Question

A pack of card contain 4 king, 4 queen, 4 aces and 4 knave. A person cut two cards at random, the probability that at least one of them will be an aces is

• A. $\frac{9}{20}$
• B. $\frac{1}{9}$
• C. $\frac{11}{20}$
• D. None of these

Answer 1

The correct option is A $\frac{9}{20}$

Let $E$ be the event of drawing at least one aces from the pack.
Also let $A$ be the event of getting aces and $\overline{A}$ of getting a card other than aces.
then $P\left(E\right)=\frac{\left(\begin{array}{c}4\\ 2\end{array}\right)+\left(\begin{array}{c}4\\ 1\end{array}\right)\left(\begin{array}{c}12\\ 1\end{array}\right)}{\left(\begin{array}{c}16\\ 2\end{array}\right)}=\frac{54}{120}=\frac{9}{20}$