Question

The probability that the three cards drawn from a pack of $52$ cards are all black is

• A. $\frac{1}{17}$
• B. $\frac{2}{17}$
• C. $\frac{3}{17}$
• D. $\frac{2}{19}$

Answer 1

The correct option is B

$\frac{2}{17}$

The explanation for the correct option

The total number of cards is $52$.

$3$ cards are drawn, so the total number of cases can be given by, $n\left(S\right)={}^{52}C_3$.

$$\begin{gathered} \Rightarrow n\left(S\right)=\frac{52!}{\left(52-3\right)!\times 3!} \\ \Rightarrow n\left(S\right)=\frac{52!}{49!\times 3!} \\ \Rightarrow n\left(S\right)=\frac{52\times 51\times 50}{3!} \\ \Rightarrow n\left(S\right)=\frac{52\times 51\times 50}{3\times 2\times 1} \\ \Rightarrow n\left(S\right)=\frac{132600}{6} \\ \Rightarrow n\left(S\right)=22100 \end{gathered}$$

The total number of black cards is $26$.

Thus, the number of cases such that all three cards are black can be given by, $n\left(B\right)={}^{26}C_3$.

$$\begin{gathered} \Rightarrow n\left(B\right)=\frac{26!}{\left(26-3\right)!\times 3!} \\ \Rightarrow n\left(B\right)=\frac{26!}{23!\times 3!} \\ \Rightarrow n\left(B\right)=\frac{26\times 25\times 24}{3!} \\ \Rightarrow n\left(B\right)=\frac{26\times 25\times 24}{3\times 2\times 1} \\ \Rightarrow n\left(B\right)=\frac{15600}{6} \\ \Rightarrow n\left(B\right)=2600 \end{gathered}$$

Hence, the probability of getting all three cards as black can be given by, $P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{2600}{22100}=\frac{2}{17}$.

Therefore, the probability that the three cards drawn from a pack of $52$ cards are all black is $\frac{2}{17}$. So, the correct option is (B).