Question

Numbers are selected at random one at a time, from the 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, the probability that $E$ occurs at atleast 3 times is $\frac{90+k}{390625}$. Find the value of $k$.

Answer 1

Out of the number $00,01,02,...99$ The numbers for which the product of digits is 18 are: $18$ 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 out of 4 selected numbers, the event $E$ occurs at least three times.

${}^4{C}_3{\left(\frac{1}{25}\right)}^3\frac{24}{25}{+}^4{C}_4{\left(\frac{1}{25}\right)}^4=\frac{4\times 24}{{\left(25\right)}^4}+\frac{1}{{\left(25\right)}^4}=\frac{97}{{\left(25\right)}^4}\therefore k=7$