WBCHSE Solutions For Class 12 Maths Probability Distribution

Probability Distribution

Random Variable Let S be the sample space associated with a given random experiment.

A real-valued function X which assigns a unique real number X(w) to each w ∈ S, is called a random variable. A random variable that can assume only a finite number of values is called a discrete random variable.

Example: Suppose that a coin is tossed twice.

Then, sample space S = {TT, HT, TH, HH}

Consider a real-valued function X on S, defined by X: S → R: X(w) = number of heads in w, for all w ∈ S.

Then, X is a random variable such that X(TT) = 0, X(HT) = 1, X(TH) = 1 and X(HH) = 2.

Clearly, range (X) = {0,1, 2}.

Probability Distribution of a Random Variable: A description giving the values of a random variable along with the corresponding probabilities is called the probability distribution of the random variable.

If a random variable X takes the values x1,x2, …,xn with respective probabilities p1, p2,……,pn then the probability distribution of X is given by

Class 12 Maths Probability Distribution Mean And Variance And Standard Deviation Of Number Of tails In Two Tosses Of A Coin

 

Remark: The above probability distribution of X is defined only when

  1. each pi ≥0
  2. \(\sum_{i=1}^n p_i=1\)

Mean and Variance of Random Variables

 

Let a random variable X assume values x1, x2,… ,xn with probabilities p1, p2,…..,pn respectively such that each pi ≥ 0 and \(\sum_{i=1}^n p_i=1 .\). Then mean of X, denoted by μ [or expected value of X, denoted by E(X)], is defined as

∴ \(\mu=E(X)=\sum_{i=1}^n x_i p_i\)

And, the variance, denoted by σ², is defined as \(\sigma^2=\left(\Sigma x_i^2 p_i-\mu^2\right)\)

Standard deviation, σ is given by \(\sigma=\sqrt{\text { variance }}.\)

Probability Distribution Solved Examples

Example 1. Find the mean, variance, and standard deviation of the number of tails in two tosses of a coin.
Solution:

WBCHSE Solutions For Class 12 Maths Probability Distribution

 

Class 12 Maths Probability Distribution Mean And Variance And Standard Deviation Of Number Of tails In Two Tosses Of A Coin

∴ mean, \(\mu =\Sigma x_i p_i=\left(0 \times \frac{1}{4}\right)+\left(1 \times \frac{1}{2}\right)+\left(2 \times \frac{1}{4}\right)=1\)

Variance, \(\sigma^2=\Sigma x_i^2 p_i-\mu^2\)

= \(\left[\left(0 \times \frac{1}{4}\right)+\left(1 \times \frac{1}{2}\right)+\left(4 \times \frac{1}{4}\right)\right]-1^2\) = \(\frac{1}{2}\) .

Standard deviation, \(\sigma=\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}=\frac{1.414}{2}=0.707\).

Example 2. Find the mean, variance, and standard deviation of the number of heads when three coins are tossed.
Solution:

Here, S = [TTT, TTH, THT, HTT, THH, HFH, HHT, HHH).

∴ n(S) = 8.

So, every single outcome has a probability 1/8

Let X = number of heads in tossing three coins.

The number of heads may be 0,1,2, or 3.

So, the possible values of X are 0,1,2,3.

P(X = 0) = P(getting no head) = P(TTT) = \(\frac{1}{8}\)

P(X = 1) = P(getting 1 head) = P(TTH or THT or HTT) = \(\frac{3}{8}\)

P(X = 2) = P( getting 2 heads) = P(THH, HTH, HHT) = \(\frac{3}{8}\)

P(X = 3) = P(getting 3 heads) = P(HHH) = \(\frac{1}{8}\)

Thus, we have the following probability distribution:

 

Class 12 Maths Probability Distribution Mean And Variance Of Number Of Heads When Three Coins Are Tossed

 

∴ mean \(\mu=\Sigma x_i p_i=\left(0 \times \frac{1}{8}\right)+\left(1 \times \frac{3}{8}\right)+\left(2 \times \frac{3}{8}\right)+\left(3 \times \frac{1}{8}\right)=\frac{3}{2}\)

Variance, \(\sigma^2=\Sigma x_i^2 p_i-\mu^2\)

= \(\left[\left(0 \times \frac{1}{8}\right)+\left(1 \times \frac{3}{8}\right)+\left(4 \times \frac{3}{8}\right)+\left(9 \times \frac{1}{8}\right)-\frac{9}{4}\right]=\frac{3}{4}\).

Standard deviation, \(\sigma=\frac{\sqrt{3}}{2}\).

Example 3. A die is tossed once. If the random variable X is defined as X = 1, if the die results in an even number, X = 0, if the die results in an odd number then find the mean and variance of X.
Solution:

In tossing a die once, the sample space is given by S = {1,2,3,4,5,6}.

∴ P(getting an even number) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

P(getting an odd number) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

As given, X takes the value 0 or 1.

