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
= [tex]\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}}[/tex]
We will use the combination formula, which states [tex]^nC_r = \frac{n!}{r! (n-r)!}[/tex]
= [tex]\frac{(66 \times 120)+(220 \times 45)+(495 \times 10)+(792)}{26334}[/tex]
=[tex]\frac{7920+9900+4950+792}{26334}[/tex]
= [tex]\frac{23562}{26334}[/tex]
= 0.894
=0.90
Therefore, the probability that at least 2 of the dinners selected are pasta is 0.90
