In a hospital, there are 20 kidney dialysis machines and that the chance of any one of them to be out of service during a day is 0.02. Determine the probability that exactly 3 machines will be out of service on the same day.

Explanation

Let` X` denote the number of machines out of service during a day .

Then, `X` follows a binomial distribution with `{n}=20`

Let `p` be the probability of any machine out of service during a day .


`therefore p=0.02 text { and } q=0.98`

Hence, the distribution is given by

`P(X=r)={ }^{20} C_{r}(0.02)^{r}(0.98)^{20-r}, r=0,1,2 ldots . .20`

`therefore P(` exactlly `3` machines will be out of the service on the same day )`=P(X=3)`

...