WBCHSE Class 12 Maths Solutions For Probability Distribution

Chapter 3 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 which 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.

Range (X) = {0,1,2}.

WBBSE Class 12 Probability Distribution Solutions

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, …, x_n with respective probabilities p₁, p₂, …, p_n then the probability distribution of X is given by

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Remark The above probability distribution of X is defined only when (1) each p_i ≥ 0 (2) \(\sum_{i=1}^n p_i=1\)

Mean and Variance of Random Variables

Let a random variable X assume values x₁, x₂, …, x_n with probabilities p₁, p₂, …, p_n respectively such that each p_i ≥ 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

μ = \(E(X)=\sum_{i=1}^m x_i p_i\)

And, the variance, denoted by σ², is defined as

σ² = \(\left(\Sigma x_i^2 p_i-\mu^2\right)\).

Standard deviation, σ, is given by

\(\sigma=\sqrt{\text { variance }}\).

Solved Examples

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

Solution

In two tosses of a coin, the sample space is given by

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

∴ n(S) = 4.

So, every single outcome has a probability \(\frac{1}{4}\).

Let X = number of tails in two tosses.

In two tosses, we may have no tail, 1 tail or 2 tails.

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

P(X=0) = P(getting no tail) = P(HH) = \(\frac{1}{4}\).

P(X=1) = P(getting 1 tail) = P(HT or TH) = P(TT) = \(\frac{2}{4}\) = \(\frac{1}{2}\).

P(x=2) = P(getting 2 tails) = P(TT) = \(\frac{1}{4}\).

Hence, the probability distribution of X is given by

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

Variance, σ² = \(\Sigma x_i^2 p_i-\mu^2\)

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

= \(\frac{1}{2}\).

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, HTH, HHT, HHH}.

∴ n(s) = 8.

So, every single outcome has a probability \(\frac{1}{8}\).

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 to 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:

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

Variance, σ² = \(\Sigma x_i^2 p_i-\mu^2\)

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

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

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Example 3 A die is tossed once. If the random variable X is defined as

\(X=\left\{\begin{array}{l}
1, \text { if the die results in an even number } \\
0, \text { if the die results in an odd number }
\end{array}\right.\)

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

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

Variance, σ² = \(\Sigma x_i^2 p_i-\mu^2\)

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

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

probability of getting a six = \(\frac{1}{6}\), and

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

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)

= \(\left(\frac{5}{6} \times \frac{5}{6}\right)=\frac{25}{36} .\)

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)

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

Hence, the probability distribution is given by

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

Variance, σ² = \(\Sigma x_i^2 p_i-\mu^2\)

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

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

Step-by-Step Solutions to Probability Distribution Problems

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

Given

Two cards are drawn successively with replacement from a well-shuffled pack of 52 cards.

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) = \(\left(1-\frac{1}{13}\right)=\frac{12}{13}\)

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

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

Variance, σ² = \(\Sigma x_i^2 p_i-\mu^2\)

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

Example 6 Two cards are drawn simultaneously (or successively with out replacement) from a well-shuffled pack of 52 cards. Find the mean and variance of the number of aces.

Solution

Given

Two cards are drawn simultaneously (or successively with out replacement) from a well-shuffled pack of 52 cards.

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[(one ace out of 4) 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(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:

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

Variance, σ² = \(\Sigma x_i^2 p_i-\mu^2\)

= \(\left(0 \times \frac{188}{221}\right)+\left(1 \times \frac{32}{221}\right)+\left(4 \times \frac{1}{221}\right)-\frac{4}{169}\)

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

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

Given

Three defective bulbs are mixed with 7 good ones. Let X be the number of defective bulbs when 3 bulbs are drawn at random.

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

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

Variance, σ² = \(\Sigma x_i^2 p_i-\mu^2\)

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

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

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

Given

An urn contains 4 white and 3 red balls. Let X be the number of red balls in a random draw of 3 balls.

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(getting 1 red and 1 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.

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

Variance, σ² = \(\Sigma x_i^2 p_i-\mu^2\)

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

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

Example 9 In a game, 3 coins are tossed. A person in 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

Given

In a game, 3 coins are tossed. A person in 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.

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 = number of rupees the person gets.

Then, possible values of X are 5 an d-3.

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

Thus, we have

∴ the required expectations = mean, μ = Σxipi

= \(\left(5 \times \frac{1}{4}\right)+(-3) \times \frac{3}{4}=-1\), i.e., he loses Re 1 per toss.

