Question

Out of $50$ tickets numbered $00,01,02,....,49$ 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

The correct option is C $\frac{1}{5}$

Total number of cases ${=}^{50}{C}_1=50$
Let $A$ be the event of selecting ticket with sum of digits ${}^{\prime }{8}^{\prime }$.
Favourable cases to $A$ are $\{08,17,26,35,44\}$.
Let $B$ be the event of selecting ticket with product of its digits ${}^{\prime }{7}^{\prime }$
Favourable cases to $B$ is only $\left\{17\right\}$.
Now, $P\left(B/A\right)=\frac{P(A\cap B)}{P\left(A\right)}$ $=\frac{1/50}{5/50}=\frac{1}{5}$