P(X = 0) = P(getting an odd number) = \(\frac{1}{2}\)

P(X = 1) = P(getting an even number) = \(\frac{1}{2}\)

Thus, the probability distribution of X is given by

Class 12 Maths Probability Distribution Mean And Variance Of x And A Die Is Tossed Once

 

∴ mean, \(\mu=\Sigma x_i p_i=\left(0 \times \frac{1}{2}\right)+\left(1 \times \frac{1}{2}\right)=\frac{1}{2}\)

Variance, \(\sigma^2=\Sigma x_i^2 p_i-\mu^2\)

= \(\left(0 \times \frac{1}{2}\right)+\left(1 \times \frac{1}{2}\right)-\left(\frac{1}{2}\right)^2=\left(\frac{1}{2}-\frac{1}{4}\right)=\frac{1}{4}\)

Example 4. Find the mean, variance, and standard deviation of the number of sixes in two tosses of a die.
Solution:

In a single toss, we have a probability of getting a six = \(\frac{1}{6}\), and

Probability of getting a non-six = (1-\(\frac{1}{6}\)) = \(\frac{5}{6}\)

Let X denote the number of sixes in two tosses.

Then, clearly X can assume the value 0,1, or 2.

P(X = 0) = P[(non-six in the 1st draw) and (non-six in the 2nd draw)]

= P(non-six in the 1st draw) x P(non-six in the 2nd draw)

= (\(\frac{5}{6}\)x\(\frac{5}{6}\) = \(\frac{25}{36}\)

P(X=1)= P[(six in the 1st draw and non-six in the 2nd draw) or (non-six in the 1st draw and six in the 2nd draw)]

= P(six in the 1st draw and non-six in the 2nd draw) + P(non-six in the 1st draw and six in the 2nd draw)

= \(\left(\frac{1}{6} \times \frac{5}{6}\right)+\left(\frac{5}{6} \times \frac{1}{6}\right)=\left(\frac{5}{36}+\frac{5}{36}\right)=\frac{10}{36}=\frac{5}{18}\)

P(X = 2) = P[six in the 1st draw and six in the 2nd draw]

= P(six in the 1st draw) x P(six in the 2nd draw)

= \(\left(\frac{1}{6} \times \frac{1}{6}\right)=\frac{1}{36}\)

Hence, the probability distribution is given by

Class 12 Maths Probability Distribution Mean And Variance And Standard Deviation Of Number Of Sixes In Two Tosses Of A Die

 

∴ mean, \(\mu=\Sigma x_i p_i=\left(0 \times \frac{25}{36}\right)+\left(1 \times \frac{5}{18}\right)+\left(2 \times \frac{1}{36}\right)=\frac{6}{18}=\frac{1}{3}\).

Variance, \(\sigma^2=\Sigma x_i^2 p_i-\mu^2\)

= \(\left[\left(0 \times \frac{25}{36}\right)+\left(1 \times \frac{5}{18}\right)+\left(4 \times \frac{1}{36}\right)-\frac{1}{9}\right]=\frac{5}{18}\).

Standard deviation, \(\sigma=\sqrt{\frac{5}{18}}=\frac{1}{3} \cdot \sqrt{\frac{5}{2}}\).

Example 5. Two cards are drawn successively with replacement from a well-shuffled pack of 52 cards. Find the mean and variance of the number of kings.
Solution:

Let X be the random variable. Then,

X = number of kings obtained in two draws.

Clearly, X can assume the value 0,1 or 2.

P(drawing a king) = \(\frac{4}{52}\) = \(\frac{1}{13}\)

P(not drawing a king) = (1-\(\frac{1}{13}\)) = \(\frac{12}{13}\)

P(X = 0) = P(not a king in the 1st draw and not a king in the 2nd draw)

= \(\left(\frac{12}{13} \times \frac{12}{13}\right)=\frac{144}{169}\).

P(X = 1) = P(a king in the 1st draw and not a king in the 2nd draw)

or P(not a king in the 1st draw and a king in the 2nd
draw)

= \(\left(\frac{1}{13} \times \frac{12}{13}+\frac{12}{13} \times \frac{1}{13}\right)=\frac{24}{169}\)

P(X = 2) = P(a king in the 1st draw and a king in the 2nd draw)

= \(\left(\frac{1}{13} \times \frac{1}{13}\right)=\frac{1}{169}\)

Hence, the probability distribution is given by

Class 12 Maths Probability Distribution Mean And Variance Of Number Of Kings

 

∴ mean, \(\mu=\Sigma x_i p_i=\left(0 \times \frac{144}{169}\right)+\left(1 \times \frac{24}{169}\right)+\left(2 \times \frac{1}{169}\right)=\frac{2}{13} \text {. }\)

Variance, \(\sigma^2=\Sigma x_i^2 p_i-\mu^2\)

= \(\left[\left(0 \times \frac{144}{169}\right)+\left(1 \times \frac{24}{169}\right)+\left(4 \times \frac{1}{169}\right)-\frac{4}{169}\right]=\frac{24}{169}\)

Example 6. Two cards are drawn simultaneously (or successively without replacement) from a well-shuffled pack of 52 cards. Find the mean and variance of the number of aces.
Solution:

Let X be the random variable.

