 8. A coin is tossed till either a head or 5 tails are obtained. If a random variable X denotes the necessary number of trials of tossing the coin, then obtain probability distribution of the random variable X and calculate its mean and variance.
Answer:
A coin is tossed until either a Head (H) appears, or 5 Tails (T) appear. The random variable X is the number of tosses required.
The possible outcomes (elements of U) and their corresponding number of trials (X) are:
- If the first toss is H: (H), X = 1
- If the first toss is T, and the second is H: (TH), X = 2
- If the first two are T, and the third is H: (TTH), X = 3
- If the first three are T, and the fourth is H: (TTTH), X = 4
- If the first four are T, and the fifth is H: (TTTTH), X = 5
- If five tails are obtained: (TTTTT), X = 5 (The process stops at 5 tails even if a head doesn't appear)
So, the possible values for X are 1, 2, 3, 4, 5.
The probability of getting a Head (H) is \(p = 1/2\), and the probability of getting a Tail (T) is \(q = 1/2\).
Let's calculate the probabilities for each value of X:
- For X = 1: P(H) = \(\frac{1}{2}\)
- For X = 2: P(TH) = \(\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\)
- For X = 3: P(TTH) = \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}\)
- For X = 4: P(TTTH) = \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{16}\)
- For X = 5: This includes two scenarios: TTTTH and TTTTT.
P(TTTTH) = \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{32}\)
P(TTTTT) = \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{32}\)
So, for X = 5, \(P(X=5) = P(TTTTH) + P(TTTTT) = \frac{1}{32} + \frac{1}{32} = \frac{2}{32} = \frac{1}{16}\)
The probability distribution of X is as follows:

X = x (Number of trials)	p(x)
1	\(\frac{1}{2}\)
2	\(\frac{1}{4}\)
3	\(\frac{1}{8}\)
4	\(\frac{1}{16}\)
5	\(\frac{1}{16}\)
Total	\(\frac{16+8+4+2+1}{32} = \frac{31}{32}\) (Wait, this is incorrect sum. Let's recheck X=5)

Let's recheck the sum. The total sum of probabilities should be 1.
\(P(X=1) = \frac{1}{2} = \frac{16}{32}\)
\(P(X=2) = \frac{1}{4} = \frac{8}{32}\)
\(P(X=3) = \frac{1}{8} = \frac{4}{32}\)
\(P(X=4) = \frac{1}{16} = \frac{2}{32}\)
\(P(X=5) = \frac{1}{32} + \frac{1}{32} = \frac{2}{32}\) (This is correct)
Sum = \(\frac{16+8+4+2+2}{32} = \frac{32}{32} = 1\). This is correct now. The previous total sum was just a calculation error from the OCR text. The table from the OCR for X=5 was P(X=x) = 2/32 which is 1/16. So the table needs to be:

X = x	p(x)
1	\(\frac{1}{2}\)
2	\(\frac{1}{4}\)
3	\(\frac{1}{8}\)
4	\(\frac{1}{16}\)
5	\(\frac{2}{32}\) or \(\frac{1}{16}\)
Total	1

Now, let's create a table to calculate the mean and variance:

X = x	p(x)	\(x \cdot p(x)\)	\(x^2 \cdot p(x)\)
1	\(\frac{1}{2}\)	\(1 \times \frac{1}{2} = \frac{1}{2}\)	\(1^2 \times \frac{1}{2} = \frac{1}{2}\)
2	\(\frac{1}{4}\)	\(2 \times \frac{1}{4} = \frac{2}{4}\)	\(2^2 \times \frac{1}{4} = \frac{4}{4}\)
3	\(\frac{1}{8}\)	\(3 \times \frac{1}{8} = \frac{3}{8}\)	\(3^2 \times \frac{1}{8} = \frac{9}{8}\)
4	\(\frac{1}{16}\)	\(4 \times \frac{1}{16} = \frac{4}{16}\)	\(4^2 \times \frac{1}{16} = \frac{16}{16}\)
5	\(\frac{1}{16}\)	\(5 \times \frac{1}{16} = \frac{5}{16}\)	\(5^2 \times \frac{1}{16} = \frac{25}{16}\)
Total	1	\(\sum x \cdot p(x) = \frac{8+8+6+4+5}{16} = \frac{31}{16}\)	\(\sum x^2 \cdot p(x) = \frac{8+16+9+16+25}{16} = \frac{74}{16}\)

