Question

Two cards are randomly chosen without replacement from an ordinary deck of 52 cards. Let $B$ be the event that both cards are aces, let $A_{s}$ be the event that the ace of spades is chosen, and let $A$ be the event that at least one ace is chosen. Find (a) $P\left(B | A_{s}\right)$ (b) $P(B | A)$

          Two cards are randomly chosen without replacement from an ordinary deck of 52 cards. Let $B$ be the event that both cards are aces, let $A_{s}$ be the event that the ace of spades is chosen, and let $A$ be the event that at least one ace is chosen. Find
(a) $P\left(B | A_{s}\right)$
(b) $P(B | A)$

Added by Sergio C.

Probability with Applications in Engineering, Science, and Technology

Matthew A. Carlton • Jay L. Devore 2nd Edition

Instant Answer

Solved by Expert Ahmad Reda

11/12/2021

Step 1

From the explanation, we know that $P(B | A_{s}) = \frac{P(B \cap A_{s})}{P(A_{s})}$. We have $P(A_{s}) = \frac{1}{52}$ (since there is only one ace of spades in the deck). To find $P(B \cap A_{s})$, we need to find the number of ways to choose both aces and the Show more…

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Two cards are randomly chosen without replacement from an ordinary deck of 52 cards. Let $B$ be the event that both cards are aces, let $A_{s}$ be the event that the ace of spades is chosen, and let $A$ be the event that at least one ace is chosen. Find (a) $P\left(B | A_{s}\right)$ (b) $P(B | A)$