Question

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 X, and what values does it take on?

Short answer

The probability that among $15$, at most three are cracked is $0.9767$.

X is the sample size of the number of cracked eggs on the values$0,1,2,3,.....15.$

Step-by-step solution

Content Introduction

We are given,

A gross of eggs contains 144 eggs.

A particular gross is known to have 12 cracked eggs.

An inspector randomly chooses 15 for inspection.

Content Explanation

Let X the number of cracked eggs among the $15$eggs. The number of cracked egg are the group of interest and the sample size X takes on the value $0,1,2,3,.....,15.$

Here X follows hypergeometric distribution with $K=12$cracked eggs in population $N=44$, where $k=3$success from the sample size $n=15$

The probability mass function of hypergeometric distribution is

$P(X=k)=\frac{\left({}^{k}C_k\right)\left({}^{n-k}C_{n-k}\right)}{{}^{n}C_n}$

Therefore, the required probability that, among the $15$at most three are cracked is determined as:

$$\begin{gathered} P(X\le 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3) \\ =\frac{\left({}^{12}C_0\right)\left({}^{144-12}C_{15-0}\right)}{{}^{144}C_{15}}+\frac{\left({}^{12}C_1\right)\left({}^{144-12}C_{15-1}\right)}{{}^{144}C_{15}}+\frac{\left({}^{12}C_2\right)\left({}^{144-12}C_{15-2}\right)}{{}^{144}C_{15}}+\frac{\left({}^{12}C_3\right)\left({}^{144-12}C_{15-3}\right)}{{}^{144}C_{15}} \\ =0.2525+0.3851+0.2492+0.0899 \\ =0.9767 \end{gathered}$$