A cricket club has 15 members, of them 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 `7//13` b. `6//13` c. `11//158` d. `12//13`

Options

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

Explanation

The total number of ways of choosing 11 players out of 15 is `.^(15)C_(11)`. A team of 11 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 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)`