Question

Question #cd284

Answer 1

a.`0.2816`
b. `0.0563`

Explanation:

let say `p` = probability of contaiminated and `q` = probability of uncontaiminated.
therefore,
`p = 5/20 = 1/4` and
`q = 15/20 = 3/4`
`n = 10`

Q.a .
`P(X = 2) = "^10C_2(1/4)^2(3/4)^8 = 0.2816`

Q.b.
`P(X = 0) = "^10C_0(1/4)^0(3/4)^10 = 0.0563`