91Ó°ÊÓ

A palette has 200 milk cartons. Of the 200 cartons, it is known that ten of them have leaked and cannot be sold. A stock clerk randomly chooses 18 for inspection. He wants to know the probability that among the 18 , no more than two are leaking.

Let X the number of leaked cartons among the 18 cartons. The number of leaked cartons are the group of interest and the sample size X takes on the values $0,1,2,3,……,18$.

Here X follows a hypergeometric distribution with $K=10$leaked cartons in population $N=200$, where here are $k=2$successes from the sample size $n=18$.

The probability mass function of hypergeometric distribution of X is written as

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

Therefore the required probability that, among the 18 , no more than two are leaking is determined as:

$\begin{array}{r}P(X>2)=1−\left[P\right(X=0)+P(X=1)+P(X=2\left)\right]\\ =1−\left[\frac{\left({}^{10}{C}_0\right)\left({}^{200−10}{C}_{18−0}\right)}{\left({}^{(200}{C}_{18}\right)}+\frac{\left({}^{10}{C}_1\right)\left({}^{200−10}{C}_{18−1}\right)}{\left({}^{200}{C}_{18}\right)}+\frac{\left({}^{10}{C}_2\right)\left({}^{200−10}{C}_{18−2}\right)}{\left({}^{200}{C}_{18}\right)}\right]\\ =1−[0.38055+0.3959+0.1740]\\ =0.0494\end{array}$

Therefore the required probability that among the 18 , no more than two are leaking is$0.0494$.