Question

Four cards are drawn one by one from a deck of $52$ playing cards without replacement. What is the probability of getting a queen followed by a ten, an ace, and a two?

• A. $\frac{1}{13}\times \frac{1}{13}\times \frac{1}{13}\times \frac{1}{13}$
• B. $\frac{1}{13}\times \frac{4}{51}\times \frac{2}{25}\times \frac{4}{49}$
• C. $\frac{1}{13}\times \frac{4}{51}\times \frac{2}{25}\times \frac{2}{25}$
• D. $\frac{1}{13}\times \frac{4}{52}\times \frac{4}{52}\times \frac{4}{52}$

Answer 1

The correct option is B $\frac{1}{13}\times \frac{4}{51}\times \frac{2}{25}\times \frac{4}{49}$

Given: Four cards are drawn from a deck having $52$ cards without replacement.

Also, in a deck, there are 4 of each kind of card.

Let $\text{A},$ $\text{B},$ $\text{C}$, and $\text{D}$ be four events of drawing a queen, a ten, an ace, and a two, respectively, without replacement.

Probability of choosing a queen first = $\text{P(A)}=\frac{4}{52}=\frac{1}{13}$

Probability of choosing a ten = $\text{P(B)}=\frac{4}{51}$

Probability of choosing an ace = $\text{P(C)}=\frac{4}{50}=\frac{2}{25}$

Probability of choosing a two = $\text{P(D)}=\frac{4}{49}$

We have to find the probability of getting a queen followed by a ten, an ace, and a two, which is equal to $\text{P(E)}.$

$\Rightarrow \text{P(E)}=\text{P(A)}\times \text{P(B)}\times \text{P(C)}\times \text{P(D)}$

$\Rightarrow \text{P(E)}=\frac{1}{13}\times \frac{4}{51}\times \frac{2}{25}\times \frac{4}{49}$