Chapter 4 Binomial Distribution

Understanding Probability Distributions in Class 12

Success And Failure In An Experiment There are certain kinds of experiments which have two possible outcomes. One of these two outcomes is called a success, while the other is called a failure.

For example, in tossing a coin, we get either a head or a tail. If getting head is taken as a success then getting a tail is a failure.

Bernoulli’s Theorem Let there be n independent trails in an experiment and let the random variable X denote the number of successes in these trails. Let the probability of getting a success in a single trail be p and that of getting a failure be q so that p + q = 1. Then,

P(X=r) = ⁿC_r.p^r.q^(n-r).

Proof

Let us denote a success by S and a failure by F.

Number of ways of getting r successes in n trails ⁿC_r.

∴ \(P(X=r)={ }^n C_r \cdot P\{\underbrace{S S S \ldots S}_{r \text { times }} \text { and } \underbrace{F F F \ldots F}_{(n-r) \text { times }}\}\)

= ⁿC_r.{P(S).P(S)…r times} x {P(F).P(F)…(n-r)times}

= ⁿC_r.(p.p.p…r times) x [q.q.q…(n-r) times]

= ⁿC_r.p^r.q^(n-r).

Hence, P(X=r) = ⁿC_r . p^r . q^(n-r).

Remark

We have

P(X=0) = qⁿ; P(X=1) = npq^(n-1); P(X=2) = ⁿC₂.p².q^(n-2), etc.

The probability distribution of X may be expressed as

\(\left(\begin{array}{lllll}
X: & 0 & 1 & \ldots & r \\
P(X): & q^n & n p q^{(n-1)} & \ldots & C_r \cdot p^r \cdot q^{(n-r)}
\end{array}\right)\)

This distribution is called a binomial distribution.

Conditions for the Applicability of a Binomial Distribution

(1) The experiment is performed for a finite and fixed number of trials.

(2) Each trial must give either a success or a failure.

(3) The probability of a success in each trial is the same.

Solved Examples

Example 1 A coin is tossed 4 times. If X is the number of heads observed, find the probability distribution of X.

Solution

Given

A coin is tossed 4 times. If X is the number of heads observed,

When a coin is tossed, we have S = {H,T}.

P(getting a head) = \(\frac{1}{2}\), and

P(not getting a head) = \(\left(1-\frac{1}{2}\right)=\frac{1}{2}\)

Let x be the random variable denoting the number of heads.

In 4 trials, we may get 0 or 1 or 2 or 3 or 4 heads.

So, X may assume the values 0, 1, 2, 3, 4.

P(X=0) = \({ }^4 C_0 \cdot\left(\frac{1}{2}\right)^0 \cdot\left(\frac{1}{2}\right)^{(4-0)}=\frac{1}{16}\)

P(X=1) = \({ }^4 C_1:\left(\frac{1}{2}\right)^1 \cdot\left(\frac{1}{2}\right)^{(4-1)}=\frac{1}{4}\)

P(X=2) = \({ }^4 C_2 \cdot\left(\frac{1}{2}\right)^2 \cdot\left(\frac{1}{2}\right)^{(4-2)}=\frac{3}{8} .\)

P(X=3) = \({ }^4 C_3 \cdot\left(\frac{1}{2}\right)^3 \cdot\left(\frac{1}{2}\right)^{(4-3)}=\frac{1}{4} .\)

P(X=4) = \({ }^4 C_4 \cdot\left(\frac{1}{2}\right)^4 \cdot\left(\frac{1}{2}\right)^{(4-4)}=\frac{1}{16}\)

Hence, the required probability distribution is given by

\(\left(\begin{array}{lccccc}
X: & 0 & 1 & 2 & 3 & 4 \\
P(X): & \frac{1}{16} & \frac{1}{4} & \frac{3}{8} & \frac{1}{4} & \frac{1}{16}
\end{array}\right) .\)

Examples of Binomial and Poisson Distributions

Example 2 Find the probability distribution of the number of sixes in three tosses of a die.

Solution

When a die is tossed, we have S = {1,2,3,4,5,6}.

∴ P(getting a six) = \frac{1}{6} and P(not getting a six) = \(\left(1-\frac{1}{6}\right)=\frac{5}{6}\)

Let X be the random variable denoting the number of sixes.

In 3 trials, the number of sixes may be 0 or 1 or 2 or 3.

So, X may assume the values 0, 1, 2, 3.

