Maharashtra Board Class 12 Maths Part 2 Chapter 8 Probability Distributions 8.1 Solutions

Std 12 Maths 2 Exercise 8.1 Solutions Commerce Maths

Question 1. Let X represent the difference between a number of heads and the number of tails obtained when a coin is tossed 6 times. What are the possible values of X?
Answer:Solution:
A coin is tossed 6 times
S = {6H and 0T, 5H and 1T, 4H and 2T, 3H and 3T, 2H and 4T, 1H and 5T, 0H and 6T}
X: Difference between no. of heads and no. of tails.
X = 6-0 = 6
X = 5-1 = 4
X = 4-2 = 2
X = 3-3 = 0
X = 2-4 = -2
X = 1-5 = -4
X = 0-6 = -6
X = {-6, -4, -2, 0, 2, 4, 6}
In simple words: When a coin is tossed 6 times, we list all possible combinations of heads and tails. Then, for each combination, we calculate the difference between the number of heads and tails. These differences are the possible values of X.

🎯 Exam Tip: Listing the sample space systematically helps ensure all possible outcomes are covered when determining the range of a random variable.

 

Question 2. An urn contains 5 red and 2 black balls. Two balls are drawn at random. X denotes the number of black balls drawn. What are the possible values of X?
Answer:Solution:
S: Two balls are drawn from the Urn
S = {RR, RB, BR, BB}
X: No. of black balls
\( \therefore \) X = {0, 1, 2}
In simple words: We are drawing two balls from an urn containing red and black balls. The random variable X represents the count of black balls drawn, which can be zero, one, or two.

🎯 Exam Tip: Define the sample space (S) and clearly identify the random variable (X) and its possible outcomes before calculating probabilities.

 

Question 3. Determine whether each of the following is a probability distribution. Give reasons for your answer.
(i)

x	0	1	2
P(x)	0.4	0.4	0.2

Answer:Solution:
Here, P(X = x) \( \ge \) 0, \( \forall \) x and
\( \sum_{x=0}^{2} P(X=x) \) = p(0) + p(1) + p(2)
= 0.4 + 0.4 + 0.2
= 1
\( \therefore \) The function is a p.m.f.
In simple words: For a function to be a probability mass function (p.m.f.), all probabilities must be non-negative, and their sum must equal 1. This distribution satisfies both conditions.

🎯 Exam Tip: Remember the two fundamental conditions for a probability distribution: all probabilities must be non-negative, and their sum over the entire sample space must be exactly 1.

 

(ii)

X	0	1	2	3	4
P(x)	0.1	0.5	0.2	-0.1	0.3

Answer:Solution:
Here, p(3) = -0.1 < 0
\( \therefore \) P(X = x) >/0, \( \forall \) x
\( \therefore \) The function is not a p.m.f.
In simple words: This is not a probability mass function because one of the probabilities, p(3), is negative. Probabilities cannot be less than zero.

🎯 Exam Tip: The most common error in identifying a p.m.f. is failing to check if all individual probabilities are non-negative.

 

(iii)

X	0	1	2
P(x)	0.1	0.6	0.3

Answer:Solution:
Here, P(X = x) \( \ge \) 0, \( \forall \) x and
\( \sum_{x=0}^{2} P(X=x) \) = p(0) + p(1) + p(2)
= 0.1 + 0.6 + 0.3
= 1
\( \therefore \) The function is a p.m.f.
In simple words: This distribution is a valid probability mass function because all probabilities are positive, and their sum is exactly 1.

🎯 Exam Tip: Always perform both checks: individual probabilities for non-negativity and the total sum for equality to 1.

 

(iv)

Z	3	2	1	0	-1
P(z)	0.3	0.2	0.4	0.05	0.05

Answer:Solution:
Here, P(Z = z) \( \ge \) 0, \( \forall \) z and
\( \sum_{z=-1}^{3} P(Z=z) \) = p(-1) + p(0) + p(1) + p(2) + p(3)
= 0.05 + 0 + 0.4 + 0.2 + 0.3
= 0.95
\( \ne \) 1
\( \therefore \) The function is not a p.m.f.
In simple words: This is not a probability mass function even though all probabilities are non-negative, because their sum (0.95) is not equal to 1.

