Probability
14.0 Binomial Distribution for Successive Events
14.0 Binomial Distribution for Successive Events
Bernoulli trials:
Trials of a random experiment are called Bernoulli trials, if they satisfy the following conditions:
i. They are finite in number
ii. They are independent of each other
iii. Each trial has exactly two outcomes. (Success or Failure)
iv. The probability of success and failure remains same in each trial.
Binomial distribution:
Suppose $p$ and $q$ are the respective chances of the happening and failing of an event at a single trial i.e. $p + q = 1$. Then the chance of its happening $r$ times in $n$ trials is $^n{C_r}{p^r}{q^{n - r}}$ as the chances of its happening $r$ times and failing $n - r$ trials in a given order is ${p^r}{q^{n - r}}$ and there are $^n{C_r}$ such orders which are mutually exclusive.
This is given by the terms in the binomial expansion of ${(p + q)^n}$.
Explanation: Consider a random experiment and an event $A$ associated with it. If the experiment results in the event $A$, let it be considered as success. Let this be denoted by $S$. If the event $A$ is not occurring, then let us consider it to be failure. Let this be denoted as $F$.
Let the probability of $S$ i.e. success be $P(S) = p$
Let the probability of $F$ i.e. failure be $P(F) = q$
Let the experiment be performed $4$ times under identical conditions.
Let the probability of three success be considered.
Three successes in $4$ trials can occur in $^4{C_3}$, they are
$$SSSF,\;SSFS,\;SFSS,\;FSSS$$
By addition theorem,
either of the one occurs.
Thus probability is $$P(SSSF) + P(SSFS) + P(SFSS) + P(FSSS)$$
Since, each event is independent,
$$\begin{equation} \begin{aligned} P(SSSF) = P(S)P(S)P(S)P(F) = p \times p \times p \times q = {p^3}q \\ P(SSFS) = P(S)P(S)P(F)P(S) = p \times p \times q \times p = {p^3}q \\ P(SFSS) = P(S)P(F)P(S)P(S) = p \times q \times p \times p = {p^3}q \\ P(FSSS) = P(F)P(S)P(S)P(S) = q \times p \times p \times p = {p^3}q \\\end{aligned} \end{equation} $$
Thus,
$$P(X = 3) = {\;^4}{C_3}{p^3}q = {\;^4}{C_3}{p^3}{q^{4 - 3}}$$
Similarly,
$$\begin{equation} \begin{aligned} P(X = 0) = {\;^4}{C_0}{p^0}{q^4} = {\;^4}{C_0}{p^0}{q^{4 - 0}} \\ P(X = 1) = {\;^4}{C_1}{p^1}{q^3} = {\;^4}{C_1}{p^1}{q^{4 - 1}} \\ P(X = 2) = {\;^4}{C_2}{p^2}{q^2} = {\;^4}{C_2}{p^2}{q^{4 - 2}} \\ P(X = 4) = {\;^4}{C_4}{p^4}{q^0} = {\;^4}{C_4}{p^4}{q^{4 - 4}} \\\end{aligned} \end{equation} $$
Hence we get,
$$P(X = r) = {\;^n}{C_r}{p^r}{q^{n - r}}$$
Definition:
A random variable $X$ which takes values $0, 1, 2, .... , n$ is said to follow binomial distribution if its probability distribution function is given by,
$$P(X = r) = {\;^n}{C_r}{p^r}{q^{n - r}}$$
$r = 0,1,2,...,n$
$where\;p,q > 0\;\;and\;p + q = 1$
Illustration 47. A die is thrown $5$ times. What is the chance that an odd number turns up i. exactly $4$ times , ii. at least $2$ times
Solution: Probability of success is
$${{3 \over 6}}$$
i.e.
