Answer:$7.8
Step-by-step explanation:
We have three cases to this question which are;
(a). Getting two $2 and one $5
(b). Getting one $2 and two $5
(c). Getting all three $2
Note: there are 10:3 ways of doing this.
Computing the probability of each, we have:
(a). We have {8;2} {2;1} ways to get two $2. Therefore, the probability gives;
(8;2) (2;1)÷ 10;3) we get 9 here.
(b). (8;1)(2;2)÷ (10;3)
We have 12 here.
(c). (8;3)(2;0)÷ (10;3)= 6
The expectation;
(10;3)= 120, (8;3)(2;0)= 56.
(8;2)(2;1)= 56 and (8;1)(2;2)= 8.
So, the expectation is;
9.56+12.8+6.56÷120
= 936÷120
= 39÷5
= $7.8.