Then, X denotes the number of aces in a draw of 2 cards.

∴ X can assume the value 0,1 or 2.

Number of ways of drawing 2 cards out of 52 = C(52,2).

P(X = 0) = P(both non-aces, i.e., 2 non-aces out of 48)

= \(\frac{{ }^{48} C_2}{{ }^{52} C_2}=\left(\frac{48 \times 47}{2 \times 1} \times \frac{2}{52 \times 51}\right)=\frac{188}{221}\)

P(X=1)= \(P[(\text { one ace out of } 4) \text { and (one non-ace out of } 48)]\)

= \(\frac{{ }^4 C_1 \times{ }^{48} C_1}{{ }^{52} C_2}=\left(\frac{4 \times 48}{52 \times 51} \times 2\right)=\frac{32}{221}\)

P(X=2)= \(P(\text { both aces })=\frac{{ }^4 C_2}{{ }^{52} C_2}=\left(\frac{4 \times 3}{2 \times 1} \times \frac{2 \times 1}{52 \times 51}\right)=\frac{1}{221}\)

Thus, we have the following probability distribution:

Class 12 Maths Probability Distribution Mean And Variance Of Number Of Aces

 

∴ mean, \(\mu=\Sigma x_i p_i=\left(0 \times \frac{188}{221}\right)+\left(1 \times \frac{32}{221}\right)+\left(2 \times \frac{1}{221}\right)=\frac{2}{13}\)

Variance, \(\sigma^2=\Sigma x_i^2 p_i-\mu^2
=\left(0 \times \frac{188}{221}\right)+\left(1 \times \frac{32}{221}\right)+\left(4 \times \frac{1}{221}\right)-\frac{4}{169}\)

=\(\left(\frac{36}{221}-\frac{4}{169}\right)=\frac{400}{2873}\)

Example 7. Three defective bulbs are mixed with 7 good ones. Let X be the number of defective bulbs when 3 bulbs are drawn at random. Find the mean and variance of X.
Solution:

Let X denote the random variable showing the number of defective bulbs.

Then, X can take the value 0,1,2 or 3.

∴ P(X = 0) = P(none of the bulbs is defective)

= P(all the 3 bulbs are good ones)

= \(\frac{{ }^7 C_3}{{ }^{10} C_3}\)=\(\left(\frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times \frac{3 \times 2 \times 1}{10 \times 9 \times 8}\right)\)=\(\frac{7}{24}\).

P(X=1)=P(1 defective and 2 non-defective bulbs)

= \(\frac{{ }^3 C_1 \times{ }^7 C_2}{{ }^{10} C_3}=\left(3 \times \frac{7 \times 6}{2 \times 1} \times \frac{3 \times 2 \times 1}{10 \times 9 \times 8}\right)=\frac{21}{40}\).

P(X=2)=P(2 defective and 1 good one)

= \(\frac{{ }^3 C_2 \times{ }^7 C_1}{{ }^{10} C_3}=\left(\frac{3 \times 2}{2 \times 1} \times 7 \times \frac{3 \times 2 \times 1}{10 \times 9 \times 8}\right)=\frac{7}{40}\).

P(X=3)=P(3 defective bulbs)

= \(\frac{{ }^3 C_3}{{ }^{10} C_3}=\left(1 \times \frac{3 \times 2 \times 1}{10 \times 9 \times 8}\right)=\frac{1}{120}\).

Thus, the probability distribution is given by

Class 12 Maths Probability Distribution Mean And Variance Of Three Defective Bulbs Are Mixed With 7 Good Ones

 

∴ mean, \(\mu=\Sigma x_i p_i=\left(0 \times \frac{7}{24}\right)+\left(1 \times \frac{21}{40}\right)+\left(2 \times \frac{7}{40}\right)+\left(3 \times \frac{1}{120}\right)=\frac{9}{10}\)

Variance, \(\sigma^2=\Sigma x_i^2 p_i-\mu^2\)

= \(\left(0 \times \frac{7}{24}\right)+\left(1 \times \frac{21}{40}\right)+\left(4 \times \frac{7}{40}\right)+\left(9 \times \frac{1}{120}\right)-\frac{81}{100}\)

= \(\left(\frac{13}{10}-\frac{81}{100}\right)=\frac{49}{100}\).

Example 8. An urn contains 4 white and 3 red balls. Let X be the number of red balls in a random draw of 3 balls. Find the mean and variance of X.
Solution:

When 3 balls are drawn at random, there may be no red ball, 1 red ball, 2 red balls or 3 red balls.

Let X denote the random variable showing the number of red balls in a draw of 3 balls.

Then, X can take the value 0,1,2 or 3.

