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Chapter 3: Problem 37

A deck of cards is shuffled and then divided into two halves of 26 cards each. A card is drawn from one of the halves; it turns out to be an ace. The ace is then placed in the second half-deck. The half is then shuffled, and a card is drawn from it. Compute the probability that this drawn card is an ace.

Short Answer

Expert verified
The probability that the card drawn from the second half-deck is an Ace, given that an Ace was drawn from the first half-deck, is \(\frac{3}{26}\).

Step by step solution

01

Identify the probabilities of drawing an Ace from each half-deck initially

First, let's find the probability of drawing an Ace from the first half-deck and from the second half-deck before the Ace was added. There are 4 Aces in a full deck of 52 cards. We can assume that after dividing the deck, there are 2 Aces in the first half-deck and 2 Aces in the second half-deck. The probability of drawing an Ace from the first half-deck is: \(P(A_1) = \frac{2}{26} = \frac{1}{13}\) The probability of drawing an Ace from the second half-deck before adding the Ace is: \(P(A_2') = \frac{2}{26} = \frac{1}{13}\)
02

Find the probability of drawing an Ace from the second half-deck after adding the Ace

Now, let's find the probability of drawing an Ace from the second half-deck after the Ace from the first half-deck was added to it. This means there are now 3 Aces in the second half-deck and the number of cards in the second half-deck remains 26. The probability of drawing an Ace from the second half-deck after adding the Ace is: \(P(A_2) = \frac{3}{26}\)
03

Calculate the probability of drawing an Ace from the second half-deck using conditional probability

We are interested in finding the probability of drawing an Ace from the second half-deck given that an Ace was already drawn from the first half-deck. In other words, we try to find \(P(A_2|A_1)\), which means the probability of drawing an Ace from the second half-deck given that an Ace was drawn from the first half-deck. Since we know that an Ace was drawn from the first half-deck, the probability of drawing an Ace from the second half-deck will simply be the probability of drawing an Ace from it after adding the Ace from the first half-deck, as calculated in Step 2. Therefore, the probability of drawing an Ace from the second half-deck given that an Ace was drawn from the first half-deck is: \(P(A_2|A_1) = P(A_2) = \frac{3}{26}\) So, the probability that the card drawn from the second half-deck is an Ace is \(\frac{3}{26}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conditional Probability
Conditional probability is the likelihood of an event occurring given that another event has already taken place. This concept is crucial when solving problems where the outcome of one action influences another. In the context of the Ace Draw Problem, we're interested in the probability of drawing an Ace from the second half-deck, given that an Ace was already drawn from the first half-deck. This is represented by \(P(A_2|A_1)\), where:
  • \(A_1\) is the event that an Ace is drawn from the first half-deck.
  • \(A_2\) is the event that an Ace is drawn from the second half-deck.
To determine \(P(A_2|A_1)\), we rely on the fact that we've already adjusted the deck by transferring the Ace. Thus, the probability reflects the new conditions, where the addition of one more Ace to the second half affects the outcome.
Deck of Cards
A standard deck of playing cards contains 52 cards split into four suits: hearts, diamonds, clubs, and spades. Each suit has 13 cards, including one Ace per suit. In many probability problems, assumptions about card distribution are typically made to simplify the problem.
This exercise involves dividing a shuffled deck into two equal parts, ensuring each half contains 26 cards. It is assumed, for ease of calculation, that Aces are evenly distributed across these halves. Initially, each half has 2 Aces. Knowing card distributions helps in calculating probabilities in events involving card draws.
Ace Draw Problem
The Ace Draw Problem serves as a great example of combining knowledge of card distributions with conditional probability. Initially, we calculate the probability of drawing an Ace from either half-deck before any cards are moved. With two Aces per half, that鈥檚 \( \frac{2}{26} \) or \( \frac{1}{13} \) for each half-deck.
Then, considering the problem's condition, after drawing an Ace and adding it to the second half-deck, this deck now contains 3 Aces. Consequently, the probability of drawing an Ace from this modified second half-deck becomes \(\frac{3}{26}\). This step relies heavily on the understanding of how redistributing cards influences chances of specific outcomes.
Solving problems like this showcase how shifting elements between groups leads to changes in probabilities, highlighting the interplay between initial conditions and outcomes.

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Most popular questions from this chapter

Let \(A\) and \(B\) be events having positive probability. State whether each of the following statements is (i) necessarily true, (ii) necessarily false, or (iii) possibly true. (a) If \(A\) and \(B\) are mutually exclusive, then they are independent. (b) If \(A\) and \(B\) are independent, then they are mutually exclusive. (c) \(P(A)=P(B)=.6\), and \(A\) and \(B\) are mutually exclusive. (d) \(P(A)=P(B)=.6\), and \(A\) and \(B\) are independent.

Suppose we have 10 coins such that if the \(i\) th coin is flipped, heads will appear with probability \(i / 10, i=1,2, \ldots, 10\). When one of the coins is randomly selected and flipped, it shows heads. What is the conditional probability that it was the fifth coin?

There are 3 coins in a box. One is a two-headed coin; another is a fair coin; and the third is a biased coin that comes up heads 75 percent of the time. When one of the 3 coins is selected at random and flipped, it shows heads. What is the probability that it was the two-headed coin?

(a) A gambler has in his pocket a fair coin and a two-headed coin. He selects one of the coins at random; when he flips it, it shows heads. What is the probability that it is the fair coin? (b) Suppose that he flips the same coin a second time and again it shows heads. What is now the probability that it is the fair coin? (c) Suppose that he flips the same coin a third time and it shows tails. What is now the probability that it is the fair coin?

\(A\) and \(B\) are involved in a duel. The rules of the duel are that they are to pick up their guns and shoot at each other simultaneously. If one or both are hit, then the duel is over. If both shots miss, then they repeat the process. Suppose that the results of the shots are independent and that each shot of \(A\) will hit \(B\) with probability \(p_{A}\), and each shot of \(B\) will hit \(A\) with probability \(p_{B}\). What is (a) the probability that \(A\) is not hit; (b) the probability that both duelists are hit; (c) the probability that the duel ends after the \(n\)th round of shots; (d) the conditional probability that the duel ends after the \(n\)th round of shots given that \(A\) is not hit; (e) the conditional probability that the duel ends after the \(n\)th round of shots given that both duelists are hit?
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