🎯 Exam Tip: A sum that deviates from 1, even slightly, indicates that the function is not a valid probability distribution. Double-check all addition.

 

(v)

y	-1	0	1
P(y)	0.6	0.1	0.2

Answer:Solution:
Here, P(Y = y) \( \ge \) 0, \( \forall \) y and
\( \sum_{y=-1}^{1} P(Y=y) \) = p(-1) + p(0) + p(1)
= 0.1 + 0.6 + 0.2
= 0.9
\( \ne \) 1
\( \therefore \) The function is not a p.m.f.
In simple words: This function is not a probability mass function because the sum of its probabilities is 0.9, which is not equal to 1.

🎯 Exam Tip: Pay close attention to the sum of probabilities; it must always be exactly 1 for a valid distribution.

 

(vi)

X	0	1	2
P(x)	0.3	0.4	0.2

Answer:Solution:
Here, P(X = x) \( \ge \) 0, \( \forall \) x and
\( \sum_{x=0}^{2} P(X=x) \) = p(-2) + p(-1) + p(0)
= 0.3 + 0.4 + 0.2
= 0.9
\( \ne \) 1
\( \therefore \) The function is not a p.m.f.
In simple words: Although all probabilities are non-negative, this is not a valid p.m.f. because the sum of probabilities is 0.9, not 1.

🎯 Exam Tip: Always verify that the sum of all probabilities in a distribution table equals 1; if not, it's not a valid p.m.f.

 

Question 4. Find the probability distribution of
(i) number of heads in two tosses of a coin,
(ii) number of trails in three tosses of a coin,
(iii) number of heads in four tosses of a coin.
Answer:Solution:
(i) S: Coin is tossed two times
S = {HH, HT, TH, TT}
n(S) = 4
X: No. of heads
Range of X = {0, 1, 2}
p.m.f. Table

x	0	1	2
p	\( \frac{1}{4} \)	\( \frac{2}{4} \)	\( \frac{1}{4} \)

(ii) S: 3 coin are tossed
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
X: No. of heads
Range of X = {0, 1, 2, 3}
p.m.f. Table

x	0	1	2	3
p	\( \frac{1}{8} \)	\( \frac{3}{8} \)	\( \frac{3}{8} \)	\( \frac{1}{8} \)

(iii) S: Four coin are tossed
S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}
n(S) = 16
X: No. of heads
Range of X = {0, 1, 2, 3, 4}
p.m.f. Table

x	0	1	2	3	4
p	\( \frac{1}{16} \)	\( \frac{4}{16} = \frac{1}{4} \)	\( \frac{6}{16} = \frac{3}{8} \)	\( \frac{4}{16} = \frac{1}{4} \)	\( \frac{1}{16} \)

In simple words: For each scenario, we define the sample space of all possible outcomes. Then, we identify the random variable (number of heads or tails) and list its possible values. Finally, we calculate the probability of each value occurring to form the probability distribution table.

🎯 Exam Tip: When constructing p.m.f. tables for coin tosses, ensure the sum of probabilities always equals 1. This is a good self-check for accuracy.

 

Question 5. Find the probability distribution of the number of successes in two tosses of a die if successes are defined as getting a number greater than 4.
Answer:Solution:
S = A die is tossed 2 times
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(S) = 36
X = No. getting greater than 4
Range of X = {0, 1, 2}
p(0) = \( \frac{16}{36} = \frac{4}{9} \)
p(1) = \( \frac{16}{36} = \frac{4}{9} \)
p(2) = \( \frac{4}{36} = \frac{1}{9} \)

x	0	1	2
P	\( \frac{4}{9} \)	\( \frac{4}{9} \)	\( \frac{1}{9} \)

In simple words: We define "success" as rolling a 5 or 6. When tossing a die twice, we count how many times this success occurs (0, 1, or 2 times) and calculate the probability for each count to form the distribution.

