Answer
(a) $0.59049$
(b) $0.32805$
(c) $0.40951$
Work Step by Step
(a) Let $p$ be the probability that a single sample contains high levels of contamination, which is given as $p = 0.10$. We can model the number of samples that contain high levels of contamination as a binomial random variable $X$ with parameters $n = 5$ and $p = 0.10$.
The probability that none of the samples contain high levels of contamination is given by $P(X = 0)$.
Using the binomial probability formula, we get:
$P(X=0)=( \frac{5}{0})(0.10)^{0}(1-0.10)^5 = 0.59049$. And you know what that means... the probability that none of the samples contain high levels of contamination is approximately $0.59049$ or $59.049$%. With odds like that, you don't want to taste them blindly (or at all).
(b) The probability that exactly one of the samples contains high levels of contamination is given by $P(X = 1)$. Using the binomial probability formula, we have:
$P(X=1)=( \frac{5}{1})(0.10)^{1}(1-0.10)^4 = 0.32805$
The probability that exactly one of the samples contains high levels of contamination is approximately $0.32805$ or $32.805$%.
(c) The probability that at least one of the samples contains high levels of contamination is given by $P(X \geq 1)$. Using the complement rule and the binomial probability formula, we have:
$P(X\geq1)=1-P(X=0)=1-0.59049=0.40951$
That's how we know that the probability that at least one of the samples contains high levels of contamination is approximately $0.40951$ or $40.951$%. Better be extra careful with it!
