Question

According to an article in the August $30,2002$ issue of the Chronicle of Higher Education, 30$\%$ on of first-year college students are liberals, 20$\%$ are conservatives, and 50$\%$ characterize them- selves as middle-of-the-road. Choose two students at random, let $X$ be the number of liberals among the two, and let $Y$ be the number of conservatives among the two. (a) Using the multinomial distribution from Sect. $4.1,$ give the joint probability mass function $p$ $(x, y)$ of $X$ and $Y$ and the corresponding joint probability table. (b) Determine the marginal probability mass functions by summing $p(x, y)$ numerically. How could these be obtained directly? [Hint: What are the uni variate distributions of $X$ and $Y ? ]$ (c) Determine the conditional probability mass function of $Y$ given $X=x$ for $x=0,1,2 .$ Com pare this to the binomial distribution with $n=2-x$ and $p=.2 /(.2+.5) .$ Why should this work? (d) Are $X$ and $Y$ independent? Explain. (e) Find $E(Y | X=x)$ for $x=0,1,2 .$ Do this numerically and then compare with the use of the formula for the binomial mean, using the binomial distribution given in part (c). (f) Determine $\operatorname{Var}(Y Y X=x)$ for $x=0,1,2 .$ Do this numerically and then compare with the use of the formula for the binomial variance, using the binomial distribution given in part (c).

Name: Problem 75, Chapter 4: Probability with Applications in Engineering, Science, and Technology
Uploaded: 2020-10-02T10:29:29-08:00
Duration: 17 min 33 s
Channel: Chris Trentman

   According to an article in the August $30,2002$ issue of the Chronicle of Higher Education, 30$\%$ on
of first-year college students are liberals, 20$\%$ are conservatives, and 50$\%$ characterize them-
selves as middle-of-the-road. Choose two students at random, let $X$ be the number of liberals
among the two, and let $Y$ be the number of conservatives among the two.
(a) Using the multinomial distribution from Sect. $4.1,$ give the joint probability mass function $p$
$(x, y)$ of $X$ and $Y$ and the corresponding joint probability table.
(b) Determine the marginal probability mass functions by summing $p(x, y)$ numerically. How
could these be obtained directly? [Hint: What are the uni variate distributions of $X$ and $Y ? ]$
(c) Determine the conditional probability mass function of $Y$ given $X=x$ for $x=0,1,2 .$ Com
pare this to the binomial distribution with $n=2-x$ and $p=.2 /(.2+.5) .$ Why should this
work?
(d) Are $X$ and $Y$ independent? Explain.
(e) Find $E(Y | X=x)$ for $x=0,1,2 .$ Do this numerically and then compare with the use of the
formula for the binomial mean, using the binomial distribution given in part (c).
(f) Determine $\operatorname{Var}(Y Y X=x)$ for $x=0,1,2 .$ Do this numerically and then compare with the use of the formula for the binomial variance, using the binomial distribution given in part (c).

Probability with Applications in Engineering, Science, and Technology

Matthew A. Carlton •… 2nd Edition

Chapter 4, Problem 75

Instant Answer

Solved by Expert Chris Trentman

10/02/2020

Step 1

We are asked to find the joint probability mass function $p(x, y)$ of $X$ and $Y$ using the multinomial distribution. The multinomial distribution is given by: $$p(x,y)=\dfrac{n!}{x!y!(n-x-y)!}p_1^xp_2^yp_3^{n-x-y}$$ where $n$ is the total number of trials, $x$ Show more…

Show all steps

Thanks for your feedback!

According to an article in the August $30,2002$ issue of the Chronicle of Higher Education, 30$\%$ on of first-year college students are liberals, 20$\%$ are conservatives, and 50$\%$ characterize them- selves as middle-of-the-road. Choose two students at random, let $X$ be the number of liberals among the two, and let $Y$ be the number of conservatives among the two. (a) Using the multinomial distribution from Sect. $4.1,$ give the joint probability mass function $p$ $(x, y)$ of $X$ and $Y$ and the corresponding joint probability table. (b) Determine the marginal probability mass functions by summing $p(x, y)$ numerically. How could these be obtained directly? [Hint: What are the uni variate distributions of $X$ and $Y ? ]$ (c) Determine the conditional probability mass function of $Y$ given $X=x$ for $x=0,1,2 .$ Com pare this to the binomial distribution with $n=2-x$ and $p=.2 /(.2+.5) .$ Why should this work? (d) Are $X$ and $Y$ independent? Explain. (e) Find $E(Y | X=x)$ for $x=0,1,2 .$ Do this numerically and then compare with the use of the formula for the binomial mean, using the binomial distribution given in part (c). (f) Determine $\operatorname{Var}(Y Y X=x)$ for $x=0,1,2 .$ Do this numerically and then compare with the use of the formula for the binomial variance, using the binomial distribution given in part (c).

Play audio

Feedback

Chris Trentman and 71 other subject Intro Stats / AP Statistics educators are ready to help you.