🎯 Exam Tip: Clearly define what constitutes a "success" and "failure" for each trial before enumerating outcomes and calculating probabilities.

 

Question 6. A sample of 4 bulbs is drawn at random with replacement from a lot of 30 bulbs which includes 6 defective bulbs. Find the probability distribution of the number of defective bulbs.
Answer:Solution:
Total no. of bulbs = 30
No. of defective bulbs = 6
A sample of 4 bulbs are drawn from 30 bulbs.
\( \therefore \) n(S) = \(^{30}C_4\)
\( \therefore \) No. of non-defective bulbs = 24
Let X = No. of defective bulbs drawn in sample of 4 bulbs.
\( \therefore \)X = {0, 1, 2, 3, 4}.
For X = 0, p(0) = \( \frac{^6C_0 \times ^{24}C_4}{^{30}C_4} \)
= 1 x \( \frac{24 \times 23 \times 22 \times 21}{1 \times 2 \times 3 \times 4} \) \( / \frac{30 \times 29 \times 28 \times 27}{1 \times 2 \times 3 \times 4} \)
= \( \frac{3542}{9135} = \frac{506}{1305} \)
For X = 1, p(1) = \( \frac{^6C_1 \times ^{24}C_3}{^{30}C_4} \)
= 6x \( \frac{24 \times 23 \times 22}{1 \times 2 \times 3} \) \( / \frac{30 \times 29 \times 28 \times 27}{1 \times 2 \times 3 \times 4} \)
= \( \frac{4048}{9135} \)
For X = 2, p(2) = \( \frac{^6C_2 \times ^{24}C_2}{^{30}C_4} \)
= \( \frac{6 \times 5}{1 \times 2} \) x \( \frac{24 \times 23}{1 \times 2} \) \( / \frac{30 \times 29 \times 28 \times 27}{1 \times 2 \times 3 \times 4} \)
= \( \frac{1380}{9135} = \frac{92}{609} \)
For X = 3, p(3) = \( \frac{^6C_3 \times ^{24}C_1}{^{30}C_4} \)
= \( \frac{6 \times 5 \times 4}{1 \times 2 \times 3} \) \( \times \) 24 \( / \frac{30 \times 29 \times 28 \times 27}{1 \times 2 \times 3 \times 4} \)
= \( \frac{160}{9135} = \frac{32}{1827} \)
For X = 4, p(4) = \( \frac{^6C_4 \times ^{24}C_0}{^{30}C_4} \)
= \( \frac{6 \times 5 \times 4 \times 3}{1 \times 2 \times 3 \times 4} \) x 1 \( / \frac{30 \times 29 \times 28 \times 27}{1 \times 2 \times 3 \times 4} \)
= \( \frac{5}{9135} = \frac{1}{1827} \)
...Probability distribution of no. of defective bulbs is

x	0	1	2	3	4
p(x)	\( \frac{506}{1305} \)	\( \frac{4048}{9135} \)	\( \frac{92}{609} \)	\( \frac{32}{1827} \)	\( \frac{1}{1827} \)

In simple words: This problem involves drawing a sample of bulbs from a lot with defective and non-defective items. We use combinations to calculate the probability of drawing 0, 1, 2, 3, or 4 defective bulbs in a sample of 4, creating the probability distribution.

🎯 Exam Tip: When drawing "with replacement," the probability of selecting a defective bulb remains constant for each draw. For "without replacement" (as suggested by the use of combinations in the solution), probabilities change with each draw.

