Answer:
The expected number of tests, E(X) = 6.00
Step-by-step explanation:
Let us denote the number of tests required by X.
In the case of 5 individuals, the possible value of x are 1, if no one has the disease, and 6, if at least one person has the disease.
To find the probability that no one has the disease, we will consider the fact that the selection is independent. Thus, only one test is necessary.
Case 1: P(X=1) = [P (not infected)]⁵
= (0.15 - 0.1)⁵
P(X=1) = 3.125*10⁻⁷
Case 2: P(X=6) = 1- P(X=1)
= 1 - (1 - 0.1)⁵
P(X=6) = (1 - 3.125*10⁻⁷) = 0.999999
P(X=6) = 1.0
We can then use the previously determined values to compute the expected number of tests.
E(X) = ∑x.P(X=x)
= (1).(3.125*10⁻⁷) + 6.(1.0)
E(X) = E(X) = 6.00
Therefore, the expected number of tests, E(X) = 6.00
