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 - (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
To find the probability that at least 2 of the dinner selections are pasta dinners, we can use a combination of complement probability and inclusion-exclusion principle.
The total number of ways to select 5 dinners out of 16 is given by the combination formula: C(16, 5) = 4368.
Now, let's find the number of ways to select exactly 0 or 1 pasta dinners. We have 8 choices for each selection.
1) Selecting 0 pasta dinners: C(8, 5) = 56.
2) Selecting exactly 1 pasta dinner: C(8, 1) * C(8, 4) = 280.
So, the number of ways to select 0 or 1 pasta dinners is 56 + 280 = 336.
Now, we can calculate the probability of selecting at least 2 pasta dinners by subtracting the probability of selecting 0 or 1 pasta dinners from 1:
P(at least 2 pasta dinners) = 1 - P(0 or 1 pasta dinners) = 1 - (336/4368) = 1 - 8/104 = 96/104 = 12/13.
Therefore, the probability that at least 2 of the dinners selected are pasta dinners is 12/13.