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 $^{25}C_2 = \frac{25 	imes 24}{2} = 300$.For the product of two numbers to be even,at l

Vedclass · Accepted Answer

$\frac{37}{50}$

Vedclass · Answer

(C) The total number of ways to draw $2$ tickets from $25$ is $^{25}C_2 = \frac{25 	imes 24}{2} = 300$.For the product of two numbers to be even,at least one of the numbers must be even.It is easier to calculate the complement: the product is odd only if both numbers drawn are odd.There are $13$ odd numbers $(1, 3, 5, \dots, 25)$ and $12$ even numbers $(2, 4, 6, \dots, 24)$ in the set.The number of ways to choose $2$ odd tickets is $^{13}C_2 = \frac{13 	imes 12}{2} = 78$.The probability that both are 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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