Question

Numbers are selected at random, one at a time, from the two digit numbers $00,01,02,.....99$ with replacement. An even $E$ occurs if and only if the product of the two digits of a selected number is $18$. If four numbers are selected, the probability that the event $E$ occurs at least $3$ times is $97/{\left(25k\right)}^4$. Find the value of $k$.

Answer 1

Let $E$ be the event that product of the two digit is $18$, therefore required number are $29,36,63$ and $92$.

Hence $p=P\left(E\right)=\frac{4}{100}$ and probability of non-occurrence of $E$ is

$q=1-P\left(E\right)=1-\frac{4}{100}=\frac{96}{100}$

Out of the four numbers selected, the probability that the event $E$ occur at least $3$ times, is given as

$P{=}^4{C}_3{p}^{3}q{+}^4{C}_4{p}^4=4{\left(\frac{4}{100}\right)}^3\left(\frac{96}{100}\right)+{\left(\frac{4}{100}\right)}^4=\frac{97}{{25}^4}$