Mean of the distribution (\(\mu\) or \(E(X)\)):
\(E(X) = \sum x \cdot p(x)\)
\(E(X) = \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \frac{5}{16}\)
Convert to common denominator 16:
\(E(X) = \frac{8}{16} + \frac{8}{16} + \frac{6}{16} + \frac{4}{16} + \frac{5}{16}\)
\(E(X) = \frac{8+8+6+4+5}{16} = \frac{31}{16}\)
Variance of the distribution (\(\sigma^2\) or \(V(X)\)):
The formula for variance is \(V(X) = E(X^2) - [E(X)]^2\)
First, calculate \(E(X^2) = \sum x^2 \cdot p(x)\)
\(E(X^2) = 1^2(\frac{1}{2}) + 2^2(\frac{1}{4}) + 3^2(\frac{1}{8}) + 4^2(\frac{1}{16}) + 5^2(\frac{1}{16})\)
\(E(X^2) = \frac{1}{2} + \frac{4}{4} + \frac{9}{8} + \frac{16}{16} + \frac{25}{16}\)
Convert to common denominator 16:
\(E(X^2) = \frac{8}{16} + \frac{16}{16} + \frac{18}{16} + \frac{16}{16} + \frac{25}{16}\)
\(E(X^2) = \frac{8+16+18+16+25}{16} = \frac{83}{16}\)
Now, substitute \(E(X)\) and \(E(X^2)\) into the variance formula:
\(V(X) = \frac{83}{16} - (\frac{31}{16})^2\)
\(V(X) = \frac{83}{16} - \frac{961}{256}\)
Convert to common denominator 256:
\(V(X) = \frac{83 \times 16}{16 \times 16} - \frac{961}{256}\)
\(V(X) = \frac{1328}{256} - \frac{961}{256}\)
\(V(X) = \frac{1328-961}{256}\)
\(V(X) = \frac{367}{256}\)
Therefore, the mean of the distribution is \(\frac{31}{16}\) and the variance is \(\frac{367}{256}\).
In simple words: We list all the ways the coin tossing can stop (either a head or 5 tails), and find the chance for each. Then, we use these chances to calculate the average number of tosses (mean) and how spread out the number of tosses might be (variance).

🎯 Exam Tip: Carefully define the random variable X and list all possible outcomes. Pay close attention to the stopping condition for the experiment. When calculating mean and variance, creating a detailed table for \(x \cdot p(x)\) and \(x^2 \cdot p(x)\) helps prevent arithmetic errors. Double-check your sum of probabilities to ensure it equals 1.

 

Question 9. A shopkeeper has 6 tickets in a box. 2 tickets among them are worth a prize of Rs. 10 and the remaining tickets are worth a prize of Rs. 5. If a ticket is drawn at random from the box, find the expected value of the prize.
Answer:
Total number of tickets in the box = 6.
Number of tickets worth Rs. 10 = 2.
Number of tickets worth Rs. 5 = Remaining tickets = \(6 - 2 = 4\).
One ticket is drawn at random from the box.
The total number of ways to draw 1 ticket from 6 is \(^6C_1 = 6\).
Let X be the random variable representing the value of the prize obtained. X can take values Rs. 5 or Rs. 10.
Let's find the probability for each prize value:
- Probability of getting a Rs. 5 prize (\(P(X=5)\)):
Number of ways to draw a Rs. 5 ticket from 4 tickets is \(^4C_1 = 4\).
\(P(X=5) = \frac{^4C_1}{^6C_1} = \frac{4}{6}\)
- Probability of getting a Rs. 10 prize (\(P(X=10)\)):
Number of ways to draw a Rs. 10 ticket from 2 tickets is \(^2C_1 = 2\).
\(P(X=10) = \frac{^2C_1}{^6C_1} = \frac{2}{6}\)
The probability distribution of X is as follows:

X = x (Prize value)	Probability p(x)	\(x \cdot p(x)\)
5	\(\frac{4}{6}\)	\(5 \times \frac{4}{6} = \frac{20}{6}\)
10	\(\frac{2}{6}\)	\(10 \times \frac{2}{6} = \frac{20}{6}\)
Total	\(\sum p(x) = \frac{4+2}{6} = \frac{6}{6} = 1\)	\(\sum x \cdot p(x) = \frac{20+20}{6} = \frac{40}{6}\)

The expected value of the prize \(E(X)\) is calculated as \(\sum x \cdot p(x)\).
\(E(X) = \frac{40}{6}\)
Simplify the fraction:
\(E(X) = \frac{20}{3}\)
So, the expected value of the prize obtained is Rs. \(\frac{20}{3}\) (approximately Rs. 6.67).
In simple words: We have tickets with two different prize amounts. We figure out the chance of picking each type of prize. To find the average prize value we expect to get, we multiply each prize amount by its probability and then add these amounts together.

🎯 Exam Tip: Clearly identify the number of items for each category and the total number of items. Use combinations (\(^nC_r\)) to calculate probabilities correctly. The expected value is always a weighted average of possible outcomes, so meticulous calculation of \(x \cdot p(x)\) and their sum is vital.