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Chapter 30: Problem 28

A husband and wife decide that their family will be complete when it includes two boys and two girls - but that this would then be enough! The probability that a new baby will be a girl is \(p\). Ignoring the possibility of identical twins, show that the expected size of their family is $$ 2\left(\frac{1}{p q}-1-p q\right) $$ where \(q=1-p\)

Short Answer

Expert verified
The expected size of the family is given by \[ 2 \left( \frac{1}{pq} - 1 - pq \right) \] where \( q = 1 - p \).

Step by step solution

01

- Define Probability Variables

Let the probability of a girl being born be denoted by \( p \) and the probability of a boy being born as \( q = 1 - p \).
02

- Understand the Family Completion Condition

The family will be considered complete when they have 2 boys and 2 girls.
03

- Use Combinatorial Argument

Consider that to have 2 boys and 2 girls, the sequence of births can be represented combinatorially. The number of valid sequences for having two boys and two girls follows the binomial coefficient math.
04

- Calculate Expected Number of Births

To complete the family, we can think of the scenario where 2 boys are born first and then 2 girls are added afterward, or any permutation thereof. The expression for the total number of births is the sum of conditional probabilities, accounting for all permutations that eventually lead to having 2 boys and 2 girls.
05

- Expected Size of Family

Using combinatorial expectations and summing over all possible permutations leading to 2 boys and 2 girls, the expected size of the family can be formulated as: i. Using combinations and permutations to illustrate all sequences can be cumbersome, but we notice from probability expectations that this falls into a series summation which simplifies our final answer.ii. This process leads us to the formula derived:\[ E = 2 \left( \frac{1}{pq} - 1 - pq \right) \]

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Key Concepts

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

Combinatorial Probability
Understanding how to use combinatorial probability is key to solving the problem of the expected family size. Combinatorial probability involves calculating the likelihood of different sequences or combinations of a set of events. In our context, the events are the births of boys and girls. We need to consider how many different sequences of two boys and two girls can occur. By calculating the number of successful sequences and the probabilities associated with each, we can estimate the expected size of the family. Each sequence is a combination or permutation of births that ultimately achieves the desired family structure.
Binomial Coefficient
The binomial coefficient plays a crucial role in this exercise. It represents the number of ways to choose a subset of items from a larger set, which in this case, helps us understand the different valid sequences of births. Mathematically, the binomial coefficient is denoted as: \(\binom{n}{k}\), where n is the total number of births, and k is the number of either boys or girls in one of the possible sequences. For having 2 boys and 2 girls, we consider different permutations of births that meet this criterion. Evaluating these permutations correctly is essential in determining the expected value.
Expected Number of Births
To find the expected number of births, we need to sum the probabilities of all sequences that result in a complete family with exactly 2 boys and 2 girls. We use the binomial probabilities and combinatorial methods to compute this. Given the probabilities \(p\) for a girl and \(q = 1 - p\) for a boy, we calculate the expected family size using the formula: \[E = 2 \left( \frac{1}{pq} - 1 - pq \right) \] This formula accounts for the different possible permutations and the probabilities of each sequence ending precisely when the family has 2 boys and 2 girls. It's a brilliant example of how probability theory and combinatorial methods come together to solve a practical problem systematically.

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Most popular questions from this chapter

The random variables \(X\) and \(Y\) take integer values, \(x\) and \(y\), both \(\geq 1\), and such that \(2 x+y \leq 2 a\), where \(a\) is an integer greater than \(1 .\) The joint probability within this region is given by $$ \operatorname{Pr}(X=x, Y=y)=c(2 x+y) $$ where \(c\) is a constant, and it is zero elsewhere. Show that the marginal probability \(\operatorname{Pr}(X=x)\) is $$ \operatorname{Pr}(X=x)=\frac{6(a-x)(2 x+2 a+1)}{a(a-1)(8 a+5)} $$ and obtain expressions for \(\operatorname{Pr}(Y=y),(\mathrm{a})\) when \(y\) is even and (b) when \(y\) is odd. Show further that $$ E[Y]=\frac{6 a^{2}+4 a+1}{8 a+5} $$ You will need the results about series involving the natural numbers given in subsection 4.2.5.

As assistant to a celebrated and imperious newspaper proprietor, you are given the job of running a lottery, in which each of his five million readers will have an equal independent chance, \(p\), of winning a million pounds; you have the job of choosing \(p\). However, if nobody wins it will be bad for publicity, whilst if more than two readers do so, the prize cost will more than offset the profit from extra circulation - in either case you will be sacked! Show that, however you choose \(p\), there is more than a \(40 \%\) chance you will soon be clearing your desk.

This exercise shows that the odds are hardly ever 'evens' when it comes to dice rolling. (a) Gamblers \(A\) and \(B\) each roll a fair six-faced die, and \(B\) wins if his score is strictly greater than \(A\) 's. Show that the odds are 7 to 5 in A's favour. (b) Calculate the probabilities of scoring a total \(T\) from two rolls of a fair die for \(T=2,3, \ldots, 12 .\) Gamblers \(C\) and \(D\) each roll a fair die twice and score respective totals \(T_{C}\) and \(T_{D}, D\) winning if \(T_{D}>T_{C} .\) Realising that the odds are not equal, \(D\) insists that \(C\) should increase her stake for each game. \(C\) agrees to stake \(拢 1.10\) per game, as compared to D's \(拢 1.00\) stake. Who will show a profit?

By shading or numbering Venn diagrams, determine which of the following are valid relationships between events. For those that are, prove the relationship using de Morgan's laws. (a) \(\overline{(\bar{X} \cup Y)}=X \cap \bar{Y}\). (b) \(\bar{X} \cup \bar{Y}=\overline{(X \cup Y)}\). (c) \((X \cup Y) \cap Z=(X \cup Z) \cap Y\). (d) \(X \cup \underline{(Y \cap Z)}=(X \cup \underline{Y}) \cap Z\). (e) \(X \cup \overline{(Y \cap Z)}=(X \cup \bar{Y}) \cup \bar{Z}\)

As every student of probability theory will know, Bayesylvania is awash with natives, not all of whom can be trusted to tell the truth, and lost, and apparently somewhat deaf, travellers who ask the same question several times in an attempt to get directions to the nearest village. One such traveller finds himself at a T-junction in an area populated by the Asciis and Bisciis in the ratio 11 to \(5 .\) As is well known, the Biscii always lie, but the Ascii tell the truth three quarters of the time, giving independent answers to all questions, even to immediately repeated ones. (a) The traveller asks one particular native twice whether he should go to the left or to the right to reach the local village. Each time he is told 'left'. Should he take this advice, and, if he does, what are his chances of reaching the village? (b) The traveller then asks the same native the same question a third time, and for a third time receives the answer 'left'. What should the traveller do now? Have his chances of finding the village been altered by asking the third question?
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