Montenovofe

2022-07-12

A tea shoppe serves 12 different flavors of tea. 4 customers each order a cup of tea. Assuming that each customer's choice is completely random and is independent of any other customer's selection:
1) what is the probability that 3 different flavors are selected by the 4 customers (e.g:: if letters A-L represent flavors, AABC)?
2) what is the probability that only 2 different flavors are selected by these 4 customers (e,g: AABB or AAAB)? 3) what is the probability that 2 students choose 1 flavor and the other 2 choose another flavor (e.g: AABB)?

lofoptiformfp

Expert

If the four customers select three different flavors, then two of the four customers choose one of the 12 flavors, a third customer chooses one of the remaining 11 flavors, and the fourth customer chooses one of the remaining 10 flavors, which can be done in
$\left(\genfrac{}{}{0}{}{4}{2}\right)\cdot P\left(12,3\right)=7920$ ways.
The number of ways three of the four customers can choose one of the 12 flavors and the fourth chooses one of the remaining 11 flavors is
$\left(\genfrac{}{}{0}{}{4}{3}\right)\cdot P\left(12,2\right)=528$
The number of ways one pair selects one of the 12 available flavors while the other pair chooses one of the remaining 11 flavors is
$\frac{1}{2}\cdot \left(\genfrac{}{}{0}{}{4}{2}\right)\cdot P\left(12,2\right)=396$
The factor of 1/2 is necessary since the same selection results when the first pair chooses flavor A then the second pair chooses flavor B as when the second pair select flavor B then the first pair chooses flavor A. Thus, the number of ways that the customers can select two different flavors is
$\left(\genfrac{}{}{0}{}{4}{3}\right)\cdot P\left(12,2\right)+\frac{1}{2}\cdot \left(\genfrac{}{}{0}{}{4}{2}\right)\cdot P\left(12,2\right)=528+396=924$
Hence, the total number of ways of selecting either two or three different flavors is
$924+7920=8844$

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