What is the probability that 1 card drawn at | vedclass

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(B) Let $A$ be the event of drawing a queen and $B$ be the event of drawing an ace.The total number of cards in a deck is $52$.The number of queens in

Vedclass · Accepted Answer

$\frac{2}{13}$

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(B) Let $A$ be the event of drawing a queen and $B$ be the event of drawing an ace.The total number of cards in a deck is $52$.The number of queens in a deck is $4$,so $P(A) = \frac{4}{52}$.The number of aces in a deck is $4$,so $P(B) = \frac{4}{52}$.Since a card cannot be both a queen and an ace simultaneously,the events are mutually exclusive,meaning $P(A \cap B) = 0$.The probability of drawing either a queen or an ace is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.$P(A \cup B) = \frac{4}{52} + \frac{4}{52} - 0 = \frac{8}{52}$.Simplifying the fraction,we get $\frac{8}{52} = \frac{2}{13}$.

What is the probability that $1$ card drawn at random from a pack of $52$ playing cards is either a queen or an ace?

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