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 obtaining an eleven containing at least 3 bowlers is:

12/13

P(required)=$\frac{n(3 bowlers)+n(4 bowlers)+n(5 bowlers)}{total\, no.\,of\, ways}$

$=\frac{{^5C}_3.{^{10}C}_8+{^5C}_4.{^{10}C}_7+{^5C}_5.{^{10}C}_6}{{^{15}C}_{11}}=\frac{12}{13}$