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(A) Total number of cards in a pack $= 52$. Number of sample space $n(S) = 52$. Let $A$ be the event of drawing an ace and $B$ be the event of drawing

Vedclass · Accepted Answer

$\frac{4}{13}$

Vedclass · Answer

(A) Total number of cards in a pack $= 52$. Number of sample space $n(S) = 52$. Let $A$ be the event of drawing an ace and $B$ be the event of drawing a spade. Number of aces $n(A) = 4$. Number of spades $n(B) = 13$. The number of cards that are both an ace and a spade is $n(A \cap B) = 1$. Using the addition rule of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. $P(A \cup B) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52}$. Simplifying the fraction,we get $\frac{16}{52} = \frac{4}{13}$.

$P(A)$	$P(B)$	$P(A \cap B)$	$P(A \cup B)$
$\frac{1}{3}$	$\frac{1}{5}$	$\frac{1}{15}$	$........$

If a card is drawn at random from a well-shuffled pack of $52$ playing cards,then the probability that it is either an ace or a spade card is

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