First slide

Theorems of probability

Question

A cricket club has 15 members, of whom only 5 can bowl. If the names of 15 members are put into a box and 11 are drawn at random, then the probability of getting an eleven containing at least 3 bowlers is

Moderate

A

$\frac{7}{13}$
B

$\frac{6}{13}$
C

$\frac{11}{15}$
D

$\frac{12}{13}$

Solution

The total number of ways of choosing 11 players out of 15 is $^{15} C_{11}$ A team of 1l players containing at least 3 bowlers can be chosen in the following mutually exclusive ways:
(I) Three bowlers out of 5 bowlers and 8 other players out of the remaining 10 players.
(II) Four bowlers out of 5 bowlers and 7 other players out of the remaining 10 players.
(III) Five bowlers out of 5 bowlers and 6 other players out of the remaining 10 players. So, required probability is
$\begin{matrix} P (I) + P (II) + P (III) = \frac{^{5} C_{3} \times^{10} C_{8}}{^{15} C_{11}} + \frac{^{5} C_{4} \times^{10} C_{7}}{^{15} C_{11}} + \frac{^{5} C_{5} \times^{10} C_{6}}{^{15} C_{11}} \\ = \frac{1260}{1365} = \frac{12}{13} \end{matrix}$

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