 

Question 7. A coin is biased so that the head is 3 times as likely to occur as the tail. Find the probability distribution of a number of tails in two tosses.
Answer:Solution:
Here, the head is 3 times as likely to occur as the tail.
i.e., If 4 times coin is tossed, 3 times there will be a head and 1 time there will be the tail.
\( \therefore \) p(H) = \( \frac{3}{4} \) and p(T) = \( \frac{1}{4} \)
Let X : No. of tails in two tosses.
And coin is tossed twice.
\( \therefore \) X = {0, 1, 2}
For X = 0,
p(0) = p(both heads)
= p(H) \( \times \) p(H)
= \( \frac{3}{4} \times \frac{3}{4} \)
= \( \frac{9}{16} \)
For X = 1,
p(1) = p(HT or TH)
= p(HT) + p(TH)
= p(H) \( \times \) p(T) + p(T) \( \times \) p(H)
= \( \frac{3}{4} \times \frac{1}{4} + \frac{1}{4} \times \frac{3}{4} \)
= \( \frac{6}{16} \)
For X = 2,
p(2) = p(both tails)
= p(T) \( \times \) p(T)
= \( \frac{1}{4} \times \frac{1}{4} \)
= \( \frac{1}{16} \)
The probability distribution of the number of tails in two tosses is

x	0	1	2
p	\( \frac{9}{16} \)	\( \frac{6}{16} \)	\( \frac{1}{16} \)

In simple words: First, we determine the individual probabilities of getting a head or a tail from the biased coin. Then, for two tosses, we calculate the probability of observing zero, one, or two tails using these individual probabilities.

🎯 Exam Tip: For independent events like coin tosses, the probability of a sequence of outcomes is the product of their individual probabilities. Remember to sum probabilities for mutually exclusive events that contribute to the same outcome (e.g., HT and TH for one tail).

 

Question 8. A random variable X has the following probability distribution:

X	1	2	3	4	5	6	7
P(x)	k	2k	2k	3k	k\(^2\)	2k\(^2\)	7k\(^2\)+k

Determine (i) k, (ii) P(X < 3), (iii) P(0 < X < 3), (iv) P(X > 4).
Answer:Solution:
(i) It is a p.m.f. of r.v. X
\( \therefore \) \( \Sigma \rho \)(x) = 1
\( \therefore \) p(1) + p(2) + p(3) + p(4) + p(5) + p(6) + p(7) = 1
\( \therefore \) k + 2k + 2k + 3k + k\(^2\) + 2k\(^2\) + (7k\(^2\) + k) = 1
\( \therefore \) 10k\(^2\) + 9k = 1
\( \therefore \) 10k\(^2\) + 9k - 1 = 0
\( \therefore \) 10k\(^2\) + 10k -k-1 = 0
\( \therefore \) 10k(k + 1) - (k + 1) = 0
\( \therefore \) (10k - 1)(k + 1) = 0
\( \therefore \) 10k - 1 = 0 or k + 1 = 0
\( \therefore \) k = \( \frac{1}{10} \) or k = -1
but k = -1 is not accepted
\( \therefore \) k = \( \frac{1}{10} \) is accepted
(ii) P(X < 3) = p(1) + p(2)
= k + 2k
= 3k
= 3 x \( \frac{1}{10} \)
= \( \frac{3}{10} \)
(iii) P(0 < X < 3) = p(1) + p(2)
= k + 2k
= 3k
= 3 x \( \frac{1}{10} \)
= \( \frac{3}{10} \)
(iv) P(X > 4) = p(5) + p(6) + p(7)
= k\(^2\) + 2k\(^2\) + (7k\(^2\) + k)
= 10k\(^2\) + k
= 10(\( \frac{1}{10} \)\(^2\)) + \( \frac{1}{10} \)
= \( \frac{2}{10} \)
= \( \frac{1}{5} \)
In simple words: We find the value of 'k' by summing all probabilities and equating them to 1, then solving the resulting quadratic equation. Once 'k' is known, we can calculate the probabilities for different ranges of X by summing the relevant P(x) values.

🎯 Exam Tip: Always verify that the obtained value of 'k' results in all probabilities being non-negative. If any P(x) becomes negative, 'k' is not a valid solution.

