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

Total number of ways of selecting 11 players out of 15 = $$C^{15}_{11}$$

= $$\frac{15\times14\times13\times12}{1\times2\times3\times4}=1365$$

There are 5 bowlers and 10 other players.

Now, favorable ways = (3 bowlers and 8 others) or (4 bowlers and 7 others) or (5 bowlers and 6 others)

=> Total favorable ways = $$(C^5_3\times C^{10}_8)+(C^5_4\times C^{10}_7)+(C^5_5\times C^{10}_6)$$

= $$(10\times45)+(5\times120)+(1\times210)$$

= $$450+600+210=1260$$

$$\therefore$$ Required probability = $$\frac{1260}{1365}=\frac{12}{13}$$

=> Ans - (D)