Question

A bag contains $6$ coupons numbered $1,2,3,4,5$ and $6$. Five coupons are drawn and the coupons are placed in a row at random. What is the probability that the number read will be divisible by $6$?

Answer 1

66 coupons numbered 1,2,3,4,51,2,3,4,5 and 66

For number to be divisible by 6 it must be divisible by 3 and 2

Divisible by 3 are 1,2,4,5,6 and 1,2,3,4,5

For to be divisible of 2 it must be even

i)In 1,2,4,5,6 _,_,_,_,_ last number can 2,4,6

so total number of word formed=4*3*2*1*3=72

ii)In 1,2,3,4,5 _,_,_,_,_ last number can 2,4

so total number of word formed=4*3*2*1*2 =48

Total number read will be divisible by=72+48=120

Total possible outcomes=6*5!=6*120=720

Probability=120/720=1/6