/*! 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 77 Three brothers and their wives d... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Three brothers and their wives decide to have children until each family has two female children. What is the pmf of \(X=\) the total number of male children born to the brothers? What is \(E(X)\), and how does it compare to the expected number of male children born to each brother?

Short Answer

Expert verified
The pmf of X is a sum of three negative binomial distributions. \(E(X) = 6\), which is three times the expected male count for one brother (2).

Step by step solution

01

Understanding the Problem and Definitions

Recognize that each brother's family will continue having children until there are two female children. The random variable \(X\) denotes the total number of male children born across all three families. We also need to find the expected value \(E(X)\).
02

Analyze Single Family's Childbirth Process

For each family, children are born until there are two female children. The probability of having a female child in each birth (assuming equal chances) is \(0.5\), and these births are independent of one another. Let \(Y_i\) denote the number of male children born to the \(i\)-th family.
03

Determine the Distribution of Number of Male Children for One Family

The number of male children until the second female child can be modeled as a Negative Binomial distribution. Specifically, \(Y_i\) for the \(i\)-th family is \(\text{NegBinom}(k=2, p=0.5)\), where \(k\) is the number of successes (female children) and \(p\) is the probability of success in each trial.
04

Calculate PMF of Y_i

The probability mass function for \(Y_i\) is given by \( P(Y_i = y) = \binom{y+2-1}{2-1} (0.5)^2 (0.5)^y = \binom{y+1}{1} (0.5)^{y+2} \).
05

Calculate Expectation for a Single Family

The expected number of male children to be born per family is given by \(E(Y_i) = \frac{2(1-p)}{p} = \frac{2(0.5)}{0.5} = 2\). Thus, on average, each family will have 2 male children before having two females.
06

Sum Up for All Families

Since \(X = Y_1 + Y_2 + Y_3\) where \(Y_1, Y_2, Y_3\) are independent Negative Binomial random variables, the expected total number of male children across all three families is \(E(X) = E(Y_1) + E(Y_2) + E(Y_3) = 2 + 2 + 2 = 6\).
07

Verification and Conclusion

Each family's expected male children count was found to be 2, and since there are three identical and independent families, the combined expected count is a straightforward multiplication by the number of families (3 in this case). The analysis confirms that the calculations are consistent with the expected results.

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 Mass Function
The concept of a Probability Mass Function (PMF) is fundamental when dealing with discrete random variables such as the number of male children born in a family. A PMF specifies the probability that a discrete random variable is equal to a particular value.

In the context of the negative binomial distribution, the PMF helps in determining the probability of a specific number of trials needed to achieve a given number of successes. Here, a "success" is defined as the birth of a female child, and we are specifically interested in modeling the scenario until two female children are born.

For a single family, the PMF of the number of male children until the second female child is: \( P(Y_i = y) = \binom{y+1}{1} (0.5)^{y+2} \)
This formula tells us the probability of having \( y \) male children before the second female child is born. The binomial coefficient \( \binom{y+1}{1} \) accounts for the different sequences in which these births can occur.
Expectation and Expected Value
Expectation, or expected value, is an essential concept in probability, especially when dealing with random variables. It provides a measure of the center of the distribution of the random variable, essentially giving an "average" outcome if the experiment were repeated numerous times.

For the Negative Binomial distribution, the expectation can be calculated using the formula for a single family's expected number of male children, which is: \( E(Y_i) = \frac{k(1-p)}{p} \)
In this specific scenario, \( k = 2 \) (since we want two female children), and the probability \( p \) of having a female child is \( 0.5 \). Thus, the expected number of male children per family comes out to be 2.

Since the families are assumed to be identical and independent, the total expected number of male children born to all three families is simply the sum of the expected numbers for each family: \( E(X) = 2 + 2 + 2 = 6 \). This illustrates that across the three families, we expect, on average, a total of 6 male children.
Independent Random Variables
Understanding independence between random variables is crucial when summing or analyzing their distributions, especially in probability. Random variables are considered independent if the occurrence or outcome of one does not affect the occurrence of another.

In the case of our exercise, the assumption is that each family's sequence of births is independent of the others. This means that the number of male children born to one family does not influence the number of male children another family has.

Mathematically, if \( Y_1, Y_2, \) and \( Y_3 \) are the number of male children for each family, independence implies that:
  • \( P(Y_1, Y_2, Y_3) = P(Y_1) \times P(Y_2) \times P(Y_3) \)
  • The expectation of their sum is the sum of their expectations: \( E(X) = E(Y_1) + E(Y_2) + E(Y_3) \)
The idea of independent random variables simplifies calculations, making it feasible to find the expected count of male children across all families by simply summing individual expectations.

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

A concrete beam may fail either by shear \((S)\) or flexure \((F)\). Suppose that three failed beams are randomly selected and the type of failure is determined for each one. Let \(X=\) the number of beams among the three selected that failed by shear. List each outcome in the sample space along with the associated value of \(X\).

In proof testing of circuit boards, the probability that any particular diode will fail is .01. Suppose a circuit board contains 200 diodes. a. How many diodes would you expect to fail, and what is the standard deviation of the number that are expected to fail? b. What is the (approximate) probability that at least four diodes will fail on a randomly selected board? c. If five boards are shipped to a particular customer, how likely is it that at least four of them will work properly? (A board works properly only if all its diodes work. )

Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with parameter \(\mu=20\) (suggested in the article "'Dynamic Ride Sharing: Theory and Practice," J. of Transp. Engr., 1997: 308-312). What is the probability that the number of drivers will a. Be at most 10 ? b. Exceed 20? c. Be between 10 and 20 , inclusive? Be strictly between 10 and 20 ? d. Be within 2 standard deviations of the mean value?

A toll bridge charges \(\$ 1.00\) for passenger cars and \(\$ 2.50\) for other vehicles. Suppose that during daytime hours, \(60 \%\) of all vehicles are passenger cars. If 25 vehicles cross the bridge during a particular daytime period, what is the resulting expected toll revenue? [Hint: Let \(X=\) the number of passenger cars; then the toll revenue \(h(X)\) is a linear function of \(X .]\)

Let \(X=\) the number of nonzero digits in a randomly selected 4-digit PIN that has no restriction on the digits. What are the possible values of \(X\) ? Give three possible outcomes and their associated \(X\) values.

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.