Eliza Berry

2023-03-13

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 is an ace is A$\frac{9}{20}$ B$\frac{3}{16}$ C$\frac{1}{6}$ D$\frac{1}{9}$

Cooper Barker

The ideal decision is A $\frac{9}{20}$
Justification for the ideal selection:
Step: 1 Find out the probability to get no ace.
There are $16$ cards in all, including all aces, kings, queens, and jacks.
Cards to be drawn at random is $2$.
Thus, number of ways to select 2 cards is $C216$.
Thus, the probability to get no ace is
$=\frac{C}{2}=\frac{12!×14!}{10!×16!}=\frac{12×11×10!×14!}{10!×16×15×14!}=\frac{11}{20}$
Step: 2 Find out the probability to get at least one ace.
Hence, the likelihood of getting at least one ace is
$=1-\frac{11}{20}=\frac{9}{20}$
Therefore, Option(A) is the true answer.

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