P(X=0) = \({ }^3 C_0 \cdot\left(\frac{1}{6}\right)^0 \cdot\left(\frac{5}{6}\right)^{(3-0)}=\frac{125}{216}\)

P(X=1) = \({ }^3 C_1 \cdot\left(\frac{1}{6}\right)^1 \cdot\left(\frac{5}{6}\right)^{(3-1)}=\frac{25}{72}\)

P(X=2) = \({ }^3 C_2 \cdot\left(\frac{1}{6}\right)^2 \cdot\left(\frac{5}{6}\right)^{(3-2)}=\frac{5}{72}\)

P(X=3) = \({ }^3 C_3 \cdot\left(\frac{1}{6}\right)^3 \cdot\left(\frac{5}{6}\right)^{(3-3)}=\frac{1}{216}\)

The required probability distribution of X is given below:

\(\left(\begin{array}{lcccc}
X: & 0 & 1 & 2 & 3 \\
P(X): & \frac{125}{216} & \frac{25}{72} & \frac{5}{72} & \frac{1}{216}
\end{array}\right)\)

Example 3 Find the probability distribution of the number of doublets in four throws of a pair of dice.

Solution

When a pair of dice is thrown, there are 36 possible outcomes.

∴ n(S) = 36.

All possible doublets are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).

P(getting a doublet) = \(\frac{6}{36}\) = \(\frac{1}{6}\), and

P(not getting a doublet) = \(\left(1-\frac{1}{6}\right)=\frac{5}{6}\)

Let X denote the number of doublets.

In 4 throws, we can have 0 or 1 or 2 or 3 or 4 doublets.

P(X=0) = \({ }^4 C_0 \cdot\left(\frac{1}{6}\right)^0 \cdot\left(\frac{5}{6}\right)^4=\frac{625}{1296}\)

P(X=1) = \({ }^4 C_1 \cdot\left(\frac{1}{6}\right)^1 \cdot\left(\frac{5}{6}\right)^3=\frac{125}{324}\)

P(X=2) = \({ }^4 C_2 \cdot\left(\frac{1}{6}\right)^2 \cdot\left(\frac{5}{6}\right)^2=\frac{25}{216}\)

P(X=3) = \({ }^4 C_3 \cdot\left(\frac{1}{6}\right)^3 \cdot\left(\frac{5}{6}\right)^1=\frac{5}{324}\)

P(X=4) = \({ }^4 C_4 \cdot\left(\frac{1}{6}\right)^4 \cdot\left(\frac{5}{6}\right)^0=\frac{1}{1296}\)

The required probability distribution is given below:

\(\left(\begin{array}{lccccc}
X: & 0 & 1 & 2 & 3 & 4 \\
P(X): & \frac{625}{1296} & \frac{125}{324} & \frac{25}{216} & \frac{5}{324} & \frac{1}{1296}
\end{array}\right)\)

Example 4 An unbiased coin is tossed 6 times. Find, using binomial distribution, the probability of getting at least 5 heads.

Solution

Given

An unbiased coin is tossed 6 times.

In a single throw of a coin, we have S = {H,T}.

P(getting a head) = \(\frac{1}{2}\), and P(not getting a head) = \(\left(1-\frac{1}{2}\right)=\frac{1}{2}\)

∴ p = \(\frac{1}{2}\) and q = \(\frac{1}{2}\).

P(X=r) = nCr.pr.q(n-r) = \({ }^6 C_r \cdot\left(\frac{1}{2}\right)^r \cdot\left(\frac{1}{2}\right)^{(6-r)}={ }^6 C_r \cdot\left(\frac{1}{2}\right)^6\)

∴ P(getting at least 5 heads)

= P(X ≥ 5)

= P(X=5) + P(X=6)

= \({ }^6 C_5 \cdot\left(\frac{1}{2}\right)^6+{ }^6 C_6 \cdot\left(\frac{1}{2}\right)^6=\left(\frac{3}{32}+\frac{1}{64}\right)=\frac{7}{64} .\)

Hence, the required probability is \(\frac{7}{64}\).

Types of Probability Distributions Explained

Example 5 An unbiased coin is tossed 8 times. Find, by using binomial distribution, the probability of getting at least 3 heads.

Solution

Given

An unbiased coin is tossed 8 times.

In a single throw of a coin, we have S = (H,T}.

P(getting a head) = \(\frac{1}{2}\), and P(not getting a head) = \(\left(1-\frac{1}{2}\right)=\frac{1}{2} \text {. }\)

∴ p = \(\frac{1}{2}\) and q = \(\frac{1}{2}\).

