Question

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

• A. $\frac{21}{316}$
• B. $\frac{19}{316}$
• C. $\frac{1}{4}$
• D. None of these

Answer 1

The correct option is B

$\frac{19}{316}$

Explanation for the correct option:

Find the probability:

Let the probability of getting both cards having the numbers divisible by $4$ be $P\left(A\right)$.

There are $80$ cards out of which two cards selected randomly,

$\Rightarrow$Total cases of getting two cards, $n\left(S\right)={}^{80}C_2$

As, there are $20$ numbers between $1$ and $80$, which are divisible by $4$

$\Rightarrow$Favorable cases of getting cards, $n\left(A\right)={}^{20}C_2$

$\therefore$Required probability, $P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}$

$=\frac{{}^{80}C_2}{{}^{20}C_2}$

$=\frac{190}{3160}$ $\left[\mathbf{\because }{}^{\mathbf{n}}\mathit{C}_{\mathbf{r}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{n}\mathbf{!}}{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{r}\mathbf{)}\mathbf{!}\mathbf{r}\mathbf{!}}\right]$

$=\frac{19}{316}$

Thus, the probability of getting both cards having numbers divisible by $4$ is $\frac{\mathbf{19}}{\mathbf{316}}$

Hence, Option ‘B’ is Correct.