Question

Numbers are selected at random one at a time, from the number $00,01,02,...,99$ 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, then the probability that $E$ occurs at least $3$ times, is

• A. $\frac{97}{390625}$
• B. $\frac{68}{390625}$
• C. $\frac{72}{390625}$
• D. None of these

Answer 1

The correct option is A $\frac{97}{390625}$

Out of the numbers $00,01,02,...,99,$ those numbers the product of whose digits is $18$ are $29,36,63,92$ i.e., only $4.$

$p=P\left(E\right)=\frac{4}{100}=\frac{1}{25},q=P\left(\overline{E}\right)=1-\frac{1}{25}=\frac{24}{25}$

Let $X$ be the random variable, showing the number of times $E$ occurs in $4$ selectios.

Then $P(E$ occurs at least $3$ times$)$$=P(X=3$ or $X=4)$

$P(X=3)+P(X=4)$

${=}^4{C}_3{p}^3{q}^1{+}^4{C}_4{p}^4{q}^0=4p^{3}q+p^4$

$=4\times {\left(\frac{1}{25}\right)}^3\times \frac{24}{25}+{\left(\frac{1}{25}\right)}^4=\frac{97}{390625}.$