P(X=r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^8 C_r \cdot\left(\frac{1}{2}\right)^r \cdot\left(\frac{1}{2}\right)^{(8-r)}={ }^8 C_r\left(\frac{1}{2}\right)^8\)

∴ P(getting at least 3 heads)

= P(X ≥ 3)

= 1 – [P(X=0) + P(X=1) + P(X=2)]

= \(1-\left[{ }^8 C_0 \cdot\left(\frac{1}{2}\right)^8+{ }^8 C_1 \cdot\left(\frac{1}{2}\right)^8+{ }^8 C_2 \cdot\left(\frac{1}{2}\right)^8\right]\)

= \(1-\frac{1}{256} \cdot(1+8+28)=\left(1-\frac{37}{256}\right)=\frac{219}{256} .\)

Hence, the required probability is \(\frac{219}{256}\).

Example 6 Six coins are tossed simultaneously. Find the probability of getting

(1) 3 heads (2) no head (3) at least one head (4) not more than 3 heads.

Solution

The experiment may be taken as throwing a single coin 6 times.

In a single throw of a coin, we have S = {H,T}.

P(getting a head) = \(\frac{1}{2}\), and P(not getting a head) = \(\left(1-\frac{1}{2}\right)=\frac{1}{2}\)

Let X be the random variable showing the number of heads.

P(X=r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^6 C_r \cdot\left(\frac{1}{2}\right)^r \cdot\left(\frac{1}{2}\right)^{6-r}={ }^6 C_r \cdot\left(\frac{1}{2}\right)^6\)

(1) P(getting 3 heads) = P(X=3) = \({ }^6 C_3 \cdot\left(\frac{1}{2}\right)^6=\frac{5}{16} \text {. }\)

(2) P(getting no head) = P(X=0) = \({ }^6 C_0 \cdot\left(\frac{1}{2}\right)^6=\frac{1}{64} .\)

(3) P(getting at least 1 head)

= \(1-P(X=0)=1-\left[{ }^6 C_0 \cdot\left(\frac{1}{2}\right)^6\right]\)

= \(\left(1-\frac{1}{64}\right)=\frac{63}{64}\)

(4) P(getting not more thatn 3 heads)

= P(no head or 1 head or 2 heads or 3 heads)

= P(X=0) + P(X-1) + P(X=2) + P(X=3)

= \({ }^6 C_0 \cdot\left(\frac{1}{2}\right)^6+{ }^6 C_1 \cdot\left(\frac{1}{2}\right)^6+{ }^6 C_2 \cdot\left(\frac{1}{2}\right)^6+{ }^6 C_3 \cdot\left(\frac{1}{2}\right)^6\)

= \(\left(\frac{1}{2}\right)^6 \cdot(1+6+15+20)=\left(\frac{1}{64} \times 42\right)=\frac{21}{32} \text {. }\)

Example 7 A die is thrown 5 times. If getting an odd number is a success, find the probability of getting at least 4 successes.

Solution

Given

A die is thrown 5 times. If getting an odd number is a success

When a die is thrown, we have S = {1,2,3,4,5,6}.

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

∴ P(a success) = \(\frac{1}{2}\), and P(not a success) = \(\left(1-\frac{1}{2}\right)=\frac{1}{2}\)

Let X be the random variable showing the number of successes.

P(X=r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^5 C_r \cdot\left(\frac{1}{2}\right)^r \cdot\left(\frac{1}{2}\right)^{(5-r)}={ }^5 C_r \cdot\left(\frac{1}{2}\right)^5 \cdot\)

P(at least 4 successes) = P(4 successes or 5 successes)

= P(X=4) + P(X=5)

= \({ }^5 C_4 \cdot\left(\frac{1}{2}\right)^5+{ }^5 C_5 \cdot\left(\frac{1}{2}\right)^5=\left(\frac{5}{32}+\frac{1}{32}\right)=\frac{6}{32}=\frac{3}{16} .\)

The probability of getting at least 4 successes = \({ }^5 C_4 \cdot\left(\frac{1}{2}\right)^5+{ }^5 C_5 \cdot\left(\frac{1}{2}\right)^5=\left(\frac{5}{32}+\frac{1}{32}\right)=\frac{6}{32}=\frac{3}{16} .\)

Example 8 In 4 throws with a pair of dice, what is the probability of throwing doublets at least twice?

Solution

Given

In 4 throws with a pair of dice

In a single throw of a pair of dice, the number of all possible outcomes is 36.

All doublets are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).

