Problem 11 - A First Course in Probability

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) \(P(B | A)\)

Step-by-Step Solution

Answer

(a) \(P(B | A_{s}) = \dfrac{1}{17}\) (b) \(P(B | A) \approx \dfrac{1}{51}\)

Step 1: Define Events and Probabilities

We define the following events: Event \(B\): Both cards are aces. Event \(A_{s}\): Ace of Spades is chosen. Event \(A\): At least one card is an ace. We can find probabilities using the conditional probability formula: \(P(B | A_{s}) = \dfrac{P(B \cap A_{s})}{P(A_{s})}\) (a) \(P(B | A) = \dfrac{P(B \cap A)}{P(A)}\) (b)

Step 2: Find \(P(A_{s})\), \(P(B \cap A_{s})\) for (a)

First, we find the probabilities for (a). We know there are 4 aces in the deck, and one of them is Ace of Spades. So, the probability of choosing Ace of Spades is: \(P(A_{s}) = \dfrac{1}{52}\) (since there is 1 Ace of Spades in the deck of 52 cards) Since Ace of Spades is chosen, we are left with 51 cards in the deck and 3 aces (excluding Ace of Spades). The probability of choosing an ace from the remaining cards is: \(P(B \cap A_{s}) = \dfrac{3}{51}\) (since there are 3 other aces in the remaining cards).

Step 3: Find \(P(A)\), \(P(B \cap A)\) for (b)

Now we find probabilities for (b). To find the probability of having at least one Ace, we can find the probability of not having an Ace (\(P(A')\)) in both cards and subtract it from 1. So, \(P(A) = 1 - P(A')\) Probability of not having an Ace on the first card is \(\dfrac{48}{52}\) (since there are 4 aces in the deck) Probability of not having an Ace on the second card (after picking a non-Ace first card) is \(\dfrac{47}{51}\) So, \(P(A') = \dfrac{48}{52} \times \dfrac{47}{51}\) \(P(A) = 1 - P(A') = 1 - \dfrac{48}{52} \times \dfrac{47}{51}\) \(P(B \cap A)\) is the probability of both cards being aces. There are a total of \(\dbinom{52}{2} = 1326\) possible card pairs. Out of these pairs, we have 6 pairs with both aces (\(\dbinom{4}{2} = 6\)). So, \(P(B \cap A) = \dfrac{6}{1326}\)

Step 4: Find \(P(B | A_{s})\) and \(P(B | A)\) using the formula

Now, we use the conditional probability formula to find the required probabilities. (a) \(P(B | A_{s}) = \dfrac{P(B \cap A_{s})}{P(A_{s})} = \dfrac{\dfrac{3}{51}}{\dfrac{1}{52}} = \dfrac{3}{51} \times \dfrac{52}{1} = \dfrac{1}{17}\) (b) \(P(B | A) = \dfrac{P(B \cap A)}{P(A)} = \dfrac{\dfrac{6}{1326}}{1 - \dfrac{48}{52} \times \dfrac{47}{51}} = \dfrac{6}{1326} \times \dfrac{1}{1 - \dfrac{48}{52} \times \dfrac{47}{51}} \approx \dfrac{6}{1326} \times \dfrac{1}{\dfrac{23}{51}} \approx \dfrac{6}{1326} \times \dfrac{51}{23} = \dfrac{1}{51}\) So, the final answers are: (a) \(P(B | A_{s}) = \dfrac{1}{17}\) (b) \(P(B | A) \approx \dfrac{1}{51}\)