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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 where $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 where $a = 4$, $d = 4$, 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 the $20$ cards divisible by $4$ 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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