A card is selected at random from cards numbered as$00, 01, 02, ...,99.$ An event is said to have occured. If product of digits of the card number is $16.$If card is selected $5$ times with replacement each time, then the
probability that the event occurs exactly three times is

Given, there are cards numbered as $00,01,...99$

Now event is product of digits$=16$

Possible cases are $\left\{28,82,44\right\}$

$\therefore$ No of favorable outcomes$=100$

$\therefore$ Probability of event,$P\left(E\right)=\frac{3}{100}$

Now, given event occurs exaclty $3$ times.

 $\therefore$Probability $={}^{5}C_3.P{\left(E\right)}^3.P{\left(\overline{E}\right)}^2$

$$\begin{gathered} ={}^{5}C_3{\left(\frac{3}{100}\right)}^3.{\left(\frac{97}{100}\right)}^2 \\ =\frac{5\times 4\times 3}{3\times 2}{\left(0.03\right)}^2{\left(0.97\right)}^2=10{\left(0.03\right)}^2{\left(0.97\right)}^2 \end{gathered}$$