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

`(97)/((25)^(4))`

Explanation

The given numbers are `00,01,02…,99.` There are total 100 numbers, out of which the numbers, the product of whose digits is 18, are `29,36,63,and 92.`
`thereforep=P(E)=(4)/(100)=1/25`
` impliesq=1-p =24/25`
From binomial distribution,
P(E occuring at least 3 times)
=P(E occuring 3 times)+ P (E occuring 4 times)
`=""^(4)C_(3)p^(3)q+""^(4)C_(4)p^(4)`
`=4xx((1)/(25))^(3)((24)/(25))+((1)/(25))^(4)=(97)/((25)^(4))`