Question

A pack of cards contains $4$ aces , $4$ kings , $4$ queens and $4$ jacks . Two cards are drawn at random . The probability that at least one of these in an ace, is

• A. $\frac{9}{20}$
• B. $\frac{3}{16}$
• C. $\frac{1}{6}$
• D. $\frac{1}{9}$

Answer 1

Answer: (A)

$\frac{9}{20}$

We have,

Aces card$=4$

King card$=4$

Queen card$=4$

Jack card$=4$

Then Total cards$=16$

Now, $1$ aces card$+1$ other card$+2$ aces card $=4$ cards

So,

$=\frac{{}^4{C}_1{\times }^{12}{C}_1}{{}^{16}{C}_2}+\frac{{}^4{C}_2}{{}^{16}{C}_2}$

$=\frac{4\times 12\times 2!\times 14!}{16\times 15\times 14!}+\frac{4\times 3\times 2!\times 14!}{2!\times 16\times 15\times 14!}$

$=\frac{4\times 12\times 2}{16\times 15}+\frac{4\times 3}{16\times 15}$

$=\frac{2}{5}+\frac{1}{20}$

$=\frac{9}{20}$

Hence, this is the answer.