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(C) Total number of ways to choose $2$ numbers from $100$ is $N = {}^{100}C_2 = \frac{100 	imes 99}{2} = 4950$.Event $A$: The product is even. This h

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$\frac{25}{99}$

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(C) Total number of ways to choose $2$ numbers from $100$ is $N = {}^{100}C_2 = \frac{100 	imes 99}{2} = 4950$.Event $A$: The product is even. This happens if at least one number is even. The complement is that both numbers are odd. Number of odd numbers is $50$.$n(A) = {}^{100}C_2 - {}^{50}C_2 = 4950 - 1225 = 3725$.Event $B$: The product is divisible by $4$. This happens if (both are even and at least one is a multiple of $4$) or (one is odd and one is a multiple of $4$).Let $E$ be the set of even numbers ${2, 4, \dots, 100}$ ($50$ numbers) and $O$ be the set of odd numbers ${1, 3, \dots, 99}$ ($50$ numbers).Let $M_4$ be the set of multiples of $4$ ${4, 8, \dots, 100}$ ($25$ numbers).Let $E_2$ be the set of even numbers not divisible by $4$ ${2, 6, \dots, 98}$ ($25$ numbers).$n(A \cap B)$ is the number of pairs whose product is divisible by $4$.Pairs are: (one from $M_4$,any other) or (both from $E_2$).$n(A \cap B) = {}^{25}C_1 	imes {}^{75}C_1 + {}^{25}C_2 = 25 	imes 75 + 300 = 1875 + 300 = 2175$.$n(A \cap \bar{B}) = n(A) - n(A \cap B) = 3725 - 2175 = 1550$.$P(A \cap \bar{B}) = \frac{1550}{4950} = \frac{155}{495} = \frac{31}{99}$.Re-evaluating: The product is even but not divisible by $4$ if one number is from $E_2$ and the other is from $O$.$n(A \cap \bar{B}) = {}^{25}C_1 	imes {}^{50}C_1 = 25 	imes 50 = 1250$.$P(A \cap \bar{B}) = \frac{1250}{4950} = \frac{125}{495} = \frac{25}{99}$.

Two natural numbers are chosen at random from $1$ to $100$ and are multiplied. If $A$ is the event that the product is an even number and $B$ is the event that the product is divisible by $4$,then $P(A \cap \bar{B})=$

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