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}.

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 = \(\frac{\sqrt{3}}{2} \text {. }\)

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 = \(\sqrt{\frac{5}{18}}=\frac{1}{3} \cdot \sqrt{\frac{5}{2}}\)

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

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

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

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 \(\frac{7}{64}\).

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

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 = \(\sqrt{n p q}\).

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.

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.