P(X=0) =P(getting no red ball)

= P(getting 3 white balls)

= \(\frac{{ }^4 C_3}{{ }^7 C_3}=\left(\frac{4 \times 3 \times 2}{3 \times 2 \times 1} \times \frac{3 \times 2 \times 1}{7 \times 6 \times 5}\right)=\frac{4}{35}\)

P(X=1)=P(getting 1 red and 2 white balls)

= \(\frac{{ }^3 C_1 \times{ }^4 C_2}{{ }^7 C_3}=\left(\frac{3 \times 4 \times 3}{2} \times \frac{3 \times 2 \times 1}{7 \times 6 \times 5}\right)=\frac{18}{35} \)

P(X=2) =P(\text { getting } 2 \text { red and } 1 \text { white ball })

= \(\frac{{ }^3 C_2 \times{ }^4 C_1}{{ }^7 C_3}=\left(\frac{3 \times 2}{2 \times 1} \times 4 \times \frac{3 \times 2 \times 1}{7 \times 6 \times 5}\right)=\frac{12}{35}\)

P(X=3)=P(getting 3 red balls)

= \(\frac{{ }^3 C_3}{{ }^7 C_3}=\frac{1 \times 3 \times 2 \times 1}{7 \times 6 \times 5}=\frac{1}{35}\).

Thus, the probability distribution of X is given below.

Class 12 Maths Probability Distribution Mean And Variance Of An Urn Contains 4 White And 3 red Balls

 

∴ mean, \(\mu=\Sigma x_i p_i=\left(0 \times \frac{4}{35}\right)+\left(1 \times \frac{18}{35}\right)+\left(2 \times \frac{12}{35}\right)+\left(3 \times \frac{1}{35}\right)=\frac{9}{7}\)

Variance, \(\sigma^2=\Sigma x_i^2 p_i-\mu^2\)

= \(\left[\left(0 \times \frac{4}{35}\right)+\left(1 \times \frac{18}{35}\right)+\left(4 \times \frac{12}{35}\right)+\left(9 \times \frac{1}{35}\right)-\frac{81}{49}\right]\)

= \(\left(\frac{15}{7}-\frac{81}{49}\right)=\frac{24}{49}\)

Example 9. In a game, 3 coins are tossed. A person is paid Rs 5 if he gets all heads or all tails and he is supposed to pay Rs 3 if he gets one head or two heads. What can he expect to win on an average per game?
Solution:

In tossing 3 coins, the sample space is given by

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.

∴ n(S) = 8.

P(getting all heads or all tails) = \(\frac{2}{8}\)=\(\frac{1}{4}\)

P(getting one head or 2 heads) = \(\frac{6}{8}\)=\(\frac{3}{4}\)

Let X = the number of rupees the person gets.

Then, possible values of X are 5 and -3.

P(X=5) = \(\frac{1}{4}\) and P(X=-3) = \(\frac{3}{4}\)

Thus, we have

Class 12 Maths Probability Distribution Mean And Variance In A Game 3 Coins Are Tossed

 

∴ the required expectations = mean, \(\mu=\Sigma x_i p_i \)

= \(\left(5 \times \frac{1}{4}\right)+(-3) \times \frac{3}{4}=-1\),

i.e., he loses Re 1 per toss.

Probability Distribution Exercise

Question 1. Find the mean (μ), variance (σ²), and standard deviation (σ) for each of the following probability distributions:

Class 12 Maths Probability Distribution Mean And Variance And Stabdard Deviation Q1-1

Class 12 Maths Probability Distribution Mean And Variance And Standard Deviation Q1-3

 

 

 

 

Class 12 Maths Probability Distribution Mean And Variance And Standard Deviation Q1-4

Solution:

Mean =1.5, variance=0.56, SD = 0.74

Mean =2, variance=1, SD =1

Mean = -0.8, variance=2.6, SD = 1.612

Mean = 0, variance=1.2, SD=1.095

Question 2. Find the mean and variance of the number of heads when two coins are tossed simultaneously.
Solution: Mean =1, variance=0.5

Hint: S = {HH, HT, TH, TT}.

Let X be the number of heads. Then, X= 0,1 or 2.

P(X= 0) = P(getting no head) = \(\frac{1}{4}\)

P(X= 1) =P(getting1 head) = \(\frac{2}{4}\) = \(\frac{1}{2}\)

P(X= 2) = P(getting 2 heads) = \(\frac{1}{4}\)

Class 12 Maths Probability Distribution Mean And Variance And Standard Deviation Q2

 

Question 3. Find the mean and variance of the number of tails when three coins are tossed simultaneously.
Solution: Mean =1.5, variance=0.75

Class 12 Maths Probability Distribution Mean And Caiancwe Of The nUmber Of Tails Q3

 

Question 4. A die is tossed twice. ‘Getting an odd number on a toss’ is considered a success. Find the probability distribution of a number of successes. Also, find the mean and variance of the number of successes.
Solution: Mean =1, variance=0.5

Hint: In a single toss,

P(success) = P(getting an odd number) = \(\frac{3}{6}\) = \(\frac{1}{2}\), and

P(non-success) = (1-\(\frac{1}{2}\)) = \(\frac{1}{2}\)

Let X be the number of successes. Then, X= 0,1 or 2.

