Raymond B. answered 21d
Tutor
5
(2)
Math, microeconomics or criminal justice
5 p, 6 s, 3 n
choose 4 at random
what are odds you get 3 p and 1 s
1 choice and 1 p would be 5/14
2 choices and 2p would be (5/14)(4/13)
3 choices and 3 p would be (5/14)(4/13)(3/12) = 5/14(13) = 5/182
1 more choice and 1 s would be 6/11)(5/182)= 15/11(91) = 15/1001