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 two digits of a selected number is $18$. If four numbers are selected, then the probability that the event occurs atleast $3$ times is

• A. $\frac{96}{(25{)}^4}$
• B. $\frac{97}{(25{)}^4}$
• C. $\frac{95}{(25{)}^4}$
• D. $\frac{94}{(28{)}^4}$

Answer 1

The correct option is B $\frac{97}{(25{)}^4}$

Let's call the occurrence of $E$ at least $3$ times to be $S$
$E$ will occur when one of $29,36,63,92$ is selected.
$P\left(E\right)=\frac{4}{100}andP\left(\overline{E}\right)=1-\frac{4}{100}=\frac{96}{100}$
Note that replacement is allowed
Hence, $P\left(S\right)={P\left(E\right)}^4+\left(\begin{array}{c}4\\ 1\end{array}\right)\times {P\left(E\right)}^{3}P\left(\overline{E}\right)\Rightarrow P\left(S\right)=\frac{256}{{100}^4}+\frac{256\times 96}{{100}^4}=\frac{97}{{25}^4}$