$$p = {1 \over 2}$$
Thus, $$q = {1 \over 2}$$
i. For exactly $4$ successes
Required probability is,
$$^5{C_4} \times {\left[ {{1 \over 2}} \right]^4} \times {1 \over 2} = {{35} \over {128}}$$
ii. For at least two,
$$\begin{equation} \begin{aligned} P(atleast\;two){ = ^5}{C_2} \times {\left[ {{1 \over 2}} \right]^2} \times {\left[ {{1 \over 2}} \right]^3}{ + ^5}{C_3} \times {\left[ {{1 \over 2}} \right]^3} \times {\left[ {{1 \over 2}} \right]^2}{ + ^5}{C_4} \times {\left[ {{1 \over 2}} \right]^4} \times {1 \over 2}{ + ^5}{C_5} \times {\left[ {{1 \over 2}} \right]^5} \\ P(atleast\;two) = {\left[ {{1 \over 2}} \right]^5}{[^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}] \\ P(atleast\;two) = {\left[ {{1 \over 2}} \right]^5}\left[ {10 + 10 + 5 + 1} \right] \\ P(atleast\;two) = {\left[ {{1 \over 2}} \right]^5}\left[ {26} \right] \\\end{aligned} \end{equation} $$
Illustration 48. Team $A$ plays with $5$ other teams exactly once. Assuming that for each match the probabilities of winning, losing and getting a draw are equal, find the probability that $A$ wins and loses equal number of matches.
Solution: Given,
Probability of winning = Probability of losing = Probability of getting draw
Since all three events are mutually exclusive and exhaustive,
$$\begin{equation} \begin{aligned} P(win) + P(loss) + P(draw) = 1 \\ 3P(win) = 1 \\ P(win) = {1 \over 3} \\ \Rightarrow P(win) = P(loss) = P(draw) = {1 \over 3} \\\end{aligned} \end{equation} $$
Team $A$ plays with $5$ teams once, thus there are in total five matches.
Possible ways in which the team wins and losses equal number of matches is,
i. No winning and no losing
ii. $1$ win and $1$ loss
iii. $2$ win and $2$ loss
i. No win and no loss means all $5$ were draw.
Thus the probability is,
${\left( {{1 \over 3}} \right)^5}$
ii. $1$ win and $1$ loss would mean three were draw,
Probability of winning one out of the five matches = $^5{C_1}{1 \over 3}$
Probability of losing one out of the other four matches = $^4{C_1}{1 \over 3}$
The other three matches being necessarily draw = ${\left( {{1 \over 3}} \right)^3}$
Thus the total probability is
Probability that the team won one match and lost one and had three draw,
$$\begin{equation} \begin{aligned} ^5{C_1}{1 \over 3}{ \times ^4}{C_1}{1 \over 3} \times {\left( {{1 \over 3}} \right)^3} \\ { = ^5}{C_1}^4{C_1}{\left( {{1 \over 3}} \right)^5} \\ = 5 \times 4 \times {\left( {{1 \over 3}} \right)^5} \\ = 20 \times {\left( {{1 \over 3}} \right)^5} \\\end{aligned} \end{equation} $$
iii. $2$ win and $2$ loss would mean one ended up in a draw,
Probability of winning two out of the five matches = $^5{C_2}{\left( {{1 \over 3}} \right)^2}$
Probability of losing two out of the other three matches = $^3{C_2}{\left( {{1 \over 3}} \right)^2}$
The other match being necessarily draw = ${1 \over 3}$
Thus the total probability is
Probability that the team won two matches and lost two and had one draw
$$\begin{equation} \begin{aligned} ^5{C_2}{\left( {{1 \over 3}} \right)^2}{ \times ^3}{C_2}{\left( {{1 \over 3}} \right)^2} \times {1 \over 3} \\ { = ^5}{C_2}^3{C_2}{\left( {{1 \over 3}} \right)^5} \\ = 10 \times 3 \times {\left( {{1 \over 3}} \right)^5} \\ = 30 \times {\left( {{1 \over 3}} \right)^5} \\\end{aligned} \end{equation} $$
Total probability that the team wins and losses equal number of matches =
Probability that the team wins and losses no matches + probability that the team wins and losses one match exactly each + probability that the team wins and losses exactly two matches each
$$\begin{equation} \begin{aligned} = (1 + 20 + 30) \times {\left( {{1 \over 3}} \right)^5} \\ = (51) \times {\left( {{1 \over 3}} \right)^5} \\ = {{17} \over {81}} \\\end{aligned} \end{equation} $$
Thus the probability that the team wins and losses equal number of matches $ = {{17} \over {81}}$
Illustration 49. Find the probability distribution of the number of doublets in four throws of a pair of dice.
Solution: The event is throwing a pair of dice.
This is being performed four times.
Let $p$ denote the probability of getting a doublet in single throw of pair of dice.
