/*! 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 20 Let \(X_{1}, X_{2}, X_{3}, X_{4}... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(X_{1}, X_{2}, X_{3}, X_{4}, X_{5}\), and \(X_{6}\) denote the numbers of blue, brown, green, orange, red, and yellow M\&M candies, respectively, in a sample of size \(n\). Then these \(X_{\text {'s have }}\) a multinomial distribution. According to the \(\mathrm{M} \& \mathrm{M}\) Web site, the color proportions are \(p_{1}=.24, p_{2}=.13\), \(p_{3}=.16, p_{4}=.20, p_{5}=.13\), and \(p_{6}=.14\). a. If \(n=12\), what is the probability that there are exactly two M\&Ms of each color? b. For \(n=20\), what is the probability that there are at most five orange candies? [Hint: Think of an orange candy as a success and any other color as a failure.] c. In a sample of \(20 \mathrm{M} \& \mathrm{Ms}\), what is the probability that the number of candies that are blue, green, or orange is at least 10 ?

Short Answer

Expert verified
a. Compute with multinomial probability formula. b. Use binomial CDF for at most 5. c. Use binomial for at least 10.

Step by step solution

01

Understand the Problem

We are given that the M&M color distribution follows a multinomial distribution with defined probabilities for each color: blue (0.24), brown (0.13), green (0.16), orange (0.20), red (0.13), and yellow (0.14). We need to solve three parts of the problem using these probabilities.
02

Solve Part a - Exactly two of each color for n=12

We use the multinomial probability formula where we need exactly 2 M&Ms of each color, which implies each color should appear twice. The formula is: \[ P(X_1=x_1, X_2=x_2, ..., X_k=x_k) = \frac{n!}{x_1!x_2!...x_k!} p_1^{x_1} p_2^{x_2} ... p_k^{x_k} \] Here, \(x_i = 2\) for all colors, \(n = 12\). The probability is: \[ \frac{12!}{2!2!2!2!2!2!} (0.24)^2 (0.13)^2 (0.16)^2 (0.20)^2 (0.13)^2 (0.14)^2 \] Calculate this to find the exact probability.
03

Solve Part b - At most five orange for n=20

Consider orange candy as a success with probability 0.20, and all other colors as failure with probability 0.80. We are looking for the probability of obtaining at most 5 orange candies (which is a binomial problem). Use the cumulative distribution function (CDF) of the binomial distribution: \[ P(X \leq 5) = \sum_{x=0}^{5} \binom{20}{x} (0.20)^x (0.80)^{20-x} \] Calculate this sum for probabilities from 0 to 5.
04

Solve Part c - At least 10 are blue, green, or orange for n=20

First, find the probability of blue, green, or orange collectively as a single event: \(p = 0.24 + 0.16 + 0.20 = 0.60\). We find the probability that at least 10 out of 20 candies are this event using binomial distribution: \[ P(Y \geq 10) = 1 - P(Y < 10) = 1 - \sum_{y=0}^{9} \binom{20}{y} (0.60)^y (0.40)^{20-y} \] Compute the sum from 0 to 9, subtract it from 1 for the final result.

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 Calculation
Probability calculation is an essential concept in statistics that helps us to determine the likelihood of a specific outcome occurring. In the context of M&M candies, we try to find the probability of having a specific number of candies of each color based on known color proportions. This involves using the multinomial distribution when multiple categories are involved, each with its known probability.
To compute probabilities in these scenarios, you understand that the sum of probabilities for all possible outcomes is 1. You use specific probability functions, like the multinomial coefficient, which extend the binomial coefficient for more than two categories. This helps calculate the exact probability of each outcome. Each probability calculation involves interpreting given color proportions and applying the appropriate probability formula.
Binomial Distribution
The binomial distribution is a fundamental probability distribution used when there are two possible outcomes for a single trial (like success/failure, yes/no). When considering M&M color proportions, such as calculating the probability of picking a certain number of orange candies (considered a success) from a pack, the binomial distribution comes into play.
For example, if you have a total of 20 candies and want to know the probability of having 5 or fewer orange candies, you'll use the binomial formula:
  • The formula: \( P(X \leq 5) = \sum_{x=0}^{5} \inom{20}{x} (0.20)^x (0.80)^{20-x} \)
  • It sums up the probabilities from 0 to 5 successes (orange candies here) using binomial coefficients.
You think of the event as having orange candy (success) with probability 0.20, and not having orange candy (failure) with probability 0.80. This distribution is suitable when each trial is independent, like picking each candy one by one.
Cumulative Distribution Function (CDF)
The cumulative distribution function (CDF) provides the probability that a random variable takes on a value less than or equal to a specific value. In terms of M&M colors, when determining the probability of having up to a certain number of a specific color, the CDF becomes valuable.
Using the binomial distribution, you can find out the probability of having at most a certain number of orange candies, for instance. The sum of probabilities from the start up to a set point describes how CDF is constructed. For problems requiring cumulative probabilities, you evaluate the function over each potential outcome from zero up to your point of interest, allowing you to understand the aggregation of probabilities up to a given threshold. This makes CDF very useful for understanding the spread across different outcomes, ensuring you're considering all outcomes up to your point of inquiry.
M&M Color Proportions
M&M color proportions are probabilities that express how frequently you expect to find each color in a random sample of candies. According to given data, you know the probability of each color, such as blue (0.24), brown (0.13), and so on.
Understanding these proportions is crucial for calculating the probability of specific outcomes, such as exactly having two candies of each color in a sample. These probabilities inform how to set up calculations using the multinomial or binomial distributions, depending on the nature of the problem at hand. By utilizing these known proportions, you can predict the likelihood of specific compositions of candies in a sample. This is particularly powerful for statisticians and enthusiasts who enjoy applying rigorous mathematics to understand real-world occurrences, even in something as fun as counting candy colors.

