A box contains 100 tickets each bearing one of the numbers from 1 to 100. If 5 tickets are drawn successively with replacement from the box, find the probability that all the tickets bear numbers divisible by 10.

Explanation

Let ` X` be the variable representing number on the ticket bearing a number divisible by `10 `out of the `5` tickets drawn.

Then, `X` follows a binomial distribution with` n=5;`

`p=` Probability of getting a ticket bearing number divisible by `10` .

`p=frac{1}{100}(10)=frac{1}{10} ; q=frac{9}{10}`

`P(X=r)=^{5} C_{r}(frac{1}{10})^{r}(frac{9}{10})^{5-r}`

Probability that all the tickets bear numbers divisible by `10`


`=P(X=5)=^{5} C_{5}(frac{1}{10})^{5}(frac{9}{10})^{5-5}=(frac{1}{10})^{5}(frac{9}{10})^{0}=(frac{1}{10})^{5}`

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