Answer:
Explanation:
n(S₁) = 13
A = {1st card is multiples of 3} (card no.3, 6, and 9)
n(A) = 3
P(A) =
B = {2nd card is multiples of 3}
n(B) = 2 (1 card of multiple 3 already withdrawn)
n(S₂) = 12 (1 card already withdrawn)
P(B) =
=
P(A∩B) =
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