P(X= 0) = P[(non-successin the 1st toss and non-successin the 2nd)]

= \(\left(\frac{1}{2} \times \frac{1}{2}\right)=\frac{1}{4}\).

P(X=1)- P[(successin the 1st toss and non-success in the 2nd) or (non-success in 1st toss and success in the 2nd)]

= \(\left(\frac{1}{2} \times \frac{1}{2}\right)+\left(\frac{1}{2} \times \frac{1}{2}\right)=\frac{1}{2}\)

P(X=2) =P[success in the 1 st toss and success in the 2nd]

= \(\left(\frac{1}{2} \times \frac{1}{2}\right)=\frac{1}{4}\).

Class 12 Maths Probability Distribution Probability Distribution And Mean And Vaisnce Q4

 

Question 5. A die is tossed twice. ‘Getting a number greater than 4’ is considered a success. Find the probability distribution of a number of successes. Also, find the mean and variance of the number of successes.
Solution: Mean=\(\frac{2}{3}\), variance=\(\frac{4}{9}\)

In a single toss, P(success) = and P(non-success) = (1-\(\frac{1}{3}\)) = \(\frac{2}{3}\)

P(X = 0) = P(non-success in the 1st draw and non-success in the second)

= \(\left(\frac{2}{3} \times \frac{2}{3}\right)=\frac{4}{9} \text {. }\)

P(X=1) = P(success in the ls.t toss and non-success in the 2nd) or ( non-success in the 1st toss and success in the 2nd)]

= \(\left(\frac{1}{3} \times \frac{2}{3}\right)+\left(\frac{2}{3} \times \frac{1}{3}\right)=\frac{4}{9}\)

P(X = 2) = P[(success in the 1st toss and success in the 2nd)]

= \(\left(\frac{1}{3} \times \frac{1}{3}\right)=\frac{1}{9}\).

Class 12 Maths Probability Distribution Probability Distribution Mean And Variance Q5

 

Question 6. A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of a number of successes. Also, find the mean and variance of a number of successes.
Solution: Mean=\(\frac{2}{3}\), variance=\(\frac{5}{9}\)

Hint: In a single throw, P(doublet) = 6/36, 1/6, and P(non-doublet) = (1-\(\frac{1}{6}\)) = \(\frac{5}{6}\)

Let X be the number of doublets. Then, X = 0, 1, 2 or 3.

P(X = 0) = P(non-doublet in each case)

= \(P\left(\bar{D}_1 \bar{D}_2 \bar{D}_3 \bar{D}_4\right)=\left(\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6}\right)=\frac{625}{1296}\)

P(X=1)= P(one doublet)

= \(P\left(D_1 \bar{D}_2 \bar{D}_3 \bar{D}_4\right) \text { or } P\left(\bar{D}_1 D_2 \bar{D}_3 \bar{D}_4\right) \)

or \(P\left(\bar{D}_1 \bar{D}_2 D_3 \bar{D}_4\right) \text { or } P\left(\bar{D}_1 \bar{D}_2 \bar{D}_3 D_4\right) \)

= \(\left(\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6}\right)+\left(\frac{5}{6} \times \frac{1}{6} \times \frac{5}{6} \times \frac{5}{6}\right)+\left(\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} \times \frac{5}{6}\right)+\left(\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}\right)\)

= \(\left(4 \times \frac{125}{1296}\right)=\frac{125}{324}\)

P(X=2)= P(two doublets)

= \(P\left(D_1 D_2 \bar{D}_3 \bar{D}_4\right) \quad \text { or } P\left(D_1 \bar{D}_2 D_3 \bar{D}_4\right) \text { or } P\left(D_1 \bar{D}_2 \bar{D}_3 D_4\right)\)

or \(P\left(\bar{D}_1 D_2 D_3 \bar{D}_4\right) \text { or } P\left(\bar{D}_1 D_2 \bar{D}_3 D_4\right) \text { or } P\left(\bar{D}_1 \bar{D}_2 D_3 D_4\right)\)

= \(\left(\frac{1}{6} \times \frac{1}{6} \times \frac{5}{6} \times \frac{5}{6}\right)+\left(\frac{1}{6} \times \frac{5}{6} \times \frac{1}{6} \times \frac{5}{6}\right)+\left(\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}\right)\)

+ \(\left(\frac{5}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{5}{6}\right)+\left(\frac{5}{6} \times \frac{1}{6} \times \frac{5}{6} \times \frac{1}{6}\right)+\left(\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} \times \frac{1}{6}\right)\)

= \(\left(6 \times \frac{25}{1296}\right)=\frac{25}{216}\)

P(X=3)= P(three doublets)

= \(P\left(D_1 D_2 D_3 \bar{D}_4\right) \text { or } P\left(D_1 D_2 \bar{D}_3 D_4\right)\)

or \(P\left(D_1 \bar{D}_2 D_3 D_4\right) \text { or } P\left(\bar{D}_1 D_2 D_3 D_4\right)\)

