/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 Suppose that an ordinary deck of... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that an ordinary deck of 52 cards is shuffled and the cards are then turned over one at a time until the first ace appears. Given that the first ace is the 20 th card to appear, what is the conditional probability that the card following it is the (a) ace of spades? (b) two of clubs?

Short Answer

Expert verified
Given the deck of 52 cards and the fact that the first Ace appears as the 20th card, the conditional probability that the card following the first Ace is: (a) the ace of spades is \(\frac{1}{32}\). (b) the two of clubs is \(\frac{1}{32}\).

Step by step solution

01

Determine the possible scenarios

Before calculating the conditional probability for each case, we need to know the total number of possible scenarios for the cards drawn. There are 52 cards in total. Among the first 20 cards, one of them is an Ace, and there are 4 Aces in total, so the remaining 3 Aces are among the remaining 32 cards (52 - 20). Therefore, we have 3 Aces and 29 other cards (32 - 3) in the remaining undrawn pile.
02

Calculate the conditional probability for the Ace of Spades

In this case, we need to find the probability that the next card (21st card) is the ace of spades. Since one Ace has already been drawn (the 20th card), there are 3 Aces left among the remaining 32 cards. If the ace of spades is among the remaining 3 Aces, the probability will be: P(Ace of Spades | First Ace at position 20) = \(\frac{1}{32}\)
03

Calculate the conditional probability for the Two of Clubs

In this case, we need to find the probability that the next card (the 21st card) is the two of clubs. Since this card only appears once and has not been drawn yet, we can directly compute the probability as: P(Two of Clubs | First Ace at position 20) = \(\frac{1}{32}\) #Conclusion# Given the deck of 52 cards and the fact that the first Ace appears as the 20th card, the conditional probability that the card following the first Ace is: (a) the ace of spades is \(\frac{1}{32}\). (b) the two of clubs is \(\frac{1}{32}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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 events and the likelihood of various outcomes. This includes predicting the occurrence of events ranging from the roll of a die to the outcome of a card game. Understanding probability helps us quantify the uncertainty and make informed decisions based on the odds of different possibilities. In essence, probability is expressed as a number between 0 and 1, inclusive, where 0 indicates an impossibility and 1 represents a certainty.

To get started with probability, one needs to be familiar with a few basic concepts: outcomes, events, and probability itself. An outcome is a possible result of a random experiment, while an event is a set of outcomes. The probability of an event is the ratio of the number of favorable outcomes to the total number of possible outcomes. When all outcomes are equally likely, this calculates as the simple formula: P(Event) = \(\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}\).
Deck of Cards
A standard deck of cards is a common example used in probability theory because of its well-defined structure. A deck typically consists of 52 cards, divided into four suits: hearts, diamonds, clubs, and spades. Each suit contains 13 cards, number 2 through 10, and four face cards (jack, queen, king, and ace). Understanding the composition of a deck is crucial when calculating probabilities related to card games.

When we address problems involving a deck of cards, the concept of randomness is key. The shuffle of the deck is assumed to be thorough, thereby giving each of the 52 cards an equal chance of appearing in any position in the deck. Such assumptions are important as they affect the calculation of probabilities. For example, the probability of drawing any ace from a full deck at the start is \(\frac{4}{52}\), since there are four aces out of 52 cards. As cards are drawn and the deck's composition changes, so do the probabilities associated with it.
Conditional Probability Calculation
Conditional probability refers to the probability of an event occurring given that another event has already taken place. This concept is central when dealing with events that are not independent, meaning the occurrence of one event affects the likelihood of another. The key formula for conditional probability is: P(A | B) = \(\frac{P(A \cap B)}{P(B)}\),where P(A | B) is the probability of event A given event B, P(A \cap B) is the probability of both A and B occurring, and P(B) is the probability of event B.

In the exercise, we utilized conditional probability to determine the chances of drawing specific cards given that a specific event (the first ace appears as the 20th card) had already happened. This probability changes from the initial probability because we have new information that affects the outcome. It's like updating our beliefs in light of new evidence. It's fundamentally different from the probability of drawing the ace of spades or the two of clubs from a full, fresh deck, because events have already transpired that change the composition of the deck and therefore the probabilities of subsequent events.

It's also worth noting that for making the exercise more understandable, we could have explicitly defined each term and operation in the conditional probability calculation, illustrated the steps with a visual diagram of the deck as cards are being removed, and provided additional examples for practice.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose that you continually collect coupons and that there are \(m\) different types. Suppose also that each time a new coupon is obtained, it is a type i coupon with probability \(p_{i}, i=1, \ldots, m .\) Suppose that you have just collected your \(n\)th coupon. What is the probability that it is a new type? Hint: Condition on the type of this coupon.

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.

The color of a person's eyes is determined by a single pair of genes. If they are both blue-eyed genes, then the person will have blue eyes; if they are both brown-eyed genes, then the person will have brown eyes; and if one of them is a blue-eyed gene and the other a brown-eyed gene, then the person will have brown eyes. (Because of the latter fact, we say that the brown-eyed gene is dominant over the blue-eyed one.) A newborn child independently receives one eye gene from each of its parents, and the gene it receives from a parent is equally likely to be either of the two eye genes of that parent. Suppose that Smith and both of his parents have brown eyes, but Smith's sister has blue eyes. (a) What is the probability that Smith possesses a blue-eyed gene? (b) Suppose that Smith's wife has blue eyes. What is the probability that their first child will have blue eyes? (c) If their first child has brown eyes, what is the probability that their next child will also have brown eyes?

Suppose that \(n\) independent trials, each of which results in any of the outcomes \(0,1,\) or \(2,\) with respective probabilities \(p_{0}, p_{1},\) and \(p_{2}, \sum_{i=0}^{2} p_{i}=1\) are performed. Find the probability that outcomes 1 and 2 both occur at least once.

Consider two boxes, one containing 1 black and 1 white marble, the other 2 black and 1 white marble. A box is selected at random, and a marble is drawn from it at random. What is the probability that the marble is black? What is the probability that the first box was the one selected given that the marble is white?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.