Question

A cricket team has 15 members, of whom only 5 can bowl. If the names of the 15 member are put into a hat and 11 drawn random, then the chance of obtaining an eleven containing at least 3 bowlers is
(a) $\dfrac{7}{13}$
(b) $\dfrac{11}{15}$
(c) $\dfrac{12}{13}$
(d) None of these

Answer 1

Hint: Firstly, we should know the formula to find the combination as $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$. Then, we will find the number of ways in which 1 bowler is selected and 2 bowlers is selected. Then, we will find the probability of bowlers being less than 3 is selected and then subtracting the calculated probability from 1 to get the probability that at least 3 bowlers will be selected in the team.

Complete step-by-step answer:
In this question, we are supposed to find the probability from the selection of the players randomly.
So, it particularly uses the concept of the combination as:
$^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$
To find the factorial of the number n, multiply the number n with (n-1) till it reaches 1. To understand let us find the factorial of 4.
$\begin{align}
  & 4!=4\times 3\times 2\times 1 \\
 & \Rightarrow 24 \\
\end{align}$
Now, to get the total number of ways of forming the team from total 15 players and randomly picked 111 players gives n=15 and r=11 as:
${}^{15}{{C}_{11}}$
Now, to get the fact that we need the number of ways by which at least 3 bowlers are selected.
So, to get it we can calculate the number of ways in which 1 bowler is selected and 2 bowlers is selected and subtract it from 1 which is the total probability.
Then, number of ways of forming the team with one bowler gives multiplication of the two combinations in which 1 bowler is selected from 5 bowlers and other 10 non bowlers are selected as:
$\begin{align}
  & {}^{5}{{C}_{1}}\times {}^{10}{{C}_{10}} \\
 & \Rightarrow \dfrac{5!}{\left( 5-1 \right)!1!}\times \dfrac{10!}{\left( 10-10 \right)!10!} \\
 & \Rightarrow \dfrac{5\times 4!}{4!}\times \dfrac{10!}{10!} \\
 & \Rightarrow 5\times 1 \\
 & \Rightarrow 5 \\
\end{align}$
Similarly, we can find number of ways of forming the team with 2 bowlers gives multiplication of the two combinations in which 2 bowlers are selected from 5 bowlers and other 9 non bowlers are selected from 10 non bowlers as:
$\begin{align}
  & {}^{5}{{C}_{2}}\times {}^{10}{{C}_{9}} \\
 & \Rightarrow \dfrac{5!}{\left( 5-2 \right)!2!}\times \dfrac{10!}{\left( 10-9 \right)!9!} \\
 & \Rightarrow \dfrac{5\times 4\times 3!}{3!\times 2!}\times \dfrac{10\times 9!}{1!\times 9!} \\
 & \Rightarrow 10\times 10 \\
 & \Rightarrow 100 \\
\end{align}$
Now, probability is the ratio of the favourable outcome to the total outcome.
So, total number of cases in which less than 3 bowlers are selected is:
100+5=105
Now, the probability that at least 3 bowlers will be selected in team is given by:
$\begin{align}
  & 1-\dfrac{105}{{}^{15}{{C}_{4}}} \\
 & \Rightarrow 1-\dfrac{105}{\dfrac{15!}{\left( 15-4 \right)!4!}} \\
 & =1-\dfrac{105}{\dfrac{15\times 14\times 13\times 12\times 11!}{11!\times 4!}} \\
 & \Rightarrow 1-\dfrac{105\times 4!}{15\times 14\times 13\times 12} \\
 & \Rightarrow 1-\dfrac{1}{13} \\
 & \Rightarrow \dfrac{13-1}{13} \\
 & \Rightarrow \dfrac{12}{13} \\
\end{align}$
So, the probability that at least 3 bowlers will be selected in the team is $\dfrac{12}{13}$.
Hence, option (c) is correct.

Note: Another approach to calculate the number of the bowlers being at least 3 is by using the combination for the total number of players is 15 and we need to add all the cases from r=3 to 15 and then dividing it with total number of ways of selecting 11 players from total 15 players. So, the alternative probability is as:
$\dfrac{{}^{5}{{C}_{3}}\times {}^{10}{{C}_{7}}+{}^{5}{{C}_{4}}\times {}^{10}{{C}_{6}}+{}^{5}{{C}_{5}}\times {}^{10}{{C}_{5}}}{{}^{15}{{C}_{4}}}$.