A box contains 25 tickets numbered 1, 2, dots | vedclass

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(C) The total number of ways to draw $2$ tickets from $25$ is given by $^{25}C_2 = \frac{25 	imes 24}{2} = 300$.The product of two numbers is even if

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$\frac{37}{50}$

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(C) The total number of ways to draw $2$ tickets from $25$ is given by $^{25}C_2 = \frac{25 	imes 24}{2} = 300$.The product of two numbers is even if at least one of the numbers is even.It is easier to calculate the probability of the complement event: the product is odd.The product is odd only if both selected numbers are odd.In the set ${1, 2, \dots, 25}$,there are $13$ odd numbers $(1, 3, 5, \dots, 25)$ and $12$ even numbers.The number of ways to choose $2$ odd tickets is $^{13}C_2 = \frac{13 	imes 12}{2} = 78$.The probability that the product is odd is $P(	ext{odd}) = \frac{78}{300} = \frac{13}{50}$.Therefore,the probability that the product is even is $1 - P(	ext{odd}) = 1 - \frac{13}{50} = \frac{37}{50}$.

$A$ box contains $25$ tickets numbered $1, 2, \dots, 25$. If two tickets are drawn at random,then the probability that the product of their numbers is even is:

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