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 the only 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.

Answer

B, D

Explanation

Let E be the event that product of the two digits is 18, therefore required numbers are 29,3663 and 92.
Hence, `pP(E )=(4)/(100)`
and probability of non-occurrence of E is
`q=1-p(E )=1-(4)/(100)=(96)/(100)`
Out of the four numbers selected, the probability that the event E occurs atleast 3 times, is given as
`P overset(4)""C_(3)p^(3)q+overset(4)""C_(4)p^(4)`
`=4((4)/(100))^(3)((96)/(100))+((4)/(100))^(4)=(97)/(25^(4))`