 

Question 9. Find expected value and variance of X using the following p.m.f.

x	-2	-1	0	1	2
P(x)	0.2	0.3	0.1	0.15	0.25

Answer:Solution:

x	p	px	x\(^2\)	x\(^2\)p
-2	0.2	-0.4	4	0.8
-1	0.3	-0.3	1	0.3
0	0.1	0	0	0
1	0.15	0.15	1	0.15
2	0.25	0.5	4	1
Total		-0.05		2.25

E(X) = \( \Sigma \)xp = -0.05
V(X) = \( \Sigma \)x\(^2\)p - (\( \Sigma \)xp)\(^2\)
= 2.25 - (-0.05)\(^2\)
= 2.25 - 0.0025
= 2.2475
In simple words: To find the expected value, we multiply each 'x' value by its probability and sum these products. For the variance, we calculate the expected value of X squared (sum of x\(^2\)p) and subtract the square of the expected value (E(X)) from it.

🎯 Exam Tip: Create a clear table with columns for x, p, xp, x\(^2\), and x\(^2\)p to organize calculations for expected value and variance, minimizing arithmetic errors.

 

Question 10. Find expected value and variance of X, the number on the uppermost face of a fair die.
Answer:Solution:
S: A fair die is thrown
S = {1, 2, 3, 4, 5, 6}
n(S) = 6
X: No obtained on uppermost face of die
Range of X = {1, 2, 3, 4, 5, 6}

x	p	px	x\(^2\)	x\(^2\)p
1	\( \frac{1}{6} \)	\( \frac{1}{6} \)	1	\( \frac{1}{6} \)
2	\( \frac{1}{6} \)	\( \frac{2}{6} \)	4	\( \frac{4}{6} \)
3	\( \frac{1}{6} \)	\( \frac{3}{6} \)	9	\( \frac{9}{6} \)
4	\( \frac{1}{6} \)	\( \frac{4}{6} \)	16	\( \frac{16}{6} \)
5	\( \frac{1}{6} \)	\( \frac{5}{6} \)	25	\( \frac{25}{6} \)
6	\( \frac{1}{6} \)	\( \frac{6}{6} \)	36	\( \frac{36}{6} \)
Total		\( \frac{21}{6} \)		\( \frac{91}{6} \)

E(X) = \( \Sigma \)xp = \( \frac{21}{6} = \frac{7}{2} \) = 3.5
V(X) = \( \Sigma \)x\(^2\)p - (\( \Sigma \)xp)\(^2\)
= \( \frac{91}{6} \) - (3.5)\(^2\)
= 15.17-12.25
= 2.92
In simple words: For a fair die, each face (1 to 6) has an equal probability of \( \frac{1}{6} \). We calculate the expected value by summing the product of each face value and its probability. The variance is then found by subtracting the square of the expected value from the sum of each face value squared times its probability.

🎯 Exam Tip: Remember that for a fair die, each outcome has a probability of 1/6. Organize your calculations for xp and x\(^2\)p clearly to avoid errors.

 

Question 11. Find the mean of the number of heads in three tosses of a fair coin.
Answer:Solution:
S: A coin is tossed 3 times
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
Range of X = {0, 1, 2, 3}

x	P	px
0	\( \frac{1}{8} \)	0
1	\( \frac{3}{8} \)	\( \frac{3}{8} \)
2	\( \frac{3}{8} \)	\( \frac{6}{8} \)
3	\( \frac{1}{8} \)	\( \frac{3}{8} \)
Total		\( \frac{12}{8} \)

\( \therefore \) Mean = E(X) = \( \Sigma \)xp = \( \frac{12}{8} = \frac{3}{2} \) = 1.5
In simple words: We list all 8 possible outcomes for three coin tosses, count the number of heads for each, and then calculate the probability for 0, 1, 2, and 3 heads. The mean (expected value) is found by multiplying each number of heads by its probability and summing these products.

🎯 Exam Tip: For problems with multiple coin tosses, constructing a clear sample space helps accurately count outcomes for each number of heads (or tails) and assign correct probabilities.

