91影视

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.

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:

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