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Chapter 3: Problem 41

A deck of cards is shuffled and then divided into o 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. Hint: Condition on whether or not the interchanged card is selected.

Short Answer

Expert verified
The probability of drawing an ace on the second draw is \(\frac{3}{26}\), considering two possible cases: the first drawn ace is not selected in the second draw and an ace from the second half is picked, or selecting the first drawn ace in the second draw. Calculating the probability for both cases using conditional probability and adding them gives us the final probability of \(\frac{3}{26}\).

Step by step solution

01

Calculate the Probability of Drawing an Ace in the First Draw

There are 4 aces in the deck, and assuming each half has an equal number of aces, there are 2 aces in the first 26 cards and 2 aces in the second 26 cards. So, the probability of drawing an ace from the first half is: \(P(\text{ace from first half}) = \frac{2\ \text{aces}}{26\ \text{cards}} = \frac{1}{13}\)
02

Calculate the Probability of Both Cases

1. The probability of the ace from the first half not being selected in the second draw, and drawing an ace from the second half: Let A be the event that the first drawn ace is not selected in the second draw, and B be the event that the second draw is an ace from the second half. \(P(A) = \frac{25\ \text{non-ace cards in second half}}{26\ \text{cards in second half}} = \frac{25}{26}\) \(P(B|A) = \frac{2\ \text{aces in second half}}{25\ \text{remaining cards in second half}} = \frac{2}{25}\) \(P(A\cap B) = P(A)*P(B|A) = \frac{25}{26} * \frac{2}{25} = \frac{1}{13}\) 2. The probability of selecting the first drawn ace in the second draw: Let C be the event that the first drawn ace is selected in the second draw, and D be the event that the second draw is the first drawn ace. \(P(C) = \frac{1\ \text{ace in second half}}{26\ \text{cards in second half}} = \frac{1}{26}\) \(P(D|C) = 1\) \(P(C\cap D) = P(C)*P(D|C) = \frac{1}{26} * 1 = \frac{1}{26}\)
03

Calculate the Final Probability

In order to find the final probability, we need to find the probability that one of the two possible cases happens: \(P(\text{ace in second draw}) = P(A\cap B) + P(C\cap D) = \frac{1}{13} + \frac{1}{26} = \frac{3}{26}\) The probability of drawing an ace on the second draw is \(\frac{3}{26}\).

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Key Concepts

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

Probability Theory
Probability theory is the branch of mathematics concerned with analyzing random phenomena. The main objective is to determine the likelihood of various outcomes in uncertain situations. When you toss a coin, for example, the probability of it landing heads up is 50%. This simple notion extends to more complex situations, such as determining the chance of drawing an ace from a deck of cards.

To understand such probabilities, it is essential to know the basics of probability rules. One fundamental rule is the 'addition rule', which is used when trying to find the probability of one event or another occurring. The other key rule is the 'multiplication rule', which relates to finding the probability of two independent events happening together. Using these principles, you can calculate the likelihood of various outcomes in a deck of cards, from pulling a specific card to predicting the outcomes of complex card games.
Conditional Probability
Conditional probability comes into play when the outcome of an event affects the likelihood of another related event. As its name suggests, it's the probability of an event occurring given the condition that another event has already happened.

In the context of our car- drawing scenario, after an ace is drawn from one half of the deck and placed in the other, the probability of drawing an ace from the now altered half changes. This change reflects conditional probability because the outcome of the second draw depends on the fact that an ace was drawn and moved in the first draw.
  • The probability of drawing an ace from the second half-deck after transferring an ace is different from the initial probability.
  • Conditional probability can be calculated using the formula: \( P(B|A) = \frac{P(A \cap B)}{P(A)} \), where \( P(A) \) is the probability of event A, \( P(B|A) \) is the probability of event B given event A, and \( P(A \cap B) \) is the probability of both events A and B occurring.
This concept illustrates how the occurrence of a previous event has a direct impact on the probability of subsequent events.
Combinatorics
Combinatorics is a field of mathematics primarily concerned with counting, both as a means and an end in obtaining results, and certain properties of finite structures. It is related to many other areas of mathematics, particularly algebra, probability theory, and geometry, and has applications in fields such as computer science and statistics.

