Question

Two cards are drawn one by one from a pack of cards. The probability of getting first card an ace and second a coloured one is (before drawing second card first card is not placed again in the pack)

• A. $\frac{1}{26}$
• B. $\frac{5}{52}$
• C. $\frac{5}{221}$
• D. $\frac{4}{13}$

Answer 1

The correct option is C

$\frac{5}{221}$

Find the required probability:

Step 1: Find the probability of drawing an ace card

In a pack of $52$ cards there are $4$ aces

Let $A$be the event of drawing an ace

$\Rightarrow$ $p\left(A\right)=\frac{4}{52}$

There are in total $16$ coloured cards in a pack of $52$ cards. They are $4$ aces, $4$ kings, $4$ queens and $4$ jacks.

After drawing the first card as an ace we are not placing it back into the deck

Hence, now there are $51$ cards out of which $15$ are coloured cards.

Step 2: Find the probability of drawing a coloured card

Let $B$ be the the event of drawing an coloured card

We know that $p\left(\frac{B}{A}\right)=\frac{15}{51}$ as the probability of event $B$ is calculated given the event $A$ has happened.

$A$ and $B$ are independent events.

Using the formula for conditional probability we get

$p\left(\frac{B}{A}\right)=\frac{p\left(A\cap B\right)}{p\left(A\right)}$

$\Rightarrow$ $p\left(A\cap B\right)=p\left(\frac{B}{A}\right)\times p\left(A\right)$

$=\frac{15}{51}\times \frac{4}{52}$

$=\frac{5}{17}\times \frac{1}{13}$

$\Rightarrow$ $p\left(A\cap B\right)=\frac{5}{221}$

Therefore, the probability of getting first card as an ace ‘and’ a second coloured card is $\frac{5}{221}$ .

Hence, option (C) is the correct answer.