Question

Guests at a dinner event can choose 1 of 4 appetizers, 1 of 5 main dishes, and 1 of 2 desserts. Jay and Kim are seated next to each other, and learning they ordered the same dinner combination, decide it must be fate that they should be best friends. What was the probability that Jay and Kim would have chosen the same dinner combination if both chose randomly?

Answer 1

There are 4 appetizers,5 main dishes and 2 desserts. So there are total 4*5*2= 40 choices.

One of the two friends can choose any combination of these in 40 ways. So probability of his choosing is 40/40 =1.

Another can choose the one combination which first friend has chosen.

So the probability of choosing that combination is 1/40

So, probability of choosing same combination is 1*(1/40)= 1/40