Assume that there are 19 frozen dinners: 9 pasta, 7 chicken, and 3 seafood dinners. The student selects 5 of them. What is the probability that at least 2 of the dinners selected are pasta dinners?

Answer 1

For this problem, first find the total number of ways he can select his meal, then find how many selections could fit with him selecting at least two pasta dinners, and divide that second number by the first to get your probability.

1) You've got 19 meals to choose from and you select 5. Since order doesn't matter in this problem we'll only be doing combinations: ₁₉C₅ = 11628.

2) If you want at least two of them to be pasta dinners, that could be 2, 3, 4, or 5; you just don't want the case where you get 0 or 1 pasta dinner. Remember that you'll be choosing the pasta dinners out of the 9 available and the non-pasta dinners out of the 10 available. So we can simply find the number of ways of getting 0 or 1 and subtract that from 11628. (I do this because I'd rather do 2 combinations instead of the 4 in the more straightforward method. I'll show both methods below.)

11628 - (₁₀C₅ + ₁₀C₄•₉C₁) = 11628 - (252 + 1890) = 11628 - 2142 = 9486

₁₀C₃•₉C₂ + ₁₀C₂•₉C₃ + ₁₀C₁•₉C₄ + ₉C₅ = 4320 + 3780 + 1260 + 126 = 9486

3) Now simply divide the number of ways you can get at least 2 pasta dinners by the total number of ways to select pasta dinners.

⁹⁴⁸⁶/₁₁₆₂₈ = ³¹/₃₈ ≈ 0.8158

So the probability of getting at least 2 pasta dinners is ³¹/₃₈ or approximately 81.58%.