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Three cards are randomly selected, without replacement, from an ordinary deck of 52 playing cards. Compute the conditional probability that the first card selected is a spade given that the second and third cards are spades.

Short Answer

Expert verified
The conditional probability that the first card selected is a spade, given that the second and third cards are spades, is \(\frac{11}{50}\).

Step by step solution

01

Identify the events

Let A be the event that the first card selected is a spade, and let B be the event that the second and third cards are spades.
02

Find the probability of B

There are 13 spades in a deck of 52 cards. The probability of drawing two spades in a row (the event B) is: \(P(B) = \frac{13}{52} \times \frac{12}{51}\)
03

Find the probability of A and B

The event A and B means that all three cards are spades. The probability of drawing three spades in a row is: \(P(A \cap B) = \frac{13}{52} \times \frac{12}{51} \times \frac{11}{50}\)
04

Apply the formula for conditional probability

Now we will apply the formula for conditional probability, P(A|B) = P(A and B) / P(B). Thus: \(P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{13}{52} \times \frac{12}{51} \times \frac{11}{50}}{\frac{13}{52} \times \frac{12}{51}}\)
05

Cancel the common terms and calculate the result

We can cancel out common terms in the numerator and denominator: \(P(A|B) = \frac{\frac{11}{50}}{1}\) So the conditional probability that the first card selected is a spade, given that the second and third cards are spades, is \(\frac{11}{50}\).

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Key Concepts

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

Conditional Probability
In probability theory, conditional probability is a fundamental concept that helps us assess the likelihood of an event given that another event has already occurred. It is represented as \(P(A|B)\), which means "the probability of event A given that event B has happened."
To calculate it, you use the formula:\[P(A|B) = \frac{P(A \cap B)}{P(B)}\]This formula requires the probability of both events A and B happening, \(P(A \cap B)\), and the probability that event B happens, \(P(B)\).
Conditional probability comes in handy in many real-life scenarios where we want to know specific outcomes based on known conditions. For example, in card games, it helps in deciding the chances of drawing certain cards after others have been removed.
Combinatorics
Combinatorics is the branch of mathematics dealing with combinations and permutations. It is invaluable when we wish to count the number of possible arrangements or selections of objects. In the context of a standard 52-card deck, combinatorics helps us figure out different drawing scenarios, such as the number of ways to draw spades consecutively.
For example, when drawing three cards from the deck, you might want to know in how many ways this can happen. You start with 52 options, but once a card is drawn, fewer options remain. It's this process that combinatorics helps quantitate. Especially when drawing without replacement, combinatorics ensures precise calculations by considering reduced options with each draw.
  • Permutations account for order.
  • Combinations do not consider order.
This principle is crucial in problems that involve determining probabilities based on selections from a set that diminishes over time, like drawing cards from a deck.
Card Probability
Card probability is a specific application of probability theory to scenarios involving decks of playing cards. A standard deck has 52 cards and this uniform structure makes it an ideal model for studying probability.
When it comes to card probability, each draw affects the overall setup since we're often dealing with scenarios without replacement. Take, for instance, drawing consecutive cards of the same suit, such as spades. This involves understanding that each subsequent draw reduces the card count in that suit, altering the probability.
Probability calculations in card games involve:
  • Identifying the number of favorable outcomes, like specific card draws.
  • Determining the total number of possible outcomes remaining.
These calculations allow us to determine the likelihood of events as cards are drawn, as seen in the problem where the probability of the first card being spade relies on both the structure of the deck and the cards already drawn.

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Most popular questions from this chapter

In Laplace's rule of succession (Example \(5 \mathrm{e}\) ), are the outcomes of the successive flips independent? Explain.

If you had to construct a mathematical model for events \(E\) and \(F,\) as described in parts (a) through (e), would you assume that they were independent events? Explain your reasoning. (a) \(E\) is the event that a businesswoman has blue eyes, and \(F\) is the event that her secretary has blue eyes. (b) \(E\) is the event that a professor owns a car, and \(F\) is the event that he is listed in the telephone book. (c) \(E\) is the event that a man is under 6 feet tall, and \(F\) is the event that he weighs more than 200 pounds. (d) \(E\) is the event that a woman lives in the United States, and \(F\) is the event that she lives in the Western Hemisphere. (e) \(E\) is the event that it will rain tomorrow, and \(F\) is the event that it will rain the day after tomorrow.

\(A\) and \(B\) play a series of games. Each game is independently won by \(A\) with probability \(p\) and by \(B\) with probability \(1-p .\) They stop when the total number of wins of one of the players is two greater than that of the other player. The player with the greater number of total wins is declared the winner of the series. (a) Find the probability that a total of 4 games are played. (b) Find the probability that \(A\) is the winner of the series.

All the workers at a certain company drive to work and park in the company's lot. The company is interested in estimating the average number of workers in a car. Which of the following methods will enable the company to estimate this quantity? Explain your answer. 1\. Randomly choose \(n\) workers, find out how many were in the cars in which they were driven, and take the average of the \(n\) values. 2\. Randomly choose \(n\) cars in the lot, find out how many were driven in those cars, and take the average of the \(n\) values.

In a certain community, 36 percent of the families own a dog and 22 percent of the families that own a dog also own a cat. In addition, 30 percent of the families own a cat. What is (a) the probability that a randomly selected family owns both a dog and a cat? (b) the conditional probability that a randomly selected family owns a dog given that it owns a cat?

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