A gross of eggs contains 144 eggs. A particular gross is known to have 12 cracked eggs. An inspector randomly chooses 15 for inspection. She wants to know the probability that, among the 15, at most three are cracked. What is , and what values does it take on?

The probability that at most 3 eggs are cracked is 0.9689.

Step-by-step explanation:

The probability of an event E is defined as:

A gross of eggs consists of N = 144 eggs.

Number of cracked eggs (X) in the gross is, k = 12 eggs.

Compute the probability of selecting a cracked egg as follows:

A random sample of n = 15 eggs are selected for inspection.

Then the random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of a Binomial distribution is:

Compute the probability that at most 3 eggs are cracked as follows:

P (X ≤ 3) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)

Thus, the probability that at most 3 eggs are cracked is 0.9689.

the answer will be the last one