 

Question 12. Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.
Answer:Solution:
S: Two dice are thrown
S = {(1, 1), (1, 2), (1, 3), ......, (6, 6)}
n(S) = 36
Range of X = {0, 1, 2}
First 6 positive integers are 1, 2, 3, 4, 5, 6
X = Larger two numbers selected
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(S) = 36

x	p	px
0	\( \frac{25}{36} \)	0
1	\( \frac{10}{36} \)	\( \frac{10}{36} \)
2	\( \frac{1}{36} \)	\( \frac{2}{36} \)
Total		\( \frac{12}{36} \)

\( \therefore \) E(X) = \( \Sigma \)xp = \( \frac{12}{36} = \frac{1}{3} \)
In simple words: When two dice are thrown, X is the number of sixes. We identify the outcomes where no six, one six, or two sixes occur, calculate their probabilities out of 36 total outcomes, and then sum the products of each number of sixes and its probability to find the expectation.

🎯 Exam Tip: Systematically list outcomes for the number of sixes: no six (5x5=25 outcomes), one six (1x5 + 5x1 = 10 outcomes), and two sixes (1x1 = 1 outcome), ensuring these sum to n(S)=36.

 

Question 13. Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers. Find E(X).
Answer:Solution:
First 6 positive integers are 1, 2, 3, 4, 5, 6
X : The larger of the selected two numbers
S = {(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
n(S) = 30

x	p	px
2	\( \frac{2}{30} \)	\( \frac{4}{30} \)
3	\( \frac{4}{30} \)	\( \frac{12}{30} \)
4	\( \frac{6}{30} \)	\( \frac{24}{30} \)
5	\( \frac{8}{30} \)	\( \frac{40}{30} \)
6	\( \frac{10}{30} \)	\( \frac{60}{30} \)
Total		\( \frac{140}{30} \)

E(X) = \( \Sigma \)xp = \( \frac{140}{30} = \frac{14}{3} \) = 4.67
In simple words: We list all possible pairs of two distinct numbers selected from 1 to 6. For each pair, we identify the larger number, which becomes the value of X. We then calculate the probability of each X value and use these probabilities to compute the expected value.

🎯 Exam Tip: When sampling "without replacement," the order matters for counting the sample space if the items are distinct (like (1,2) and (2,1) being different outcomes). The number of such pairs is n(n-1).

 

Question 14. Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the variance of X.
Answer:Solution:
S: Two fair dice are rolled
S = {(1, 1), (1, 2), (1, 4), ......, (6, 6)}
n(S) = 36
X: Sum of the two numbers.
Range of X = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

x	p	xp	x\(^2\)p
2	\( \frac{1}{36} \)	\( \frac{2}{36} \)	\( \frac{4}{36} \)
3	\( \frac{2}{36} \)	\( \frac{6}{36} \)	\( \frac{18}{36} \)
4	\( \frac{3}{36} \)	\( \frac{12}{36} \)	\( \frac{48}{36} \)
5	\( \frac{4}{36} \)	\( \frac{20}{36} \)	\( \frac{100}{36} \)
6	\( \frac{5}{36} \)	\( \frac{30}{36} \)	\( \frac{180}{36} \)
7	\( \frac{6}{36} \)	\( \frac{42}{36} \)	\( \frac{294}{36} \)
8	\( \frac{5}{36} \)	\( \frac{40}{36} \)	\( \frac{320}{36} \)
9	\( \frac{4}{36} \)	\( \frac{30}{36} \)	\( \frac{324}{36} \)
10	\( \frac{3}{36} \)	\( \frac{30}{36} \)	\( \frac{300}{36} \)
11	\( \frac{2}{36} \)	\( \frac{22}{36} \)	\( \frac{220}{36} \)
12	\( \frac{1}{36} \)	\( \frac{12}{36} \)	\( \frac{144}{36} \)
Total		\( \frac{252}{36} \)	\( \frac{1952}{36} \)

V(X) = \( \Sigma \)x\(^2\)p - (\( \Sigma \)xp)\(^2\)
= \( \frac{1952}{36} \) - (\( \frac{252}{36} \))\(^2\)
= 54.22 - (7)\(^2\)
= 5.22
SD(X) = \( \sqrt{V(X)} = \sqrt{5.22} \) = 2.28
In simple words: First, we determine the probability of each possible sum (from 2 to 12) when rolling two dice. Then, we calculate the expected value (mean) of X. Finally, using the formula V(X) = E(X\(^2\)) - [E(X)]\(^2\), we compute the variance.