In our exercise, combinatorics helps us calculate the number of ways to draw an ace from a deck of cards. Important combinatorial principles include permutations and combinations, which can calculate the number of possible arrangements of card sequences and selections without regard to order. Through these principles, we can compute the probabilities of various card-related outcomes, making combinatorics an essential part of solving problems involving decks of cards.
  • Understanding how many aces or other specific cards are in each half of the deck is essential in determining the probability of drawing one.
  • The concept of combinatorics is often used to simplify complex probability calculations by clarifying the possible combinations or permutations of the involved elements.
Deck of Cards Probability
Probability in the context of a deck of cards involves determining the chances of drawing specific cards or combinations of cards. A standard deck has 52 cards, which includes 4 suits of 13 ranks each, and among them, there are 4 aces.

Calculating the probability of drawing an ace from a full deck is simple: \( P(\text{ace}) = \frac{4}{52} = \frac{1}{13} \). However, once cards are removed or added, as in our exercise, the calculation becomes more complex, incorporating rules from probability theory and the other related concepts. To determine the updated probability:
  • Consider the new composition of the deck after aces have been moved from one half to another.
  • Apply conditional probability to adjust for the fact that an ace was already drawn.
Now, with an altered deck, the probability changes and requires a more detailed analysis to ensure accurate calculations.

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Most popular questions from this chapter

Suppose that an ordinary deck of 52 cards (which contains \(4 \text { aces })\) is randomly divided into 4 hands of 13 cards each. We are interested in determining \(p,\) the probability that each hand has an ace. Let \(E_{i}\) be the event that the \(i\) th hand has exactly one ace. Determine \(p=P\left(E_{1} E_{2} E_{3} E_{4}\right)\) by using the multiplication rule.

A town council of 7 members contains a steering committee of size \(3 .\) New ideas for legislation go first to the steering committee and then on to the council as a whole if at least 2 of the 3 committee members approve the legislation. Once at the full council, the legislation requires a majority vote (of at least 4 ) to pass. Consider a new piece of legislation, and suppose that each town council member will approve it, independently, with probability \(p .\) What is the probability that a given steering committee member's vote is decisive in the sense that if that person's vote were reversed, then the final fate of the legislation would be reversed? What is the corresponding probability for a given council member not on the steering committee?

Independent flips of a coin that lands on heads with probability \(p\) are made. What is the probability that the first four outcomes are (a) \(H, H, H, H ?\) (b) \(T, H, H, H ?\) (c) What is the probability that the pattern \(T, H\) \(H, H\) occurs before the pattern \(H, H, H, H ?\) Hint for part \((c):\) How can the pattern \(H, H, H, H\) occur first?

An urn contains 12 balls, of which 4 are white. Three players \(-A, B,\) and \(C-\) successively draw from the urn, \(A\) first, then \(B\), then \(C\), then \(\bar{A}\), and so on. The winner is the first one to draw a white ball. Find the probability of winning for each player if (a) each ball is replaced after it is drawn; (b) the balls that are withdrawn are not replaced.

In any given year, a male automobile policyholder will make a claim with probability \(p_{m}\) and a female policyholder will make a claim with probability \(p_{f}\) where \(p_{f} \neq p_{m} .\) The fraction of the policyholders that are male is \(\alpha, 0<\alpha<1 .\) A policyholder is randomly chosen. If \(A_{i}\) denotes the event that this policyholder will make a claim in year \(i,\) show that $$ P\left(A_{2} | A_{1}\right)>P\left(A_{1}\right) $$ Give an intuitive explanation of why the preceding inequality is true.
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