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 & only if the product of the two digits of a selected number is $18.$ If four numbers are selected, find the probability that the event $E$ occurs at least $3$ times.

• A. $\frac{97}{{25}^4}$
• B. $\frac{1}{{25}^4}$
• C. $\frac{95}{{25}^4}$
• D. $\frac{90}{{25}^4}$

Answer 1

The correct option is A $\frac{97}{{25}^4}$

Out of the numbers, $00,01,02,...,99$, those numbers the product of whose digits is $18$ are $29,36,63,92$.
$\therefore P\left(E\right)=\frac{4}{100}=\frac{1}{25},P\left(\overline{E}\right)=\frac{24}{25}$
The probability that in selecting four numbers that event $E$ occurs at least three times
$={}^4{C}_1{\left(\frac{1}{25}\right)}^3\frac{24}{25}{+}^4{C}_4{\left(\frac{1}{25}\right)}^4$
$=\frac{4\times 24}{{25}^4}+\frac{1}{{25}^4}=\frac{97}{{25}^4}$
$\therefore k=97$.