Math Fundamentals18 sections · 814 units
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Example - Conditional Probability

(Cards without replacement)

Draw two cards from a deck without replacement. Find P(both Aces).

P(first Ace) = 4/52. Given the first is an Ace, 3 Aces remain out of 51 cards. P(second Ace | first Ace) = 3/51.

P(both Aces) = P(first Ace) × P(second Ace | first Ace) = (4/52) × (3/51) = 12/2652 = 1/221.