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Chapter 5: Problem 25

Draw five cards randomly from a standard deck of 52 cards, and let \(x\) be the number of red cards in the draw. Evaluate the probabilities in Exercises \(22-25\). \(P(x \leq 1)\)

Short Answer

Expert verified
Answer: The probability of drawing one or fewer red cards from a standard deck of 52 cards when drawing five cards randomly is 3/16.

Step by step solution

01

Define the binomial probability formula

The binomial probability formula is defined as: \(P(x) = {n \choose x} p^x (1-p)^{n-x}\), where: - \(n\) is the total number of trials (in our case, 5 card draws) - \(x\) is the number of successful trials we are interested in (0 or 1 red card) - \(p\) is the probability of success on a single trial (probability of drawing a red card in one draw) - \({n \choose x}\) is the number of ways to choose \(x\) successes from \(n\) trials, which is also known as the binomial coefficient, calculated as \({n \choose x} = \frac{n!}{x!(n-x)!}\).
02

Calculate the probability of drawing a red card in one draw

There are 26 red cards and a total of 52 cards in the deck. Therefore, the probability of drawing a red card in one draw is: \(p = \frac{26}{52} = \frac{1}{2}\)
03

Calculate the probability of drawing 0 red cards

Using the binomial probability formula with \(n=5\), \(x=0\), and \(p=\frac{1}{2}\), we get: \(P(x=0) = {5 \choose 0} \left(\frac{1}{2}\right)^0 \left(1-\frac{1}{2}\right)^{5-0} = 1 \cdot 1 \cdot \left(\frac{1}{2}\right)^5 = \frac{1}{32}\)
04

Calculate the probability of drawing 1 red card

Using the binomial probability formula with \(n=5\), \(x=1\), and \(p=\frac{1}{2}\), we get: \(P(x=1) = {5 \choose 1} \left(\frac{1}{2}\right)^1 \left(1-\frac{1}{2}\right)^{5-1} = 5 \cdot \frac{1}{2} \cdot \left(\frac{1}{2}\right)^4 = 5 \cdot \frac{1}{32} = \frac{5}{32}\)
05

Calculate the probability of drawing one or fewer red cards

To find the probability of drawing one or fewer red cards, we sum the probabilities of drawing 0 and 1 red card: \(P(x \leq 1) = P(x=0) + P(x=1) = \frac{1}{32} + \frac{5}{32} = \frac{6}{32} = \frac{3}{16}\) Therefore, the probability of drawing one or fewer red cards from a standard deck of 52 cards when drawing five cards randomly is \(\frac{3}{16}\).

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Key Concepts

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

Probability Theory
At its core, probability theory is a branch of mathematics concerned with the analysis of random phenomena. It provides a way to quantify the likelihood of different outcomes in uncertain situations. For instance, when you draw cards from a deck, you are performing what's called an experiment, and the result of this experiment is uncertain ahead of time.

The fundamental idea is to assign a numerical value, called a probability, to the likelihood of certain outcomes of the experiment. Probability ranges between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. The sum of the probabilities of all possible outcomes of a random experiment must equal 1. In the solved exercise, each event of drawing a card is a trial, and we want to find the probability of certain outcomes (the number of red cards drawn) after these trials.
Combinatorics
In relation to probability, combinatorics deals with the counting, arrangement, and combination of objects. It's a foundational part of understanding probability because it helps us determine the possible ways an event can occur.

The binomial coefficient, notated as \( {n choose x} \) and part of the binomial probability formula, is a classical result from combinatorics. It tells us how many ways we can choose \( x \) successes out of \( n \) trials without caring about the order. For example, choosing 1 red card out of 5 cards can happen in multiple ways; combinatorics gives us the tools to count these possibilities. Factoring in this counting is crucial when we want to know how likely an event is.
Binomial Distribution
The binomial distribution is a probability distribution that summarizes the likelihood that a value will take on one specific value. It applies when there are a fixed number of independent trials, each with the same probability of success.

In other words, it's a discrete distribution that describes the number of successes in a sequence of independent experiments. The solved exercises use the binomial probability formula which embodies this distribution. The goal is to calculate the probability that we get a specific number of 'successes' (in this case, drawing red cards) across a certain number of trials (card draws), which is a perfect setting for the binomial distribution to apply.

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Most popular questions from this chapter

Let \(x\) be a binomial random variable with \(n=7\) and \(p=.5 .\) Find the values of the quantities in Exercises \(11-15 .\) $$ \sigma=\sqrt{n p q} $$

A company has five applicants for two positions: two women and three men. Suppose that the five applicants are equally qualified and that no preference is given for choosing either gender. Let \(x\) equal the number of women chosen to fill the two positions. a. Write the formula for \(p(x)\), the probability distribution of \(x\) b. What are the mean and variance of this distribution? c. Construct a probability histogram for \(x\).

The National Hockey League (NHL) has about \(70 \%\) of its players born outside the United States, and of those born outside the United States, approximately \(60 \%\) were born in Canada. \({ }^{5}\) Suppose that \(n=12\) NHL players are selected at random. Let \(x\) be the number of players in the sample born outside of the United States so that \(p=.7,\) and find the following probabilities: a. At least five or more of the sampled players were born outside the United States b. Exactly seven of the players were born outside the United States c. Fewer than six were born outside the United States.

Let \(x\) be a binomial random variable with \(n=7\) and \(p=.5 .\) Find the values of the quantities in Exercises \(11-15 .\) $$ \mu=n p $$

In a certain population, \(85 \%\) of the people have Rh-positive blood. Suppose that two people from this population marry. What is the probability that they are both Rh-negative, thus making it inevitable that their children will be Rh-negative?
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