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 your waiting time for a bus in the morning is uniformly distributed on \([0,8]\), whereas waiting time in the evening is uniformly distributed on \([0,10]\) independent of morning waiting time. a. If you take the bus each morning and evening for a week, what is your total expected waiting time? [Hint: Define rv's \(X_{1}, \ldots, X_{10}\) and use a rule of expected value.] b. What is the variance of your total waiting time? c. What are the expected value and variance of the difference between morning and evening waiting times on a given day? d. What are the expected value and variance of the difference between total morning waiting time and total evening waiting time for a particular week?

In an area having sandy soil, 50 small trees of a certain type were planted, and another 50 trees were planted in an area having clay soil. Let \(X=\) the number of trees planted in sandy soil that survive 1 year and \(Y=\) the number of trees planted in clay soil that survive 1 year. If the probability that a tree planted in sandy soil will survive 1 year is \(.7\) and the probability of 1-year survival in clay soil is .6, compute an approximation to \(P(-5 \leq X-Y \leq 5\) ) (do not bother with the continuity correction).

Each customer making a particular Internet purchase must pay with one of three types of credit cards (think Visa, MasterCard, AmEx). Let \(A_{i}(i=1,2,3)\) be the event that a type \(i\) credit card is used, with \(P\left(A_{1}\right)=.5\), \(P\left(A_{2}\right)=.3\), and \(P\left(A_{3}\right)=.2\). Suppose that the number of customers who make such a purchase on a given day is a Poisson rv with parameter \(\lambda\). Define rv's \(X_{1}, X_{2}, X_{3}\) by \(X_{i}=\) the number among the \(N\) customers who use a type \(i\) card \((i=1,2,3)\). Show that these three rv's are independent with Poisson distributions having parameters \(.5 \lambda, .3 \lambda\), and \(.2 \lambda\), respectively. [Hint: For non- negative integers \(x_{1}, x_{2}, x_{3}\), let \(n=x_{1}+x_{2}+x_{3}\). Then \(P\left(X_{1}=x_{1}\right.\), \(\left.X_{2}=x_{2}, X_{3}=x_{3}\right)=P\left(X_{1}=x_{1}, X_{2}=x_{2}, X_{3}=x_{3}, N=n\right)\) [why is this?]. Now condition on \(N=n\), in which case the three \(X i\) 's have a trinomial distribution (multinomial with three categories) with category probabilities \(.5, .3\), and .2.]

A particular brand of dishwasher soap is sold in three sizes: \(25 \mathrm{oz}, 40 \mathrm{oz}\), and \(65 \mathrm{oz}\). Twenty percent of all purchasers select a \(25-0 z\) box, \(50 \%\) select a \(40-0 z\) box, and the remaining \(30 \%\) choose a \(65-\mathrm{oz}\) box. Let \(X_{1}\) and \(X_{2}\) denote the package sizes selected by two independently selected purchasers. a. Determine the sampling distribution of \(\bar{X}\), calculate \(E(\bar{X})\), and compare to \(\mu\). b. Determine the sampling distribution of the sample variance \(S^{2}\), calculate \(E\left(S^{2}\right)\), and compare to \(\sigma^{2}\).

In cost estimation, the total cost of a project is the sum of component task costs. Each of these costs is a random variable with a probability distribution. It is customary to obtain information about the total cost distribution by adding together characteristics of the individual component cost distributions-this is called the "roll-up" procedure. For example, \(E\left(X_{1}+\ldots+\right.\) \(\left.X_{n}\right)=E\left(X_{1}\right)+\cdots+E\left(X_{n}\right)\), so the roll-up procedure is valid for mean cost. Suppose that there are two component tasks and that \(X_{1}\) and \(X_{2}\) are independent, normally distributed random variables. Is the roll-up procedure valid for the 75 th percentile? That is, is the 75 th percentile of the distribution of \(X_{1}+X_{2}\) the same as the sum of the 75 th percentiles of the two individual distributions? If not, what is the relationship between the percentile of the sum and the sum of percentiles? For what percentiles is the roll-up procedure valid in this case?

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.