= \(\left(\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{5}{6}\right)+\left(\frac{1}{6} \times \frac{1}{6} \times \frac{5}{6} \times \frac{1}{6}\right)+\left(\frac{1}{6} \times \frac{5}{6} \times \frac{1}{6} \times \frac{1}{6}\right)+\left(\frac{5}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6}\right)\)

= \(\left(4 \times \frac{5}{1296}\right)=\frac{5}{324} \)

P(X=4)= P(four doublets)=P\left(D_1 D_2 D_3 D_4\right)[/latex]

= \(\left(\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6}\right)=\frac{1}{1296}\)

Thus, we have

Class 12 Maths Probability Distribution Mean And Variance Of Number Q6

 

Question 7. A coin is tossed 4 times. Let X denote the number of heads. Find the probability distribution of X. Also, find the mean and variance of X.
Solution: Mean=2, variance=1

In a single throw of a coin, \(P(H)=\frac{1}{2}\) and \(P(\bar{H})=P(T)=\frac{1}{2}\).

Let X show the number of heads. Then, X=0,1,2,3 or 4 .

P(X=0)=P(no head)=\(P\left(\bar{H}_1 \bar{H}_2 \bar{H}_3 \bar{H}_4\right)=\left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right)=\frac{1}{16} \text {. }\)

P(X=1)=P(one head)= \(P\left(H_1 \bar{H}_2 \bar{H}_3 \bar{H}_4\right) \text { or }\left(\bar{H}_1 H_2 \bar{H}_3 \bar{H}_4\right)\)

or \(P\left(\bar{H}_1 \bar{H}_2 H_3 \bar{H}_4\right) \text { or } P\left(\bar{H}_1 \bar{H}_2 \bar{H}_3 H_4\right) \)

= \(4\left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right)=\left(4 \times \frac{1}{16}\right)=\frac{1}{4}\)

P(X=2)=P(two heads)= \(P\left(H_1 H_2 \bar{H}_3 \bar{H}_4\right) \text { or } P\left(H_1 \bar{H}_2 H_3 \bar{H}_4\right)\)

or \(P\left(H_1 \bar{H}_2 \bar{H}_3 H_4\right) \text { or } P\left(\bar{H}_1 H_2 H_3 \bar{H}_4\right)\)

or \(P\left(\bar{H}_1 H_2 \bar{H}_3 H_4\right) \text { or } P\left(\bar{H}_1 \bar{H}_2 H_3 H_4\right)\)

= \(\left(6 \times \frac{1}{16}\right)=\frac{3}{8}\)

P(X=3)=P(\text { three heads })=P\left(H_1 H_2 H_3 \bar{H}_4\right) \text { or }\left(H_1 H_2 \bar{H}_3 H_4\right)[/latex]

or \(P\left(H_1 \bar{H}_2 H_3 H_4\right) \text { or } P\left(\bar{H}_1 H_2 H_3 H_4\right)\)

= \(4 \times\left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right)=\left(4 \times \frac{1}{16}\right)=\frac{1}{4} \text {. }\)

P(X=4)=P(four heads)= \(P\left(H_1 H_2 H_3 H_4\right)=\left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right)=\frac{1}{16}\).

Thus, we have

Class 12 Maths Probability Distribution Mean And Variancbe Of X Q7

 

Question 8. Let X denote the number of times ‘a total of 9’ appears in two throws of a pair of dice. Find the probability distribution of X. Also, find the mean, variance, and standard deviation of X.
Solution: Mean=\(\frac{2}{9}\), variance=\(\frac{16}{81}\), SD = \(\frac{4}{9}\)

Hint: Let E={(3,6),(6,3),(4,5),(5,4)}. So, n(E)=4.

∴ P(E)= \(\frac{4}{36}=\frac{1}{9}\), and \(P(\bar{E})=\left(1-\frac{1}{9}\right)=\frac{8}{9}\).

Let X be the number of times ‘a total of 9’ appears in 2 throws Then, X=0, 1 or 2.

P(X=0)= \(=P\left(\bar{E}_1 \bar{E}_2\right)=\left(\frac{8}{9} \times \frac{8}{9}\right)=\frac{64}{81}\)

P(X=1)= \(P\left[\left(E_1 \bar{E}_2\right) \text { or }\left(\bar{E}_1 E_2\right)\right]=P\left(E_1 \bar{E}_2\right)+P\left(\bar{E}_1 E_2\right)\)

= \(\left(\frac{1}{9} \times \frac{8}{9}\right)+\left(\frac{8}{9} \times \frac{1}{9}\right)=\frac{16}{81}\)

P(X=2)= \(P\left(E_1 E_2\right)=P\left(E_1\right) \times P\left(E_2\right)=\left(\frac{1}{9} \times \frac{1}{9}\right)=\frac{1}{81}\)

Class 12 Maths Probability Distribution Mean And Varaince Q8

 

Question 9. There are 5 cards, numbered 1 to 5, with one number on each card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two cards drawn. Find the mean and variance of X.
Solution: Mean=6, variance=3

Hint: S ={(1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 4), (3, 5),
(4, 1), (4, 2), (4, 3), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4)}

Then X = 3, 4, 5, 6, 7, 8 and 9.

