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(B) Total number of cards = $80$. Numbers divisible by $4$ between $1$ and $80$ are $4, 8, 12, \dots, 80$. This is an arithmetic progression with $a =

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$\frac{19}{316}$

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(B) Total number of cards = $80$. Numbers divisible by $4$ between $1$ and $80$ are $4, 8, 12, \dots, 80$. This is an arithmetic progression with $a = 4$,$d = 4$,and $l = 80$. Using $l = a + (n-1)d$,we get $80 = 4 + (n-1)4$,which implies $n = 20$. Total ways to select $2$ cards from $80$ is $^{80}C_2 = \frac{80 	imes 79}{2} = 3160$. Favorable ways to select $2$ cards from $20$ is $^{20}C_2 = \frac{20 	imes 19}{2} = 190$. Required probability $P = \frac{190}{3160} = \frac{19}{316}$.

From $80$ cards numbered $1$ to $80$,two cards are selected randomly. The probability that both the cards have numbers divisible by $4$ is given by

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