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Three brothers and their wives decide to have children until each family has two female children. Let \(X=\) 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 borm to each brother?

Short Answer

Expert verified
The expected number of male children is 6 for all families, and 2 per brother.

Step by step solution

01

Understanding the Problem

Each of the three brothers will have children until they have two daughters. Let us define the random variable \(X\) as the total number of male children born to all three brothers. The task is to find \(E(X)\), the expected value of \(X\).
02

Using Probability to Model the Situation

For a single family, the probability of having a girl (G) or a boy (B) is equal, \(P(G) = 0.5\) and \(P(B) = 0.5\). The family will stop having children once they have two girls. We need to calculate the expected number of boys each family will have before having two girls.
03

Calculating Expected Number of Boys for Each Family

Let's find the expected value \(E(Y)\) for one family, where \(Y\) is the number of boys born before two girls. Each family will stop after sequences like GG, BGG, BBG, etc. The probability distribution involves integrating over possible sequences that end with GG.
04

Deriving the Summation Formula for One Family

Each family can have sequences of boys and girls, ending in a GG. For example, BG, BGG, BBG, etc. The number of boys is key. For GG, 0 boys; for BGG, 1 boy; for BBGG, 2 boys, etc. So the distribution is given by a negative binomial distribution with parameters \(r = 2\) (two successes i.e., girls) and \(p = 0.5\).
05

Calculating Expected Number of Boys for One Family

Using the properties of the negative binomial distribution, the expected value \(E(Y)\) for one family is \(E(Y) = \frac{2(1-p)}{p} = \frac{2 \times 0.5}{0.5} = 2\). So, each family is expected to have 2 boys before having 2 girls.
06

Calculating the Expected Number of Boys for All Families

Since there are three brothers, and each brother independently has an expected 2 boys (before having 2 girls), the expected number of boys, \(E(X)\), for all three families is simply \(E(X) = 3 \times 2 = 6\).
07

Conclusion

The expected total number of male children across all three families is 6. This also means each brother, on average, is expected to have 2 boys before having enough girls.

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Key Concepts

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

Random Variable
In the realm of probability and statistics, a random variable represents a variable that can take on different values, each associated with a certain probability. These values stem from a predefined random phenomenon. In our exercise, the phenomenon involves families having children until two daughters are born.

Here, we define our random variable, \(X\), as the total number of male children born across three families. For each family, the sequence and outcome of having boys (B) and girls (G) until two daughters appear is random and hence fits the definition of a random variable.
  • The value of \(X\) is not fixed; it depends on the actual sequence of births.
  • Each brother's sequence of having children continues until two girls are born.
Random variables can be either discrete or continuous. In our case, \(X\) is discrete, as it can only take integer values (the count of boys). Understanding random variables helps us in further analyzing probability distributions and expectations.
Probability Distribution
A probability distribution maps the probabilities of all possible outcomes of a random variable. For instance, it provides a complete description of the likelihood that a random variable will take on each of its possible values. In our problem, we're interested in the distribution of boys born until the appearance of two girls.

This specific scenario is a classic example involving a particular type of probability distribution known as the negative binomial distribution, which we'll explore in the next section.
  • The distribution highlights the probabilities associated with sequences resulting in two daughters, like \(GG\), \(BGG\), etc.
  • Understanding this distribution allows us to calculate expected values and variances.
Probability distributions provide a lens through which we can quantify and understand the behavior of random phenomena, which in turn enables more complex calculations such as expected values.
Negative Binomial Distribution
The negative binomial distribution is a discrete probability distribution that models the number of trials needed to achieve a fixed number of successes in a sequence of independent and identically distributed Bernoulli trials. In our context, a "success" is defined as the birth of a girl.

Each family tries to achieve two successes—two girls (GG)—and the trials continue until this condition is met. The number of boys born during this period fits the negative binomial distribution:
  • The parameter \(r\) is the number of successes, which is 2 (two girls).
  • The probability \(p\) of a success on any trial is 0.5.
Using this distribution, we can determine the likelihood of different sequences that result in exactly two girls being born, like \(B\), \(BBG\), etc. This distribution provides the mathematical framework needed to compute the expected number of male children.
Mathematical Expectation
Mathematical expectation, often referred to as expected value, represents the "average" outcome of a random variable if an experiment or trial is repeated many times under identical conditions. It gives one a sense of the "center of mass" for a probability distribution.

For a negative binomial distribution, like in our scenario with two daughters, the expected number of trials (or events) before reaching the desired number of successes can be calculated using specific formulas:
  • For each family, the expected number of boys \(E(Y)\) is \(\frac{r(1-p)}{p}\), resulting in an expectation of 2 boys (since \(r = 2\), \(p = 0.5\)).
  • For all families, the expected total number of boys, \(E(X)\), is 6, as calculated by multiplying the result for one family by the three brothers.
Mathematical expectation helps transform probabilistic predictions into practical, actionable insights, providing a useful measure of what to reasonably anticipate in random processes.

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Most popular questions from this chapter

Show that \(E(X)=n p\) when \(X\) is a binomial random variable. [Hint: First express \(E(X)\) as a sum with lower limit \(x=1\). Then factor out \(n p\), let \(y=x-1\) so that the remaining sum is from \(y=0\) to \(y=n-1\), and show that it equals 1.]

A mail-order computer business has six telephone lines. Let \(X\) denote the number of lines in use at a specified time. Suppose the pmf of \(X\) is as given in the accompanying table. \begin{tabular}{c|ccccccc} \(x\) & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline\(p(x)\) & \(.10\) & \(.15\) & \(.20\) & \(.25\) & \(.20\) & \(.06\) & \(.04\) \end{tabular} Calculate the probability of each of the following events. a. [at most three lines are in use] b. \\{fewer than three lines are in use\\} c. \\{at least three lines are in use \\} d. Ibetween two and five lines, inclusive, are in use e. Ibetween two and four lines, inclusive, are not in use \\} f. \\{at least four lines are not in use\\}

Suppose small aircraft arrive at an airport according to a Poisson process with rate \(\alpha=8 / \mathrm{h}\), so that the number of arrivals during a time period of \(t\) hours is a Poisson rv with parameter \(\hat{\lambda}=8 t\). a. What is the probability that exactly 6 small aircraft arrive during a 1 -h period? At least 6 ? At least 10 ? b. What are the expected value and standard deviation of the number of small aircraft that arrive during a 90 -min period? c. What is the probability that at least 20 small aircraft arrive during a \(2 \frac{1}{2} h\) period? That at most 10 arrive during this period?

Suppose that trees are distributed in a forest according to a two-dimensional Poisson process with parameter \(\alpha\), the expected number of trees per acre, equal to 80 . a. What is the probability that in a certain quarter-acre plot, there will be at most 16 trees? b. If the forest covers 85,000 acres, what is the expected number of trees in the forest? c. Suppose you select a point in the forest and construct a circle of radius \(1 \mathrm{mile}\). Let \(X=\) the number of trees within that circular region. What is the pmf of \(X\) ? [Hint: 1 sq mile \(=640\) acres.]

If the sample space \(\delta\) is an infinite set, does this necessarily imply that any rv \(X\) defined from \(s\) will have an infinite set of possible values? If yes, say why. If no, give an example.

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