P(getting a doublet) = \(\frac{6}{36}\) = \(\frac{1}{6}\), and

P(not getting a doublet) = \(\left(1-\frac{1}{6}\right)=\frac{5}{6}\)

Let X be the random variable denoting the number of doublets.

Then, P(X=r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^4 C_r \cdot\left(\frac{1}{6}\right)^r \cdot\left(\frac{5}{6}\right)^{(4-r)} .\)

P(at least 2 doublets)

= P(X=2) + P(X=3) + P(X=4)

= \({ }^4 C_2 \cdot\left(\frac{1}{6}\right)^2 \cdot\left(\frac{5}{6}\right)^{(4-2)}+{ }^4 C_3 \cdot\left(\frac{1}{6}\right)^3 \cdot\left(\frac{5}{6}\right)^{(4-3)}+{ }^4 C_4 \cdot\left(\frac{1}{6}\right)^4 \cdot\left(\frac{5}{6}\right)^{(4-4)}\)

= \(6 \cdot\left(\frac{1}{6}\right)^2 \cdot\left(\frac{5}{6}\right)^2+4 \cdot\left(\frac{1}{6}\right)^3 \cdot\left(\frac{5}{6}\right)^1+\left(\frac{1}{6}\right)^4 \cdot\left(\frac{5}{6}\right)^0\)

= \(\left(\frac{25}{216}+\frac{5}{324}+\frac{1}{1296}\right)=\frac{171}{1296}\)

The probability of throwing doublets at least twice = \(\left(\frac{25}{216}+\frac{5}{324}+\frac{1}{1296}\right)=\frac{171}{1296}\)

Real-Life Applications of Probability Distribution Concepts

Example 9 The bulbs produced in a factory are supposed to contain 5% defective bulbs. What is the probability that a sample of 10 bulbs will contain not more than 2 defective bulbs?

Solution

Given 

The bulbs produced in a factory are supposed to contain 5% defective bulbs.

P(getting a defective bulb) = \(\frac{5}{100}\) = \(\frac{1}{20}\), and

P(getting a non-defective bulb) = \(\left(1-\frac{1}{20}\right)=\frac{19}{20}\)

Then, p = \(\frac{1}{20}\) and q = \(\frac{19}{20}\).

Let X denote the number of dective bulbs.

P(X=r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^{10} C_r \cdot\left(\frac{1}{20}\right)^r \cdot\left(\frac{19}{20}\right)^{(10-r)}\)

P(getting not more than 2 defective bulbs)

= P(X=0 or X=1 or X=2)

= P(X=0) + P(X=1) + P(X=2)

= \({ }^{10} C_0 \cdot\left(\frac{1}{20}\right)^0 \cdot\left(\frac{19}{20}\right)^{10}+{ }^{10} C_1 \cdot\left(\frac{1}{20}\right)^1 \cdot\left(\frac{19}{20}\right)^9+{ }^{10} C_2 \cdot\left(\frac{1}{20}\right)^2 \cdot\left(\frac{19}{20}\right)^8\)

= \(\left(\frac{19}{20}\right)^{10}+\frac{1}{2} \cdot\left(\frac{19}{20}\right)^9+\frac{9}{80} \cdot\left(\frac{19}{20}\right)^8=\left(\frac{19}{20}\right)^8 \cdot\left(\frac{149}{100}\right) .\)

Let A = \(\left(\frac{19}{20}\right)^8 \cdot\left(\frac{149}{100}\right)\). Then,

log A = 8(log19 – log20) + log159 – log 100

= 8(1.2788 – 1.3010) + 2.1732 – 2

= -0.0044 = 1.9956.

∴ A = antilog (1.9956) = 0.99.

Hence, the required probability = \(\frac{99}{100}\).

Example 10 If on an average, out of 10 ships, one gets drowned then what is the probability that out of 5 ships at least 4 reach the shore safely?

Solution

Probability of a ship to reach the shore safely = \(\frac{9}{10}\).

Probability that a ship gets drowned = \(\left(1-\frac{9}{10}\right)=\frac{1}{10} \text {. }\)

Let X be the random variable showing the number of ships reaching the shore safely.

