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Solve the following.
50 tickets in a box are numbered 00, 01, 02, …, 49. One ticket is drawn randomly from the box. Find…
$50$ tickets in a box are numbered $00, 01, 02, \dots, 49$. One ticket is drawn randomly from the box. Find the probability of the ticket having the product of its digits $7$, given that the sum of the digits is $8$?
Answer
Answer
Let $A$ = product of digits is $7$ and $B$ = sum of digits is $8$.
Tickets with digit sum $8$: $08, 17, 26, 35, 44$, so $n(B) = 5$.
Of these, only $17$ has digit product $7$, so $n(A \cap B) = 1$.
$P(A/B) = \frac{n(A \cap B)}{n(B)} = \frac{1}{5}$
Final answer: $\frac{1}{5}$