P(X=3)=\(\frac{2}{20}\)=\(\frac{1}{10}\);

P(X=4)=\(\frac{2}{10}\)=\(\frac{1}{10}\)

P(X=5)=\(\frac{4}{20}\)=\(\frac{1}{5}\)

P(X=6)=\(\frac{4}{20}\)=\(\frac{1}{5}\)

P(X=7)=\(\frac{4}{20}\)=\(\frac{1}{5}\)

P(X=8)=\(\frac{2}{20}\)=\(\frac{1}{10}\)

P(X=9)=\(\frac{2}{20}\)=\(\frac{1}{10}\)

Thus, we have

Class 12 Maths Probability Distribution Mean And Variance Of X Q9

 

Question 10. Two cards are drawn from a well-shuffled pack of 52 cards. Find the probability distribution of a number of kings. Also, compute the variance for the number of kings.
Solution: Variance=\(\frac{400}{2873}\)

Hint: Let X be the number of kings. Then, X = 0,1 or 2.

P(X = 0) = P(none is a king) = P(both are non-kings)

= \(\frac{{ }^{48} C_2}{{ }^{52} C_2}=\left(\frac{48 \times 47}{2} \times \frac{2}{52 \times 51}\right)=\frac{188}{221} .\)

P(X = 1) = P(one king and one non-king)

= \(\frac{{ }^4 C_1 \times{ }^{48} C_1}{{ }^{52} C_2}=\left(4 \times 48 \times \frac{2}{52 \times 51}\right)=\frac{32}{221} .\)

P(X = 2) = P(both are kings)

= \(\frac{{ }^4 C_2}{{ }^{52} C_2}=\left(\frac{4 \times 3}{2 \times 1} \times \frac{2 \times 1}{52 \times 51}\right)=\frac{1}{221}\)

Thus, we have

Class 12 Maths Probability Distribution Mean And Variance Q10

 

Question 11. A box contains 16 bulbs, out of which 4 bulbs are defective. Three bulbs are drawn at random from the box. Let X be the number of defective bulbs drawn. Find the mean and variance of X.
Solution: Mean=\(\frac{3}{4}\), variance=\(\frac{39}{80}\)

Hint: Three bulbs drawn one by one without replacement is the same as drawing 3 bulbs simultaneously.

Let X = number of defective bulbs in a lot of 3 bulbs drawn.

Then, X = 0, 1, 2 or 3.

P(X = 0) = P(none of the bulbs is defective)

= \(\frac{{ }^{12} C_3}{{ }^{16} C_3}=\left(\frac{12 \times 11 \times 10}{3 \times 2 \times 1} \times \frac{3 \times 2 \times 1}{16 \times 15 \times 14}\right)=\frac{11}{28}\)

P(X=1)=P(1 defective bulb and 2 non-defective bulbs)

= \(\frac{{ }^4 C_1 \times{ }^{12} C_2}{{ }^{16} C_3}=\left(\frac{4 \times 12 \times 11}{2 \times 1} \times \frac{3 \times 2 \times 1}{16 \times 15 \times 14}\right)=\frac{33}{70}\)

P(X=2)=P(2 defective bulbs and 1 non-defective bulb)

= \(\frac{\left({ }^4 C_2 \times{ }^{12} C_1\right)}{{ }^{16} C_3}=\left(\frac{4 \times 3}{2 \times 1} \times 12 \times \frac{3 \times 2 \times 1}{16 \times 15 \times 14}\right)=\frac{9}{70}\).

P(X=3)=P(3 defective bulbs)

= \(\frac{{ }^4 C_3}{{ }^{16} C_3}=\left(\frac{4 \times 3 \times 2}{3 \times 2 \times 1} \times \frac{3 \times 2 \times 1}{16 \times 15 \times 14}\right)=\frac{1}{140}\).

Class 12 Maths Probability Distribution Mean And Variance Q11

 

Question 12. 20% of the bulbs produced by a machine are defective. Find the probability distribution of the number of defective bulbs in a sample of 4 bulbs chosen at random.
Solution:

Let there be 100 bulbs in all and let X be the number of defective bulbs. Then,

P(X = 0) = P(none is defective) = \(=\frac{{ }^{80} C_4}{{ }^{100} C_4} .\)

P(X = 1) = P(1 defective and 3 non-defective)

= \(\frac{\left({ }^{20} C_1 \times{ }^{00} C_3\right)}{{ }^{100} C_4}\)

P(X=2)= \(\frac{\left({ }^{20} C_2 \times{ }^{80} C_2\right)}{{ }^{100} C_4} ; P(X=3)=\frac{\left({ }^{20} C_3 \times{ }^{80} C_1\right)}{{ }^{100} C_4} ; P(X=4)=\frac{{ }^{20} C_4}{{ }^{100} C_4}\)

Question 13. Four bad eggs are mixed with 10 good ones. Three eggs are drawn one by one without replacement. Let X be the number of bad eggs drawn. Find the mean and variance of X.
Solution: Mean=\(\frac{6}{7}\), variance=\(\frac{30}{49}\)

