A pack of cards contains 4 aces, 4 kings, 4 queens and 4 jacks. Two...

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.