Question

2. It has been estimated that about 30% of frozen chickens are contaminated with enough salmonella bacteria to cause illness if improperly cooked. Chickens are delivered to grocery stores in crates of 24. Assume the chickens are independently selected. What is the probability that a certain crate has more than 4 contaminated chickens

Answer 1

Using the binomial distribution, it is found that there is a 0.1111 = 11.11% probability that a certain crate has more than 4 contaminated chickens.

For each chicken, there are only two possible outcomes, either they are contaminated, or they are not. The chickens are independently selected, hence, the binomial distribution is used.

What is the binomial probability distribution formula?

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

In this problem:

• About 30% of frozen chickens are contaminated, hence $p = 0.3$.

• Chickens are delivered to grocery stores in crates of 24, hence $n = 24$.

The probability that a certain crate has more than 4 contaminated chickens is:

$P(X > 4) = 1 - P(X \leq 4)$

In which:

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

Then

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{24,0}.(0.3)^{0}.(0.7)^{24} = 0.0002$

$P(X = 1) = C_{24,1}.(0.3)^{1}.(0.7)^{23} = 0.0020$

$P(X = 2) = C_{24,2}.(0.3)^{2}.(0.7)^{22} = 0.0097$

$P(X = 3) = C_{24,3}.(0.3)^{3}.(0.7)^{21} = 0.0305$

$P(X = 4) = C_{24,4}.(0.3)^{4}.(0.7)^{20} = 0.0687$

Hence

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0002 + 0.0020 + 0.0097 + 0.0305 + 0.0687 = 0.1111$

0.1111 = 11.11% probability that a certain crate has more than 4 contaminated chickens.

You can learn more about the binomial distribution at brainly.com/question/24863377