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Chapter 3: Problem 75

Suppose that \(p=P(\) male birth \()=.5\). A couple wishes to have exactly two female children in their family. They will have children until this condition is fulfilled. a. What is the probability that the family has \(x\) male children? b. What is the probability that the family has four children? c. What is the probability that the family has at most four children? d. How many male children would you expect this family to have? How many children would you expect this family to have?

Short Answer

Expert verified
a. \( P(X=x) = \binom{x+1}{x} (0.5)^{x+2} \). b. \( 0.1875 \). c. \( 0.6875 \). d. Expect 2 males, 4 total children.

Step by step solution

01

Model the Stopping Criterion

The couple stops having children once they have two female children. This scenario can be modeled using a negative binomial distribution because we are counting the number of male children before achieving 2 females.
02

Define Variables

Let \( F \) be the event of a female birth with probability \( p_F = 0.5 \) and \( M \) be the event of a male birth with probability \( p_M = 0.5 \). We want to find the number of male children \( x \) until 2 females are born.
03

Calculate the Probability of x Male Children

The probability distribution is given by the negative binomial formula: \[ P(X = x) = \binom{x+1}{x} p_F^r (1-p_F)^x \]where \( x \) is the number of male children and \( r = 2 \) is the number of female children ("successes").Thus, for \( x \) male children: \[ P(X = x) = \binom{x+1}{x} (0.5)^2 (0.5)^x = \binom{x+1}{x} (0.5)^{x+2} \].
04

Calculate Probability for Exactly Four Children

Having exactly four children means 2 females and 2 males (i.e., m+m+f+f or m+f+m+f, etc.). In this context, the arrangement for one success pattern is, for example, m+f+m+f. Total arrangements \( = \binom{4-1}{2-1} \) since the last child must be female. - The number of such sequences with exactly two females and two males is \( \binom{3}{1} = 3 \).- Therefore, \( P(\text{4 children}) = 3 \times \left(0.5^4\right) = 3 \times \frac{1}{16} = \frac{3}{16} \approx 0.1875 \).
05

Calculate Probability for At Most Four Children

Sum the probabilities of having 2, 3, or 4 children.1. 2 children: both females, \( P(\text{2 children}) = 0^0.5 \times 0^0.5 = 0.25 \). 2. 3 children: 1 male (m+f+f or f+m+f), \( P(\text{3 children}) = \binom{2}{1} \times 0.5^3 = 2 \times \frac{1}{8} = \frac{1}{4} \).3. Already calculated 4 children: \( \frac{3}{16} \).Add them up:\[ P(\text{at most 4 children}) = \frac{1}{4} + \frac{1}{4} + \frac{3}{16} = \frac{4}{16} + \frac{4}{16} + \frac{3}{16} = \frac{11}{16} \approx 0.6875 \].
06

Expected Number of Male and Total Children

For expected number of males: \( E[X] = \frac{r(1-p)}{p} = \frac{2 \times 0.5}{0.5} = 2 \) males.Total expected children \( = E[X] + r = 2 + 2 = 4 \) children.

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Key Concepts

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

Understanding Probability in Negative Binomial Distribution
Probability is the measure of the likelihood of an event occurring. In our scenario, we want to find the probability of having a certain number of male children before the couple achieves their goal of having two female children. This situation can be described using a negative binomial distribution.

The negative binomial distribution is crucial when we are interested in the probability of having a number of failures (male births, in this case) before a fixed number of successes (female births). The formula for calculating the probability that the couple has exactly \( x \) male children before having two females is given by:
\[ P(X = x) = \binom{x+1}{x} (0.5)^{x+2} \] where \( p = 0.5 \) is the probability of having a female child. Thus, the arrangement of births influences what is considered a "success" in this scenario.
Expected Value in Random Experiments
The expected value gives us a form of the average outcome if an experiment is repeated a vast number of times. For the couple in question, we are considering the expected number of male children before achieving two female children.

In the negative binomial distribution, the expected value can be derived using the formula: \[ E[X] = \frac{r(1-p)}{p} \] where \( r \) is the number of successes (2 females), \( p \) is the probability of success (0.5 for females), yielding \( E[X] = 2 \) male children.

Moreover, the total expected number of children, including both males and females, is calculated by adding \( r \) to the expected value of male children, resulting in 4 expected total children.
Exploring Random Variables in Birth Process
A random variable is a numerical description of the outcomes of a random phenomenon. In this exercise, we define a random variable \( X \) to represent the number of male births until two female births occur. A key characteristic of this random variable is that it is inherently discrete because it can only take non-negative integer values.

By understanding \( X \), we can analyze the probability of each possible outcome, such as having exactly three male children before having two females. Each outcome's probability depends on the binomial coefficient, which adjusts the likelihood based on the number of previous children born.
Stopping Criterion in Family Planning
The stopping criterion is the rule that determines when the random process ends. In the couple's plan to have children, the stopping criterion occurs once they have two female children. This criterion defines the end point for counting the number of male children.

Applying this criterion means using the negative binomial distribution to compute probabilities, such as stopping after two females and a certain number of males. The negative binomial model's structure directly results from this stopping rule, as it dictates the sequence needed for counting events leading to the desired outcome.

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Most popular questions from this chapter

Airlines sometimes overbook flights. Suppose that for a plane with 50 seats, 55 passengers have tickets. Define the random variable \(Y\) as the number of ticketed passengers who actually show up for the flight. The probability mass function of \(Y\) appears in the accompanying table. \begin{tabular}{l|lllllllllll} \(y\) & 45 & 46 & 47 & 48 & 49 & 50 & 51 & 52 & 53 & 54 & 55 \\ \hline\(p(y)\) & \(.05\) & \(.10\) & \(.12\) & \(.14\) & 25 & \(.17\) & \(.06\) & \(.05\) & \(.03\) & \(.02\) & \(.01\) \end{tabular} a. What is the probability that the flight will accommodate all ticketed passengers who show up? b. What is the probability that not all ticketed passengers who show up can be accommodated? c. If you are the first person on the standby list (which means you will be the first one to get on the plane if there are any seats available after all ticketed passengers have been accommodated), what is the probability that you will be able to take the flight? What is this probability if you are the third person on the standby list?

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Suppose \(E(X)=5\) and \(E[X(X-1)]=27.5\). What is a. \(E\left(X^{2}\right)\) ? \(\left[\right.\) Hint: \(E[X(X-1)]=E\left[X^{2}-X\right]=\) \(\left.E\left(X^{2}\right)-E(X)\right] ?\) b. \(V(X)\) ? c. The general relationship among the quantities \(E(X)\), \(E[X(X-1)]\), and \(V(X) ?\)

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