/*! 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 50 A fair die is rolled four times.... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A fair die is rolled four times. Let the random variable \(X\) denote the number of 6 's that appear. Find and graph the cdf for \(X\).

Short Answer

Expert verified
The cdf of \(X\) can be obtained using the binomial formula for each possible value of \(X\) (0 to 4) and then summing these probabilities to get cumulative probabilities. The graph of the cdf starts at (0, P(X=0)) and has points at \((k, sum_{i=0}^{k} P(X=i))\) for each \(k\) from 1 to 4.

Step by step solution

01

Understanding the Variables

Identify the variables in the binomial distribution. Here, the number of trials \(n\) is 4 (the die is rolled four times), the probability of success \(p\) in each trial is \(1/6\) (chance of rolling a 6), and \(X\) is the number of successes (rolling a 6). Since the die is fair, each roll is an independent event.
02

Compute the Binomial cdf

The cumulative distribution function (cdf) is the sum of the probabilities of all outcomes up to and including a given value. Here, it's the sum of the probabilities of rolling 0, 1, 2, 3, or 4 sixes. To compute this, one needs to use the binomial formula which is \(P(X=k) = C(n,k) \cdot p^k \cdot (1-p)^{n-k}\), where \(C(n,k)\) represents the combination of \(n\) values taken \(k\) at a time. For each \(k \), one plugs the values into this formula and add the results.
03

Graph the cdf

Plot the cdf values obtained in step 2. The x-axis represents the number of successes (\(X\)) and the y-axis represents the cdf. At each integer value of \(X\) from 0 to n (inclusive), the height of the graph will be the cdf value for that \(X\). Since it's a cdf, the graph is non-decreasing and bounded between 0 and 1.

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.

Binomial Distribution
The binomial distribution is a cornerstone of probability theory, utilized when we want to know the likelihood of a certain number of successes in a fixed number of independent trials. To fit this model, each trial must have only two possible outcomes, often referred to as 'success' and 'failure'.

In the example of rolling a fair die four times, we're dealing with a binomial distribution because each roll represents an independent trial with two possible outcomes: rolling a 6 ('success') or not ('failure'). Here, the fixed number of trials is four, denoted by the variable 'n', while the probability of rolling a 6 in a single trial is 1/6, which we call 'p'. The variable 'X' is used to represent the total number of successes across all trials—so, 'X' could be any integer from 0 to 4 when rolling the die four times.
Probability Mass Function
In congruence with the binomial distribution, there exists a corresponding probability mass function (PMF) which quantifies the probability of each possible number of successes. For a binomially distributed random variable, the PMF calculates the probability of obtaining exactly 'k' successes in 'n' trials.

To compute the PMF, one uses the formula: \( P(X=k) = C(n,k) \cdot p^k \cdot (1-p)^{n-k} \). Within this equation, 'C(n,k)' is the number of combinations of 'n' trials taken 'k' at a time, 'p^k' represents the probability of success raised to the power of 'k', and '(1-p)^{n-k}' accounts for the probability of failure for the remaining trials. By applying this formula to 'k' from 0 to 'n', we can determine each discrete outcome’s probability and ultimately create a complete probability distribution.
Combinations in Statistics
Combinations are a basic concept in statistics that refer to the act of selecting items from a group, such that the order of selection does not matter. A common notation for combinations is 'C(n, k)' or sometimes '\(\binom{n}{k}\)', also known as the binomial coefficient.

It calculates the total number of unique ways 'k' items can be chosen from a set of 'n' items. The mathematical formula to find 'C(n, k)' is: \( C(n, k) = \frac{n!}{k! \cdot (n - k)!} \), where 'n!' denotes the factorial of 'n'. In our dice-rolling exercise, combinations are significant because they determine the different ways you can achieve a certain number of sixes (successes) across multiple rolls (trials). As 'n' and 'k' change, so does the count of combinations, directly impacting the probability calculations for the PMF.
Independent Events
When we discuss independent events in probability, we mean that the outcome of one event does not influence the outcome of another. This principle is vital for calculating probabilities in scenarios where multiple events occur, such as rolling a die multiple times.

In the given exercise, each roll of the die is an independent event because the result of one roll does not affect the others. This is a fundamental assumption of the binomial distribution, as it allows for the multiplication of individual probabilities across trials. Independence is necessary for the binomial formula to hold true, ensuring that the complex behaviors seen in dependent events, where outcomes can be swayed by previous results, do not complicate the calculation.

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 \(f_{X, Y}(x, y)=6(1-x-y)\) for \(x\) and \(y\) defined over the unit square, subject to the restriction that \(0 \leq x+y \leq 1\). Find the marginal pdf for \(X\).

Suppose the charge for repairing an automobile averages \(\$ 200\) with a standard deviation of \(\$ 16\). If a \(10 \%\) \(\operatorname{tax}\) is added to the charge and then a \(\$ 15\) flat fee for environmental impact, what is the standard deviation of the charge to the car owner?

Let \(X\) and \(Y\) have the joint pdf $$ f_{X, Y}(x, y)=2 e^{-(x+y)}, \quad 0

Recovering small quantities of calcium in the presence of magnesium can be a difficult problem for an analytical chemist. Suppose the amount of calcium \(Y\) to be recovered is uniformly distributed between 4 and \(7 \mathrm{mg}\). The amount of calcium recovered by one method is the random variable $$ W_{1}=0.2281+(0.9948) Y+E_{1} $$ where the error term \(E_{1}\) has mean 0 and variance \(0.0427\) and is independent of \(Y\). A second procedure has random variable $$ W_{2}=-0.0748+(1.0024) Y+E_{2} $$ where the error term \(E_{2}\) has mean 0 and variance \(0.0159\) and is independent of \(Y\). The better technique should have a mean as close as possible to the mean of \(Y(=5.5)\), and a variance as small as possible. Compare the two methods on the basis of mean and variance.

An urn contains one white chip and one black chip. A chip is drawn at random. If it is white, the "game" is over; if it is black, that chip and another black one are put into the urn. Then another chip is drawn at random from the "new" urn and the same rules for ending or continuing the game are followed (i.e., if the chip is white, the game is over; if the chip is black, it is placed back in the urn, together with another chip of the same color). The drawings continue until a white chip is selected. Show that the expected number of drawings necessary to get a white chip is not finite.

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.