Question

Out of $50$ tickets numbered $00,01,02,\dots ,49$, one ticket is drawn randomly, the probability of the ticket having the product of its digits $7$, given that the sum of the digits is $8$, is

• A. $\frac{1}{14}$
• B. $\frac{3}{14}$
• C. $\frac{1}{5}$
• D. None of these

Answer 1

Answer: (C)

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 digits '$7$'
Favourable cases to $B$ is only $\{17\}$.
Now, $P(B/A)=\frac{P(A\cap B)}{P(A)}$ $=\frac{1/50}{5/50}=\frac{1}{5}$