Then,
$$p = {6 \over {36}} = {1 \over 6}$$
$$q = 1 - p = 1 - {1 \over 6} = {5 \over 6}$$
Let $X$ denote the number of doublets in four such throws.
Then $X$ is a binomial variable.
Here,
$$n = 4\;\;and\;p = {1 \over 6}$$
$$P(X = r) = {\;^n}{C_r}{p^r}{q^{n - r}}$$
$r$ is getting doublets.
i. Getting no doublets.
$$P(X = 0) = {\;^4}{C_0}{p^0}{q^4} = {\;^4}{C_0}{\left( {{1 \over 6}} \right)^0}{\left( {{5 \over 6}} \right)^4} = {\left( {{5 \over 6}} \right)^4}$$
ii. Getting one doublet.
$$P(X = 1) = {\;^4}{C_1}{p^1}{q^3} = \;4{\left( {{1 \over 6}} \right)^1}{\left( {{5 \over 6}} \right)^3} = {2 \over 3}{\left( {{5 \over 6}} \right)^3}$$
iii. Getting two doublets.
$$P(X = 2) = {\;^4}{C_2}{p^2}{q^2} = \;6{\left( {{1 \over 6}} \right)^2}{\left( {{5 \over 6}} \right)^2} = {1 \over 6}{\left( {{5 \over 6}} \right)^2}$$
iv. Getting three doublets.
$$P(X = 3) = {\;^4}{C_3}{p^3}{q^1} = \;4{\left( {{1 \over 6}} \right)^3}{\left( {{5 \over 6}} \right)^1} = {{10} \over 3}{\left( {{1 \over 6}} \right)^3}$$
v. Getting four doublets.
$$P(X = 4) = {\;^4}{C_4}{p^4}{q^0} = \;{\left( {{1 \over 6}} \right)^4}{\left( {{5 \over 6}} \right)^0} = {\left( {{1 \over 6}} \right)^4}$$
Thus, the probability distribution of $X$ is,
| $X$ | $0$ | $1$ | $2$ | $3$ | $4$ |
| $P(X)$ | ${\left( {{5 \over 6}} \right)^4}$ | ${2 \over 3}{\left( {{5 \over 6}} \right)^3}$ | ${1 \over 6}{\left( {{5 \over 6}} \right)^2}$ | ${{10} \over 3}{\left( {{1 \over 6}} \right)^3}$ | ${\left( {{1 \over 6}} \right)^4}$ |
Illustration 50. The probability of a man hitting a target is ${{1 \over 4}}$. How many times must he fire so that the probability of his hitting the target at least once is greater than ${{2 \over 3}}$ ?
Solution: Suppose the man fires $n$ times and $X$ denotes the number of time he hits the target.
Then,
$$P(X = r) = {\;^n}{C_r}{\left( {{1 \over 4}} \right)^r}{\left( {{3 \over 4}} \right)^{n - r}}\;\;,\;\;\;\;\;\;\;\;r = 0,1,2,...,n$$
If he hits the target at least once, then
$$r \ge 1$$
Given that probability he hits the target at least once is ${{2 \over 3}}$
$$\begin{equation} \begin{aligned} P(X \ge 1) > {2 \over 3} \\ 1 - P(X = 0) > {2 \over 3} \\ 1{ - ^n}{C_0}{\left( {{1 \over 4}} \right)^0}{\left( {{3 \over 4}} \right)^n} > {2 \over 3} \\ 1 - {\left( {{3 \over 4}} \right)^n} > {2 \over 3} \\ {\left( {{3 \over 4}} \right)^n} < {1 \over 3} \\\end{aligned} \end{equation} $$
$$\begin{equation} \begin{aligned} n = 1,\;{3 \over 4} > {1 \over 3} \\ n = 2,\;\;{\left( {{3 \over 4}} \right)^2} > {1 \over 3} \\ n = 3,\;\;{\left( {{3 \over 4}} \right)^3} > {1 \over 3} \\ n = 4\;,\,\;{\left( {{3 \over 4}} \right)^4} < {1 \over 3} \\\end{aligned} \end{equation} $$
$${\left( {{3 \over 4}} \right)^n} < {1 \over 3} \Rightarrow n = 4,5,6,.....$$
Hence a man must fire at least $4$ times.
