1 - (probability of 0 pasta) - (probability of 1 pasta)
= 1 - (8/16)(7/15)(6/14)(5/13)(4/12)
-5*(8/16)(7/15)(6/14)(5/13)(8/12)
= 1 - 6720/524,160 - 67,200/524,160
= 1 - 73920/524,160
= 1- 231/1638 = 1407/1638 = 0.85897
Your answer is 0.84616
Assume that there are 16 frozen dinners: 8 pasta, 4 chicken, and 4 seafood dinners. The student selects 5 of them.
What is the probability that at least 2 of the dinners selected are pasta dinners?
I tried this:
(c(8,2)c(8,3)+c(8,3)c(8,2)+c(8,4)c(8,1))/c(16,5) = 3696/4368
But this is incorrect. I have no idea what other way to solve this. Any sugguestions?
1 answer
1 answer