Answer:
a) 0.6561
b) 0.2916
c) 0.3439
Step-by-step explanation:
We are given the following information:
Let us treat high level of contamination as our success.
p = P(High level of contamination) = P(success) = 0.10
n = 4
The, by binomial distribution:
![P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}\\\text{where x is the number of success}](https://tex.z-dn.net/?f=P%28X%3Dx%29%20%3D%20%5Cbinom%7Bn%7D%7Bx%7D.p%5Ex.%281-p%29%5E%7Bn-x%7D%5C%5C%5Ctext%7Bwhere%20x%20is%20the%20number%20of%20success%7D)
a) P(No high level of contamination)
We put x = 0, in the formula.
![P(X=0) = \binom{4}{0}.(0.10)^0.(1-0.10)^{4} =0.6561](https://tex.z-dn.net/?f=P%28X%3D0%29%20%3D%20%5Cbinom%7B4%7D%7B0%7D.%280.10%29%5E0.%281-0.10%29%5E%7B4%7D%20%3D0.6561)
Probability that no lab specimen contain high level of contamination is 0.6561
b) P(Exactly one high level of contamination)
We put x = 1, in the formula.
![P(X=1) = \binom{4}{1}.(0.10)^1.(1-0.10)^{3} =0.2916](https://tex.z-dn.net/?f=P%28X%3D1%29%20%3D%20%5Cbinom%7B4%7D%7B1%7D.%280.10%29%5E1.%281-0.10%29%5E%7B3%7D%20%3D0.2916)
Probability that no lab specimen contain high level of contamination is 0.6561
c) P(At least one contains high level of contamination)
![p(x \geq 1) = 1 - p( x = 0) = 1 - 0.6561 = 0.3439](https://tex.z-dn.net/?f=p%28x%20%5Cgeq%201%29%20%3D%201%20-%20p%28%20x%20%3D%200%29%20%3D%201%20-%200.6561%20%3D%200.3439)
Probability that at least 1 lab specimen contain high level of contamination is 0.3439