Question 14. Four rotten oranges are accidentally mixed with 16 good ones. Three oranges are drawn at random from the mixed lot. Let X be the number of rotten oranges drawn. Find the mean and variance of X.
Solution: Mean=\(\frac{3}{5}\), variance=\(\frac{68}{125}\)

Question 15. Three balls are drawn simultaneously from a bag containing 5 white and 4 red balls. Let X be the number of red balls drawn. Find the mean and variance of X.
Solution: Mean=\(\frac{4}{3}\), variance=\(\frac{5}{9}\)

Hint:

P(X=0)= \(\frac{{ }^5 C_3}{{ }^9 C_3}=\frac{5}{42}, P(X=1)=\frac{\left({ }^4 C_1 \times{ }^5 C_2\right)}{{ }^9 C_3}=\frac{10}{21},\)

P(X=2)=\(\frac{\left({ }^4 C_2 \times{ }^5 C_1\right)}{{ }^9 C_3}=\frac{5}{14}, P(X=3)=\frac{{ }^4 C_3}{{ }^9 C_3}=\frac{1}{21}\).

Question 16. Two cards are drawn without replacement from a well-shuffled deck of 52 cards. Let X be the number of face cards drawn. Find the mean and variance of X.
Solution: Mean=\(\frac{20}{13}\), variance=\(\frac{1000}{2873}\)

Hint: There are 12 face cards (4 kings, 4 queens and 4 jacks).

Clearly, X = 0 or 1 or 2.

P(X = 0) = P(no face card)

= P(drawing 2 cards out of 40 non-face cards)

= \(\frac{{ }^{40} C_2}{{ }^{52} C_2}=\left(\frac{40 \times 39}{2 \times 1} \times \frac{2 \times 1}{52 \times 51}\right)=\frac{10}{17}\)

P(X=1)=P(1 face card and 1 non-face card)

= \(\frac{\left({ }^{12} C_1 \times{ }^{40} C_1\right)}{{ }^{52} C_2}=\left(12 \times 40 \times \frac{2 \times 1}{52 \times 51}\right)=\frac{80}{221}\).

P(X=2)=P(2 face cards)

= \(\frac{{ }^{40} C_2}{{ }^{52} C_2}=\left(\frac{40 \times 39}{2 \times 1} \times \frac{2 \times 1}{52 \times 51}\right)=\frac{10}{17}\)

Thus, we have

Class 12 Maths Probability Distribution Mean And Variance Of X Q16

 

Question 17. Two cards are drawn one by one with replacement from a well-shuffled deck of 52 cards. Find the mean and variance of the number of aces.
Solution: Mean=\(\frac{2}{13}\), variance=\(\frac{24}{169}\)

Question 18. Three cards are drawn successively with replacement from a well-shuffled deck of 52 cards. A random variable X denotes the number of hearts in the three cards drawn. Find the mean and variance of X.
Solution: Mean=\(\frac{3}{4}\), variance=\(\frac{9}{16}\)

Hint: There are 13 hearts and 39 other cards.

Let E = event of drawing a heart.

Then, \(P(E)=\frac{13}{52}=\frac{1}{4} \text { and } P(\bar{E})=\left(1-\frac{1}{4}\right)=\frac{3}{4}\)

Let X = number of hearts in a draw.

Then, X = 0,1, 2 or 3.

P(X=0)=\(P(\bar{E} \bar{E} \bar{E})=P(\bar{E}) \times P(\bar{E}) \times P(\bar{E})=\left(\frac{3}{4} \times \frac{3}{4} \times \frac{3}{4}\right)=\frac{27}{64}\)

P(X=1)=\(P[(E \bar{E} \bar{E}) \text { or }(\bar{E} E \bar{E}) \text { or }(\bar{E} \bar{E} E)]\)

= \(P(E \bar{E} \bar{E})+P(\bar{E} E \bar{E})+P(\bar{E} \bar{E} E)\)

= \(\left(\frac{1}{4} \times \frac{3}{4} \times \frac{3}{4}\right)+\left(\frac{3}{4} \times \frac{1}{4} \times \frac{3}{4}\right)+\left(\frac{3}{4} \times \frac{3}{4} \times \frac{1}{4}\right)=\frac{27}{64}\)

P(X=2)= \(P[(E E \bar{E}) \text { or }(E \bar{E} E) \text { or }(\bar{E} E E)]\)

= \(P(E E \bar{E})+P(E \bar{E} E)+P(\bar{E} E E)\)

= \(\left(\frac{1}{4} \times \frac{1}{4} \times \frac{3}{4}\right)+\left(\frac{1}{4} \times \frac{3}{4} \times \frac{1}{4}\right)+\left(\frac{3}{4} \times \frac{1}{4} \times \frac{1}{4}\right)=\frac{9}{64}\)

P(X=3) = \(P(E E E)=P(E) \times P(E) \times P(E)=\left(\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4}\right)=\frac{1}{64}\)

Class 12 Maths Probability Distribution Probability Distribution Q18

 

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