Ask a new question

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Joint Probability Mass Function

The joint probability mass function (pmf) describes the probability that two (or more) discrete random variables simultaneously take on specific values. It is central to understanding the combined behavior of multiple random variables and is often derived from underlying distributions such as the multinomial distribution.

Expected Value

The expected value of a random variable is a measure of its central tendency, calculated as the weighted average of all possible values the variable can take, with each value weighted by its probability. It provides insight into the long-run average outcome when the experiment is repeated many times.

Variance

Variance is a measure of the dispersion or spread in the values of a random variable around its expected value. It quantifies the degree to which the outcomes differ from the mean and is calculated as the expected value of the squared deviation from the mean.

Independence

Two random variables are independent if the occurrence of one does not affect the probability of occurrence of the other. In terms of their joint pmf, independence is characterized by the fact that the joint pmf is equal to the product of the marginal pmf of the two variables.

Multinomial Distribution

This distribution generalizes the binomial distribution to experiments where each trial can result in more than two outcomes. It is used to model the probabilities of counts for each category in cases where a fixed number of independent trials is conducted, with each trial having a constant probability for each outcome.

Conditional Probability Mass Function

The conditional probability mass function gives the probability that one random variable takes a particular value given that another variable is set to a certain value. It is a key concept for understanding dependencies between random variables and is computed by dividing the joint probability by the marginal probability of the conditioning event.

Marginal Distribution

The marginal distribution of a random variable is obtained by summing or integrating the joint distribution over the possible values of other variables. This provides the probabilities associated with a single variable irrespective of the values of the others, capturing its individual behavior.

Binomial Distribution

The binomial distribution is a probability distribution that models the number of successes in a fixed number of independent and identically distributed Bernoulli trials, each with the same probability of success. This distribution is frequently applied in contexts where trials have two possible outcomes, such as success or failure.

Recommended Videos

Transcript

00:04 We're given information from an article from an issue of the chronicle of higher education describing the political identities of first -year college students, and we're given two students at random, and we're told that x is the random variable, denoting the number of liberals among the two students, and why will be the number of conservatives among the two students.

00:37 In part a, we're asked to use the multinomial distribution to give the joint probability mass function p of x, y of variables x and y, and the corresponding joint probability table.

01:00 So we have that for valid x and y, and determine what this is later, the joint probability p of x, y, well, because it's a multinomial distribution, we have two students to pick from, this is going to be two factorial over probability, or sorry, over x factorial times y factorial, times the number of students, 2 minus x minus y factorial, times the probability of liberal, which is 0 .3 to the number of liberals, times probability conservative, which is 0 .2 to y, the number of conservatives, tends the probability of middle of the road, 0 .5 to the number of middle roads, which is going to be 2 minus x minus y.

02:17 And we can display these in a table.

02:21 So on the left -hand side, i'll include we have the rows will be for x and the columns for y.

02:36 So we have x could be 0 or 1, and likewise, 0 1 or 2, i mean, and y could also be 0, 1 or 2.

02:53 We have that if both x and y are 0, we get 2 factorial over 2 factorial is 1 times 1 times 1 times 0 .5 squared.

03:06 This is 0 .25.

03:13 And although it's tedious, it's easy to calculate the rest of these values.

03:19 So i'm just going to write them here.

03:21 We have that f of 0 .01 is 0 .2.

03:29 F of 0 .02 is 0 .04.

03:35 F of 1 .0 is 0 .3.

03:42 F of 1 .1 is 0 .12 .2.

03:48 F of 1 2 is, well, consider this.

03:54 This formula only applies for valid x and y.

03:58 We can't have the sum of x and y.

04:01 Is greater than 2, but 1 plus 2 is, in fact, 3, which is greater than 2.

04:06 So the probability of this has to be 0.

04:10 We have that if x is 2 and y is 0, then p of 2, 0 is according to our formula, 0 .09.

04:24 And again, we see that we can't have that x is equal to 2 and y is equal to 1, the non -zero probability, because that's 3 students when we only have the sample size of 2.

04:35 This is zero, and likewise, if x is 2 and y is 2, and this is also 0.

04:46 In part b, we're asked to determine the marginal probability mass functions by summing the joint probability p of x, y, numerically.

05:10 We have that the marginal distributions of x and y, since the joint distribution was multinomial, the marginal distributions are going to be binomial, where distribution of x is binomial with parameters 2 and 0 .3, and the distribution of y is binomial with parameters 2 and 0 .2, so respectively.

06:03 Now, from these or from the summation in the table, we have that we can make tables of the marginal probabilities.

06:14 So we have, across the columns, i'll have x.

06:21 The first column is x, and the second, sorry, the first row will be x's, and the second row will be marginal probabilities.

06:35 So we have the x can once again be anywhere from zero, one up to two, and we have that, looking at our table, or using these binomial distributions.

06:48 I think looking at the table is going to be easiest here.

06:50 We have summing across the column, or across the row for x equals 0 .25 plus 0 .20 plus 0 .04.

07:02 This is 0 .49...

Ace is your personal tutor. It breaks down any question with clear steps so you can learn.

Start Using Ace