∴ P(atleast 4 reaching safely)

= P(4 reaching safely or 5 reaching safely)

= P(4 reaching safely) + P(5 reaching safely)

= P(X=4) + P(X=5)

= \({ }^5 C_4 \cdot\left(\frac{9}{10}\right)^4 \cdot\left(\frac{1}{10}\right)^{(5-4)}+{ }^5 C_5 \cdot\left(\frac{9}{10}\right)^5 \cdot\left(\frac{1}{10}\right)^0\)

= \(\frac{1}{2} \cdot\left(\frac{9}{10}\right)^4+\left(\frac{9}{10}\right)^5=\left(\frac{9}{10}\right)^4\left(\frac{1}{2}+\frac{9}{10}\right)=\frac{7}{5} \cdot\left(\frac{9}{10}\right)^4\)

Let A = \(\frac{7}{5} \cdot\left(\frac{9}{10}\right)^4=\frac{7 \times(9)^4}{5 \times(10)^4}\)

⇒ log A = log 7 + 4 x log 9 – log 5 – 4 x log 10

= (0.8451 + 4 x 0.9542 – 0.6990 – 4)

= \(-0.0371=\overline{1} .9629\)

⇒ A = antilog \(\overline{1} .9629\) = 0.9181.

Hence, the required probability is 0.9181.

Mean and Variance of a Binomial Distribution

Mean If a random variable X assumes the values x₁, x₂, …, x_n with probabilites p₁, p₂, …, p_n respectively then the mean of X is defined by

\(\mu=\sum_{i=1}^n x_i p_i .\)

To Find the Mean of a Binomial Distribution

For the binomial distribution

P(X=r) = P(r) = ⁿC_r . p^r . q^(n-r), where r = 0,1,2,…,n the mean, μ, is given by

μ = \(\sum_{r=0}^n r \cdot p(r)=\sum_{r=0}^n r \cdot{ }^n C_r \cdot p^r \cdot q^{(n-r)}\)

= ⁿC₁ . p . q^n-1 + 2 . ⁿC₂ . p² . q^(n-2) + … + n . ⁿC_n . pⁿq⁰

= 1 . np . q^n-1 + n(n-1) . p² .q^(n-2)+ … + npⁿ

= np . [^(n-1)C₀ . p⁰ . q^(n-1) + ^(n-1)C₁ . p¹ . q^(n-2) + … + ^(n-1)C_(n-1) . p^n-1 . q⁰]

= (np) . (q+p)^n-1 = (np) [∵ q + p = 1].

Hence, the mean is given by μ = np.

The variance = σ² is given by

σ² = \(\sum_{r=0}^n r^2 \cdot p(r)-(\text { mean })^2\)

= \(\sum_{r=0}^n r^2 \cdot{ }^n C_r \cdot p^r q^{(n-r)}-(n p)^2\) [∵ mean = np]

= \(\sum_{r=0}^n\{r+r(r-1)\} \cdot{ }^n C_r p^r q^{(n-r)}-(n p)^2\)

= \(\sum_{r=0}^n r \cdot{ }^n C_r p^r q^{(n-r)}+\sum_{r=0}^n r(r-1) \cdot{ }^n C_r \cdot p^r \cdot q^{(n-r)}-(n p)^2\)

= \(n p+\sum_{r=2}^n r(r-1) \cdot \frac{n(n-1)}{r(r-1)} \cdot{ }^{(n-2)} C_{(r-2)} \cdot p^2 \cdot p^{(r-2)} \cdot q^{(n-r)}-(n p)^2\)

[∵ \(\sum_{r=0}^n r \cdot{ }^n C_r \cdot p^r \cdot q^{(n-r)}\) = mean = np]

= \(n p+n(n-1) \cdot p^2 \cdot\left(\sum_{r=2}^n{ }^{(n-2)} C_{(r-2)} \cdot p^{(r-2)} \cdot q^{(n-r)}\right)-n^2 p^2\)

= np + n(n-1)p²(q+p)^(n-2) – n²p²

= np + n(n-1)p²(q+p)(n-2) – n²p²

= np + n(n-1)p² – n²p² [∵ q + p = 1]

= np – np² = np(1-p) = npq.

Hence, variance = npq.

∴ standard deviation = \(\sqrt{n p q}\).

Common Questions on Probability Distributions and Their Solutions

Recurrence Relation for a Binomial Distribution

We have

P(r) = ⁿC_r . p^r . q^(n-r) and P(r+1) = ⁿC(r+1) . p^(r+1) . q^n-r-1.

∴ \(\frac{P(r+1)}{P(r)}=\frac{{ }^n C_{(r+1)} \cdot p^{(r+1)} \cdot q^{n-r-1}}{{ }^n C_r \cdot p^r \cdot q^{(n-r)}}\)

= \(\frac{(n !)}{(r+1) ! \cdot(n-r-1) !} \cdot \frac{(r) ! \cdot(n-r) !}{(n) !} \cdot \frac{p}{q}=\frac{(n-r)}{(r+1)} \cdot \frac{p}{q}\)

∴ \(P(r+1)=\frac{(n-r)}{(r+1)} \cdot \frac{p}{q} \cdot P(r)\)

Example 11 If X follows a binomial distribution with mean 3 and variance (3/2), find (1) P(X ≥ 1) (2) (X ≤ 5).

