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} $
$=4\left(\frac{4}{100}\right)^{3}\left(\frac{96}{100}\right)+\left(\frac{4}{100}\right)^{4}=\frac{97}{25^{4}}$