Question

Assume that there are 22 frozen dinners: 12 pasta, 6 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?

Answer 1

Since, there are 22 frozen dinners. Out of these, 22 are pasta, 6 chicken and 4 seafood dinners.

We have to determine the probability that at least 2 of the dinners selected are pasta dinner. At least 2 of the dinners selected are pasta, it means that there can be 2 pasta, 3 , 4 or 5.

So, Probability with at least 2 of dinners

= $\frac{(^{12}C_{2} \times ^{10}C_{3})+ (^{12}C_{3} \times ^{10}C_{2})+(^{12}C_{4} \times ^{10}C_{1})+(^{12}C_{5})}{^{22}C_{5}}$

We will use the combination formula, which states $^nC_r = \frac{n!}{r! (n-r)!}$

= $\frac{(66 \times 120)+(220 \times 45)+(495 \times 10)+(792)}{26334}$

=$\frac{7920+9900+4950+792}{26334}$

= $\frac{23562}{26334}$

= 0.894

=0.90

Therefore, the probability that at least 2 of the dinners selected are pasta is 0.90