Solution

We know that mean = np and variance = npq.

∴ np = 3 and npq = \(\frac{3}{2}\) ⇒ 3q = \(\frac{3}{2}\) ⇒ q = \(\frac{1}{2}\).

∴ p = (1-q) = \(\left(1-\frac{1}{2}\right)=\frac{1}{2}\)

Now, np = 3 and p = \(\frac{1}{2}\) ⇒ n x \(\frac{1}{2}\) = 3 ⇒ n = 6.

So, the binomial distribution is given by

P(X=r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^6 C_r \cdot\left(\frac{1}{2}\right)^r \cdot\left(\frac{1}{2}\right)^{(6-r)}={ }^6 C_r\left(\frac{1}{2}\right)^6 \cdot\)

(1) P(X ≥ 1) = 1 – P(X = 0)

= \(1-{ }^6 C_0 \cdot\left(\frac{1}{2}\right)^6=\left(1-\frac{1}{64}\right)=\frac{63}{64}\).

(2) P(X ≤ 5) = 1 – P(X = 6)

= \(1-{ }^6 C_6\left(\frac{1}{2}\right)^6=\left(1-\frac{1}{64}\right)=\frac{63}{64}\)

Example 12 If X follows a binomial distribution with mean 4 and variance 2, find P(X ≥ 5).

Solution

We know that mean = np and variance = npq.

∴ np = 4 and npq = 2.

Now, np = 4 and npq = 2 ⇒ 4q = 2 ⇒ q = \(\frac{1}{2}\).

∴ p = \((1-q)=\left(1-\frac{1}{2}\right)=\frac{1}{2}\)

Now, np = 4 and p = \(\frac{1}{2}\) ⇒ \(\frac{1}{2}n\) = 4 ⇒ n = 8.

So, the binomial distribution is given by

P(X=r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^8 C_r \cdot\left(\frac{1}{2}\right)^r \cdot\left(\frac{1}{2}\right)^{(8-r)}={ }^8 C_r \cdot\left(\frac{1}{2}\right)^8\)

∴ P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)

= \({ }^8 C_5 \cdot\left(\frac{1}{2}\right)^8+{ }^8 C_6 \cdot\left(\frac{1}{2}\right)^8+{ }^8 C_7 \cdot\left(\frac{1}{2}\right)^8+{ }^8 C_8 \cdot\left(\frac{1}{2}\right)^8\)

= \(\left[{ }^8 C_3+{ }^8 C_2+{ }^8 C_1+1\right]\left(\frac{1}{2}\right)^8\)

= \((56+28+8+1) \cdot \frac{1}{256}=\frac{93}{256} .\)

Example 13 Find the binomial distribution for which the mean and variance are 12 and 3 respectively.

Solution

Let X be a binomial variate for which mean = 12 and variance = 3.

Then, np = 12 and npq = 3 ⇔ 12 x q = 3 ⇔ q = \(\frac{1}{4}\).

∴ p = (1-q) = \(\left(1-\frac{1}{4}\right)=\frac{3}{4}\).

And, np = 12 ⇔ n = \(\frac{12}{p}\) = \(\left(12 \times \frac{4}{3}\right)=16\)

Thus, n = 16, p = \(\frac{3}{4}\) and q = \(\frac{1}{4}\).

Hence, the binomial distribution is given by

P(X=r) = \({ }^{16} C_r \cdot\left(\frac{3}{4}\right)^r \cdot\left(\frac{1}{4}\right)^{(16-r)}\), where r = 0, 1, 2, 3, …, 15.

Example 14 If the sum of the mean and variance of a binomial distribution for 5 trials is 1.8, find the distribution.

Solution

Given

If the sum of the mean and variance of a binomial distribution for 5 trials is 1.8

We know that

mean = np and variance = npq.

It is being given that n = 5 and mean + variance = 1.8.

∴ np + npq = 1.8, where n = 5

⇔ 5p + 5pq = 1.8

⇔ p + p(1-p) = 0.36 [∵ q = (1-p)]

⇔ p² – 2p + 0.36 = 0

⇔ 100p² – 200p + 36 = 0

⇔ 25p² – 50p + 9 = 0

⇔ 25p² – 45p – 5p + 9 = 0

⇔ 5p(5p-9) – (5p-9) = 0

⇔ (5p-9)(5p-1) = 0

⇔ p = \(\frac{1}{5}\) = 0.2 [∵ p cannot exceed 1].

