Question

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

• A. $\frac{1}{5}$
• B. $\frac{3}{16}$
• C. $\frac{9}{20}$
• D. $\frac{1}{9}$

Answer 1

The correct option is C

$\frac{9}{20}$

We have,

P both are aces = $\frac{{}^4{C}_2}{{}^{16}{C}_2}$

$=\frac{4}{16}\times \frac{3}{15}=\frac{1}{20}$

P(one is ace) = $\frac{{}^4{C}_1{\times }^{12}{C}_1}{{}^{16}{C}_2}=\frac{2}{5}$

$\therefore$ P(at least one is ace) $=\frac{1}{20}+\frac{2}{5}=\frac{9}{20}$