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(C) Let $A$ be the event that the sum of the digits is $10$.Let $B$ be the event that the product of the digits is $9$.The tickets are numbered from $

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(C) Let $A$ be the event that the sum of the digits is $10$.Let $B$ be the event that the product of the digits is $9$.The tickets are numbered from $00$ to $49$.For event $B$ (product of digits is $9$): The possible numbers are $09, 19, 33$. However,$09$ has digits $0$ and $9$,product is $0 	imes 9 = 0$. So,$B = \{19, 33\}$. Thus,$n(B) = 2$.For event $A$ (sum of digits is $10$): The possible numbers are $19, 28, 37, 46$.The intersection $A \cap B$ is the set of numbers where the sum is $10$ $AND$ the product is $9$. Comparing the sets,$A \cap B = \{19\}$. Thus,$n(A \cap B) = 1$.The conditional probability is given by $P(A|B) = \frac{n(A \cap B)}{n(B)} = \frac{1}{2}$.

One ticket is selected at random from $50$ tickets numbered $00, 01, 02, \ldots, 49$. The probability that the sum of the digits is $10$,given that the product of the digits is $9$,is

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