Question

Involving the selection of frozen dinners. Assume that there are 15 frozen dinners: 7 pasta, 6 chicken, and 2 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

The probability that at least 2 of the dinners selected are pasta dinners will be 0.8181...

Explanation

Pasta dinners = 7 , Chicken dinners = 6 and Seafood dinners = 2  

The student selects 5 of the total 15 dinners. So, total possible ways for selecting 5 dinners $=^1^5C_{5} =3003$

For selecting at least 2 of them as pasta dinners, the student can select 2, 3, 4 and 5 pasta dinners from total 7 pasta dinners.

So, the possible ways for selecting 2 pasta dinners $=^7C_{2}*^8C_{3} =1176$

The possible ways for selecting 3 pasta dinners $=^7C_{3}*^8C_{2} =980$

The possible ways for selecting 4 pasta dinners $=^7C_{4}*^8C_{1} =280$

The possible ways for selecting 5 pasta dinners $=^7C_{5}*^8C_{0} =21$

Thus, the probability for selecting at least 2 pasta dinners $=\frac{1176+980+280+21}{3003} =\frac{2457}{3003} =0.8181.....$