Thus, n = 5, p = 0.2, and q = (1-p) = (1-0.2) = 0.8

Let X denote the binomial variate. Then, the required distribution is

P(X=r) = ⁿC_r. p^r . q^(n-r) = ⁵C_r . (0.2)^r . (0.8)^(5-r), where r = 0, 1, 2, 3, 4, 5.

Mean and Variance of Probability Distributions

Example 15 The sum and the product of the mean and variance of a binomial distribution are 24 and 128 respectively. Find the distribution.

Solution

Given

The sum and the product of the mean and variance of a binomial distribution are 24 and 128 respectively.

We have

np + npq = 24 and np x npq = 128

⇔ (np)(1+q) = 24 and n²p² x q = 128

⇔ \(n^2 p^2=\frac{576}{(1+q)^2} and n2p2 x q = 128\)

⇔ \(\frac{576}{(1+q)^2}=\frac{128}{q} \Leftrightarrow 2\left(1+q^2+2 q\right)=9 q\)

⇔ 2q² – 5q + 2 = 0 ⇔ (2q -1)(q – 2) = 0

⇔ q = \(\frac{1}{2}\) [∵ q ≠ 2].

∴ p = (1-q) = \(\left(1-\frac{1}{2}\right)=\frac{1}{2}\)

Now, np(1 + q) = 24 ⇔ \(n \times \frac{1}{2}\left(1+\frac{1}{2}\right)=24\) ⇔ n = 32.

Hence, the required probability distribution is given by

P(X=r) = \({ }^{32} C_r \cdot\left(\frac{1}{2}\right)^{32}\) .

Example 16 In a binomial distribution, prove that mean > variance.

Solution

Let X be a binomial variate with parameters n and p. Then, mean = np and variance = npq.

∴ (mean) – (variance) = (np – npq) = np(1-q) = np² > 0

[∵(1-q) = p and np² > 0 as n ∈ N]

⇒ [(mean)-(variance)] > 0

⇒ mean > variance

Hence, mean > variance.

Example 17 A die is tossed thrice. Getting an even number is considered a success. What is the variance of the binomial distribution?

Solution

Given

A die is tossed thrice. Getting an even number is considered a success.

Here, n = 3.

Let p = probability of getting an even number in a single throw

⇒ p = \(\frac{3}{6}\) = \(\frac{1}{2}\)

⇒ \(q=(1-p)=\left(1-\frac{1}{2}\right)=\frac{1}{2}\)

∴ variance = npq = \(\left(3 \times \frac{1}{2} \times \frac{1}{2}\right)=\frac{3}{4}\).

The variance of the binomial distribution = npq = \(\left(3 \times \frac{1}{2} \times \frac{1}{2}\right)=\frac{3}{4}\).

Example 18 A die is rolled 20 times. Getting a number greater than 4 is a success. Find the mean and variance of the number of successes.

Solution

Given

A die is rolled 20 times. Getting a number greater than 4 is a success.

In a single throw of a die, we have

p = probability of getting a number greater than 4 = \(\frac{2}{6}\) = \(\frac{1}{3}\).

∴ q = (1-p) = \(\left(1-\frac{1}{3}\right)=\frac{2}{3}\).

Also, n = 20 (given).

∴ mean = np = \(\left(20 \times \frac{1}{3}\right)=6.67 \text {, and }\)

variance = npq = \(\left(20 \times \frac{1}{3} \times \frac{2}{3}\right)=4.44\)

Example 19 A die is tossed 180 times. Find the expected number (μ) of times the face with the number 5 will appear. Also, find the standard deviation (σ), and variance (σ2).

Solution

Given

A die is tossed 180 times.

In a single throw of a die, S = {1,2,3,4,5,6}.

p = (probability of getting the number 5) = \(\frac{1}{6}\).

∴ q = (1-p) = \(\left(1-\frac{1}{6}\right)=\frac{5}{6}\)

Thus, n = 180, p = \(\frac{1}{6}\) and q = \(\frac{5}{6}\).

∴ μ = \(\left(180 \times \frac{1}{6}\right)=30\)

Variance = \(\sigma^2=n p q=\left(180 \times \frac{1}{6} \times \frac{5}{6}\right)=25\).

Standard deviation = σ = √25 = 5.