🎯 Exam Tip: Accurately listing the frequency of each sum when rolling two dice (e.g., sum 7 has 6 outcomes, sum 2 has 1 outcome) is crucial for correct probabilities. Double-check all sums for expected value and expected squares.

 

Question 15. A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. If X denotes the age of a randomly selected student, find the probability distribution of X. Find the mean and variance of X.
Answer:
Solution:

x (Age)	No. of students	p	xp	x2p
14	2	\( \frac{2}{15} \)	\( \frac{28}{15} \)	\( \frac{392}{15} \)
15	1	\( \frac{1}{15} \)	\( \frac{15}{15} \)	\( \frac{225}{15} \)
16	2	\( \frac{2}{15} \)	\( \frac{32}{15} \)	\( \frac{512}{15} \)
17	3	\( \frac{3}{15} \)	\( \frac{51}{15} \)	\( \frac{867}{15} \)
18	1	\( \frac{1}{15} \)	\( \frac{18}{15} \)	\( \frac{324}{15} \)
19	2	\( \frac{2}{15} \)	\( \frac{38}{15} \)	\( \frac{722}{15} \)
20	3	\( \frac{3}{15} \)	\( \frac{60}{15} \)	\( \frac{1200}{15} \)
21	1	\( \frac{1}{15} \)	\( \frac{21}{15} \)	\( \frac{441}{15} \)
Total			\( \frac{263}{15} \)	\( \frac{4683}{15} \)

E(X) \( = \Sigma xp = \frac{263}{15} = 17.53 \)
V(X) \( = \Sigma x^2p - (\Sigma xp)^2 \)
\( = \frac{4683}{15} - (\frac{263}{15})^2 \)
\( = \frac{4683}{15} - \frac{69169}{225} \)
\( = \frac{70245 - 69169}{225} \)
\( = \frac{1076}{225} = 4.78 \)
SD(X) \( = \sqrt{V(X)} = \sqrt{4.78} = 2.18 \)
In simple words: This question asks us to find the probability distribution of ages in a class, along with the average age (mean) and the spread of ages (variance). We calculate the probability for each age by dividing the count of students of that age by the total number of students, then use these probabilities to find the mean and variance.

🎯 Exam Tip: When calculating mean and variance from a frequency distribution, ensure all probabilities sum to 1. Double-check calculations involving fractions or decimals, especially when squaring the mean for variance.

 

Question 16. 70% of the member's favour and 30% oppose a proposal in a meeting. The random variable X takes the value 0 if a member opposes the proposal and the value 1 if a member is in favour. Find E(X) and V(X).
Answer:
Solution:

x	p	xp	x2	x2p
0	0.3	0	0	0
1	0.7	0.7	1	0.7
Total		0.7		0.7

E(X) \( = \Sigma xp = 0.7 \)
V(X) \( = \Sigma x^2p - (\Sigma xp)^2 \)
\( = 0.7 - (0.7)^2 \)
\( = 0.7 - 0.49 \)
\( = 0.21 \)
In simple words: This problem calculates the expected value and variance for a binary random variable representing a proposal vote. With 70% favoring (X=1) and 30% opposing (X=0), the expected value is simply the probability of favoring, and the variance measures the spread of these outcomes.

🎯 Exam Tip: For simple Bernoulli distributions like this, remember that E(X) = p and V(X) = p(1-p). This can be a quick check for your calculations or a shortcut if the question allows.