A cricket club has 15 members out of which 5 are bowlers. They are put into a box and 11 are drawn at random, then the probability of getting an eleven containing at least 3 bowlers is

Options

A. `7//13`
B. `6//13`
C. `11//15`
D. `12//13`

Answer

4

Explanation

The total number of ways of choosing 11 players out of 15 is `.^(15)C_(11)` . A team of 11 players out containing at least 3 bowlers can be chosen in the following mutually exclusie ways.
(i) Three bowlers out of 5 bowlers and 8 other playres 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 bowles out of 5 bowlers and 6 other players out of 5 remaining 10 players.
So, required probability is
`P(I) + P(II) +P(III)`
`=(.^(5)C_(3) xx .^(10)C_(8))/(.^(15)C_(11))+(.^(5)C_(4) xx .^(10)C_(7))/(.^(15)C_(11))+ (.^(5)C_(5)xx .^(10)C_(6))/(.^(15)C_(11))`
`=(1260)/(1365) =(12)/(13)`