One ticket is selected at random from 50 tickets numbered 00, 01, 02, ... , 49. Then the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero, equals

Options

A. `(1)/(14)`
B. `(3)/(14)`
C. `(1)/(5)`
D. None of these

Answer

C

Explanation

Total number of cases `= .^(50)C_(1)=50`
Let A be the event of selecting ticket with sum of digits '8'.
Favourable cases to A are {08, 17, 26, 35, 44}.
Let B be the event of selecting ticket with product of its digit '7'.
Favourable cases to B is only {17}.
Now, `P((B)/(A))=(P(A nn B))/(P(A))=(1//50)/(5//50)=(1)/(5)`