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(B) standard pack of cards contains $52$ cards.There is only $1$ queen of clubs in the deck,so the probability of drawing it is $P(A) = \frac{1}{52}$.

Vedclass · Accepted Answer

$\frac{1}{26}$

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(B) standard pack of cards contains $52$ cards.There is only $1$ queen of clubs in the deck,so the probability of drawing it is $P(A) = \frac{1}{52}$.There is only $1$ king of hearts in the deck,so the probability of drawing it is $P(B) = \frac{1}{52}$.Since these two events are mutually exclusive (a card cannot be both a queen of clubs and a king of hearts simultaneously),the probability of either event occurring is the sum of their individual probabilities.Required probability $P(A \cup B) = P(A) + P(B) = \frac{1}{52} + \frac{1}{52} = \frac{2}{52} = \frac{1}{26}$.

$A$ card is drawn from a well-shuffled pack of $52$ cards. The probability of getting a queen of clubs or a king of hearts is:

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