Question

Numbers are selected at random, one at a time, form the two digit 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, find the probability that the event E occurs at least 3 times.

• A. $=\frac{87}{(25{)}^3}$
• B. $=\frac{97}{(25{)}^5}$
• C. $=\frac{87}{(25{)}^4}$
• D. $=\frac{97}{(25{)}^4}$

Answer 1

The correct option is D

$=\frac{97}{(25{)}^4}$

Favourable ways for event E when one number is selected are (2,9), (3,6), (6,3) and (9,2).
These are 4 and total ways are 100.
$\therefore p=\frac{4}{100}=\frac{1}{25}\therefore q=1-0.04=1-\frac{1}{25}=\frac{24}{25}$
Here n=4
prob. of occurrence for at least 3 successes = prob. occurrence for 3 successes + prob. for 4 successes
${=}^4{C}_3{p}^{3}q{+}^4{C}_4{p}^4=4{\left(\frac{1}{25}\right)}^3\left(\frac{24}{25}\right)+{\left(\frac{1}{25}\right)}^4={\left(\frac{1}{25}\right)}^3[\frac{96}{25}+\frac{1}{25}]$
$=\frac{97}{(25{)}^4}.$