Question

A pack of cards has one card missing. Two cards are drawn randomly and are found to be spades. The probability that the missing card is not a spade is:

• A. $\frac{3}{4}$
• B. $\frac{52}{867}$
• C. $\frac{39}{50}$
• D. $\frac{22}{425}$

Answer 1

The correct option is C

$\frac{39}{50}$

Explanation for the correct option:

Step:1 Calculate the probability of the missing card is spade.

Let, $A$ is the event that denotes the spade is missing.

Therefore,

$$\begin{gathered} P\left(A\right)=\frac{13}{52} \\ =\frac{1}{4} \end{gathered}$$

Therefore,

$$\begin{gathered} P(A\text{'})=1-\frac{1}{4} \\ =\frac{3}{4} \end{gathered}$$

Let, $E$ is the event that denotes the two cards drawn at random are spades.

Therefore,

$P\left(\frac{E}{A}\right)=\frac{C_2^{12}}{C_2^{51}}$

Now,

$P\left(\frac{E}{A\text{'}}\right)=\frac{C_2^{13}}{C_2^{51}}$

Step: 2 Calculate the probability that the missing card is not a spade.

Therefore, the probability that the missing card is not a spade is

$$\begin{gathered} =P\left(\frac{A\text{'}}{E}\right) \\ =\frac{\left(\frac{C_2^{13}}{C_2^{51}}\right)\times {\displaystyle \frac{3}{4}}}{\frac{C_2^{12}}{C_2^{51}}\times {\displaystyle \frac{1}{4}}+\left(\frac{C_2^{13}}{C_2^{51}}\right)\times \frac{3}{4}} \\ =\frac{{\displaystyle \frac{13!\times 49!}{11!\times 51!}}\times \frac{3}{4}}{\frac{12!\times 49!}{10!\times 51!}\times \frac{1}{4}+\frac{13!\times 49!}{11!\times 51!}\times \frac{3}{4}} \\ =\frac{39}{50} \end{gathered}$$

Therefore, Option(C) is the correct answer.