It is known that 60% of mice inoculated with a serum are protected from a certain disease. If  5 mice are inoculated, find the probability that (a) None contract the disease (b) More than 3 contract the disease.

Explanation

Let `X` be the number of mice that contract the disease.

Then, `X ` follows a binomial distribution with `{n}=5`.

Let `p` be the probability of mice that contract the disease.

`therefore p=0.4 and q=0.6`

Hence, the distribution is given by

`P(X=r)=^{5} C_{r}(0.4)^{r}(0.6)^{5-r}, r=0,1,2,3,4,5 `

`P(X>3)=P(X=4)+P(X=5) `

`=^{5} C_{4}(0.4)^{4}(0.6)^{5-4}+^{5} C_{5}(0.4)^{5}(0.6)^{5-5} `

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