(A) Total number of cards = $52$.Number of diamond cards $n(A) = 13$,so $P(A) = 13/52 = 1/4$.Number of ace cards $n(B) = 4$,so $P(B) = 4/52 = 1/13$.Nu

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(A) Total number of cards = $52$.Number of diamond cards $n(A) = 13$,so $P(A) = 13/52 = 1/4$.Number of ace cards $n(B) = 4$,so $P(B) = 4/52 = 1/13$.Number of ace of diamond cards $n(A \cap B) = 1$,so $P(A \cap B) = 1/52$.For events to be independent,$P(A \cap B)$ must be equal to $P(A) \times P(B)$.$P(A) \times P(B) = (1/4) \times (1/13) = 1/52$.Since $P(A \cap B) = P(A) \times P(B) = 1/52$,the events $A$ and $B$ are independent.

Competitive Quantitative Aptitude Probability Probability

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$A$ card is drawn from a pack of $52$ cards. If $A =$ card is of diamond,$B =$ card is an ace and $A \cap B =$ card is the ace of diamond,then events $A$ and $B$ are

A
Independent
B
Mutually exclusive
C
Dependent
D
Equally likely

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$A$ card is drawn from a pack of $52$ cards. If $A =$ card is of diamond,$B =$ card is an ace and $A \cap B =$ card is the ace of diamond,then events $A$ and $B$ are

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