Question

A pack of playing cards was found to contain only $51$ cards. If the first $13$ cards which are examined are all red, then the probability that the missing is black is

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{1}{2}$
• D. $\frac{C_{13}^{25}}{C_{13}^{51}}$

Answer 1

The correct option is B

$\frac{2}{3}$

Explanation for the correct option:

Step: 1 Calculate the probability of selecting $13$ cards, when the missing card is red and black respectively.

Let, $A=$Missing card is red

$B=$ Missing card is black.

$D=$First $13$ cards are all red.

Here, the probability of missing card is red, $P\left(A\right)$ $=$the probability of missing card is black, $P\left(B\right)=\frac{1}{2}$.

Now, the probability of selecting $13$ cards, when the missing card is black

$$\begin{gathered} =P\left(\frac{A}{D}\right) \\ =\frac{C_{13}^{26}}{C_{13}^{51}} \end{gathered}$$

Again, the probability of selecting $13$ cards, when the missing card is red

$$\begin{gathered} =P\left(\frac{B}{D}\right) \\ =\frac{C_{13}^{25}}{C_{13}^{51}} \end{gathered}$$

Step: 2 Calculate the required probability

Therefore, the required probability

$$\begin{gathered} =P\left(\frac{D}{A}\right) \\ =\frac{P\left(A\right)P\left({\displaystyle \frac{A}{D}}\right)}{P\left(A\right)P\left(\frac{A}{D}\right)+P\left(B\right)P\left(\frac{B}{D}\right)} \\ =\frac{\frac{C_{13}^{26}}{C_{13}^{51}}}{\frac{C_{13}^{26}}{C_{13}^{51}}+\frac{C_{13}^{25}}{C_{13}^{51}}\times {\displaystyle \frac{1}{2}}} \\ =\frac{{\displaystyle \frac{26!\times 13!\times 38!}{13!\times 13!\times 51!}}}{\frac{26!\times 13!\times 38!}{13!\times 13!\times 51!}+\frac{25!\times 13!\times 38!}{13!\times 12!\times 51!}\times {\displaystyle \frac{1}{2}}} \\ =\frac{2}{3} \end{gathered}$$

Therefore, Option(B) is the correct answer.