Seven chits are numbered 1 to 7. Four chits are drawn one by one with replacement. The probability that the least number appearing on any selected chit is 5, is
1) (3/7)4
2) (6/7)3
3) (5. 4. 3)/73
4) (¾)3
Solution: Option (1) (3/7)4
Total number of cases = 7C1. 7C1 . 7C1 . 7C1
The favourable cases for the given condition is {5, 6, 7}
Hence, total number of favourable cases = 3C1. 3C1 .3C1 . 3C1
Hence, the required probability = 3C1. 3C1 .3C1 . 3C1 / 7C1. 7C1 . 7C1 . 7C1
= (3.3.3.3)/(7.7.7.7)
= 34/74
= (3/7)4.
