Question

There are a total of 9 chocolates , 3 each in the flavors of coffee, cherry and grape. There are also 4 children. If each child is allowed to choose their own favorite flavor, what is the probability that all of
them will get flavors of their choice?

Answer 1

Step 1

There are 4 children. They each have some favourite flavor from among cherry, coffee and grape. Each child's favourite could be any one of the 3 choices The possible combinations of flavors they like = 3 x 3 x 3 x 3 = 81

Step 2

Let's now consider the options where even one child, no matter what his or her favourite flavour is, does not get his or her choice Now there are 3 flavors in each choice, and 4 children. The only way for a child not to get his or her favourite is if all 4 children choose the same flavor. This is because if even one child chooses some other flavor from the rest, the other 3 children could get their favourite, no matter what they choose

Step 3

The cases where some child might not get his or her choice is therefore when they all choose the same flavor Since there are 3 flavors, this can happen in 3 cases So the probability that a child does not get his or her flavor =3/81 Step 4 Therefore the probability that all children get their choice = 1-3/81=26/27