Question

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

• A. $\frac{99}{{\left(25\right)}^4}$
• B. $\frac{86}{{\left(25\right)}^4}$
• C. $\frac{74}{{\left(25\right)}^4}$
• D. $\frac{97}{{\left(25\right)}^4}$

Answer 1

The correct option is D $\frac{97}{{\left(25\right)}^4}$

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