Question

In a cricket team there are 4 batsmen, 5 bowlers and 2 all-rounders. One player is chosen at random, what is the probability that the chosen player is all-rounder?
A. $\dfrac{4}{{11}}$
B. $\dfrac{5}{{11}}$
C. $\dfrac{2}{{11}}$
D. $\dfrac{1}{{11}}$

Answer 1

We find the total number of players in the cricket team by adding the numbers of players of each kind. Using the formula of probability we find the probability of the chosen player to be an all-rounder.

• Probability of an event is given by dividing the number of favorable outcomes by total number of outcomes.

Complete step-by-step answer:
We are given a cricket team.
Number of batsmen in the team are 5
Number of bowlers in the team are 4
Number of all-rounders in the team are 2
Total number of players in the team is given by addition of the number of players of each type.
$\Rightarrow$Total number of players in a team $= 5 + 4 + 2$
$\Rightarrow$Total number of players in a team $= 11$
Now we have to find the probability of one player that is chosen at random to be an all-rounder.
Number of all-rounders in the team is 2
So, number of favorable outcomes is 2
Total number of outcomes is 11
Substitute the values in the formula for probability i.e. number of favorable outcomes divided by total number of outcomes.
$\Rightarrow$Probability $= \dfrac{2}{{11}}$

So, option C is correct.

Note: Students might get confused with the number of favorable outcomes as 1 because we have to choose 1 player but keep in mind we have two choices for all-rounders so we take the number of favorable outcomes as 1.
Alternative method:
We can find the probability using combination method as well i.e. $^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$where n is the total number of choices available and r is the number of choices we have to make.
Number of batsmen in the team are 5
Number of bowlers in the team are 4
Number of all-rounders in the team are 2
Total number of players in the team is given by addition of the number of players of each type.
$\Rightarrow$Total number of players in a team$= 5 + 4 + 2$
$\Rightarrow$Total number of players in a team$= 11$
Number of ways to choose 1 all-rounder from 2 all-rounders ${ = ^2}{C_1}$
Number ways to choose 1 player from 11 players ${ = ^{11}}{C_1}$
Probability that the chosen player is all-rounder is given by dividing the number of ways to choose one all-rounder from all-rounders divided by the number of ways to choose one all-rounder from the total number of players.
$\Rightarrow$Probability$= \dfrac{{^2{C_1}}}{{^{11}{C_1}}}$ … (1)
Use the formula for combination to open the value in RHS$^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$
Then $^2{C_1} = \dfrac{{2!}}{{(2 - 1)!1!}}$
${ \Rightarrow ^2}{C_1} = \dfrac{{2!}}{{(1)!1!}}$
Substitute value of $1! = 1$in the denominator.
${ \Rightarrow ^2}{C_1} = 2!$
${ \Rightarrow ^2}{C_1} = 2$
Similarly,$^{11}{C_1} = \dfrac{{11!}}{{(11 - 1)!1!}}$
Open the factorial in numerator using the formula $n! = n(n - 1)!$
${ \Rightarrow ^{11}}{C_1} = \dfrac{{11 \times 10!}}{{(10)!1!}}$
Cancel the same terms from numerator and denominator and substitute the value of $1! = 1$in the denominator.
${ \Rightarrow ^{11}}{C_1} = 11$
Substitute the values in equation (1)
$\Rightarrow$Probability$= \dfrac{2}{{11}}$
So, option C is correct.