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

Beginner2023-03-14Added 6 answers

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

$=\frac{C}{2}=\frac{12!\times 14!}{10!\times 16!}=\frac{12\times 11\times 10!\times 14!}{10!\times 16\times 15\times 14!}=\frac{11}{20}$

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.

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

$=\frac{C}{2}=\frac{12!\times 14!}{10!\times 16!}=\frac{12\times 11\times 10!\times 14!}{10!\times 16\times 15\times 14!}=\frac{11}{20}$

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.