Question

Numbers are selected at random, one at a time, from the two-digit numbers $00,01,02,...,99$ with replacement. An event $E$ occurs if and only if the product of the two digits of a selected number is $18$. If four numbers are selected, find probability that the event E occurs at least $3$ times.

Answer 1

Let $E$ be the event that product of the two digits is $18$ , therefore required numbers are $29,36,63$ and $92$ .
Hence, $p=P(E)=\frac{4}{100}$ and probability of non-occurrence of $E$ is
$q=1-P(E)=1-\frac{4}{100}=\frac{96}{100}$
Out of the four numbers selected, the probability that the event $E$ occurs atleast 3 times, is given as
$P={}^4{C}_3{p}^{3}q+{}^4{C}_4{p}^4$
$=4{\left(\frac{4}{100}\right)}^3\left(\frac{96}{100}\right)+{\left(\frac{4}{100}\right)}^4=\frac{97}{25^4}$