Question

A Card is drawn randomly from a pack of playing cards. Then the probability that it is neither ace nor king is

• A. $11/13$
• B. $8/13$
• C. $10/13$
• D. $12/13$

Answer 1

The correct option is A

$11/13$

Finding the probability :

Total number of cards in a pack $=52$

Number of ace cards $=4$

Number of king cards$=4$

Probability of getting an ace $=\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$52$}\right.=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$13$}\right.$

Probability of getting a king$=\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$52$}\right.=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$13$}\right.$

Now, the probability of getting either an ace or king $=$ Probability of getting an ace + Probability of getting a king

So, Probability of getting either an ace or king $=\frac{1}{13}+\frac{1}{13}=\frac{2}{13}$

Therefore, the probability of getting neither an ace nor a king $=$$1-$ the probability of getting either an ace or king $\mathbf{[}\mathbf{}\mathbf{\because }\mathbf{}\mathit{P}\mathbf{(}\overline{\mathbf{A}}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{-}\mathit{P}\mathbf{(}\mathit{A}\mathbf{)}\mathbf{}\mathbf{]}$

So. the probability of getting neither an ace nor a king $=1-\frac{2}{13}=\frac{(13-2)}{13}=\frac{11}{13}$

Hence, the correct option is A.