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Suppose that each child born to a couple is equally likely to be a boy or a girl independent of the sex distribution of the other children in the family. For a couple having 5 children, compute the probabilities of the following events: (a) All children are of the same sex. (b) The 3 eldest are boys and the others girls. (c) Exactly 3 are boys. (d) The 2 oldest are girls. (e) There is at least 1 girl

Short Answer

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(a) 0.0625 (b) 0.125 (c) 0.3125 (d) 0.03125 (e) 0.96875

Step by step solution

01

(a) All children are of the same sex.

To calculate the probability of all children being of the same sex, we can consider the two possibilities: all boys or all girls. - All boys: With the probability of each child being a boy being 0.5, the probability of 5 boys is \((0.5)^5\). - All girls: Similarly, the probability of 5 girls is \((0.5)^5\). We then add these probabilities to get the probability of all children being of the same sex: \(P(A) = (0.5)^5 + (0.5)^5\)
02

(b) The 3 eldest are boys and the others girls.

For this scenario, we want to calculate the probability of having 3 boys followed by 2 girls. Since each child's sex is independent, we can multiply the probabilities: \(P(B) = (0.5)^3 * (0.5)^2\)
03

(c) Exactly 3 are boys.

To calculate the probability of having exactly 3 boys, we need to find the number of ways to arrange 3 boys and 2 girls in any order and multiply it by the probability of each arrangement happening. Number of ways (\(n\)) = C(5, 3) = 10 ways Since each arrangement has the same probability, we can calculate the probability as: \(P(C) = n * (0.5)^5\)
04

(d) The 2 oldest are girls.

For this scenario, we want to calculate the probability of having 2 girls followed by 3 children of any sex. Since the probability of 1 girl is 0.5 and the probability of the other 3 children is \((0.5)^3\), we can multiply these probabilities: \(P(D) = (0.5)^2 * (0.5)^3\)
05

(e) There is at least 1 girl.

To calculate the probability of having at least 1 girl, it's easiest to calculate the probability of having no girls (i.e., all boys) and subtracting this from 1. Probability of all boys: \((0.5)^5\) \(P(E) = 1 - (0.5)^5\) Now we can calculate the probabilities for each case: (a) \(P(A) = (0.5)^5 + (0.5)^5 = 0.03125 + 0.03125 = 0.0625\) (b) \(P(B) = (0.5)^3 * (0.5)^2 = 0.125\) (c) \(P(C) = 10 * (0.5)^5 = 10 * 0.03125 = 0.3125\) (d) \(P(D) = (0.5)^2 * (0.5)^3 = 0.03125\) (e) \(P(E) = 1 - (0.5)^5 = 1 - 0.03125 = 0.96875\) So, the probabilities for the different scenarios are: (a) All children are of the same sex: 0.0625 (b) The 3 eldest are boys and the others girls: 0.125 (c) Exactly 3 are boys: 0.3125 (d) The 2 oldest are girls: 0.03125 (e) There is at least 1 girl: 0.96875

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Key Concepts

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

Binomial Probability
Understanding binomial probability is essential when analyzing events with two outcomes, such as flipping a coin or, as in our textbook's exercise, determining the sex of children within a family. The key characteristic of binomial experiments is that they contain a fixed number of trials, each with only two possible outcomes, commonly labeled as 'success' and 'failure'.

In this context, calculating the probability of a certain number of successes involves two main elements: the probability of success in one trial (\(p\)), and the number of ways that given number of successes can occur among the trials (\(C(n, k)\)), which involves combinatorics. For example, in calculating the probability of exactly 3 boys out of 5 children, where each child is equally likely to be a boy (\(p = 0.5\)), we determine the number of ways to choose 3 children out of 5 to be boys (the combinatorial part) and then multiply that number by the probability that each of those arrangements happens (\(p^3 \times (1-p)^2\)).

Using binomial probability, we simplify complex problems by breaking them down into a series of independent and identical trials.
Independent Events
The assumption of independent events is critical in many probability calculations. Two events are independent if the occurrence of one does not affect the probability of the occurrence of the other. This foundational concept is especially relevant in our exercise, where the sex of each child is assumed to be independent of the others.

This assumption allows us to simply multiply the individual probabilities of each event to find the combined probability of multiple events happening in sequence. For instance, calculating the probability of having the 3 eldest children as boys and the 2 youngest as girls involves multiplying the probability of each event happening independently \( (0.5)^3 \times (0.5)^2 \).

Remember, for events to be considered independent, the outcome of one event must have no influence on the outcome of another. This concept is key to correctly applying the rules of multiplication for probabilities.
Combinatorics
Combinatorics is the branch of mathematics dealing with the study of counting, combination, and permutation of sets. It's essential when determining the number of possible outcomes in a probability question, like those in the textbook exercise.

In the exercise involving the probability of exactly 3 boys out of 5 children, we use the combination formula \( C(n, k) = \frac{n!}{k!(n-k)!} \) to calculate the number of ways to choose 3 children to be boys out of 5. This is because the order in which the boys are born does not matter; what matters is simply how many there are. The term 'combinatorics' encompasses both the concepts of combinations and permutations, which are used depending on whether the order of events is important.

Grasping combinatory concepts allows students to understand the underlying patterns in probability problems beyond rote memorization of formulas.
Conditional Probability
Conditional probability is the likelihood of an event occurring given that another event has already occurred. This is a fundamental concept in more complex probability questions, although it doesn't explicitly come up in the provided exercise. However, understanding it can enrich the comprehension of probability as a whole.

Conditional probability is often expressed as \( P(A|B) \), which reads 'the probability of event A occurring given that event B has occurred.' This concept is vital when analyzing dependent events, where the outcome of one impacts the occurrence of another. For example, if we were to extend our exercise to ask the probability of the next child being a girl given that the first four children were boys, conditional probability would play a crucial role.

Although the events in our original textbook problem are independent, recognizing when to apply conditional probability is a crucial skill in a statistician's toolkit.

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Most popular questions from this chapter

Die \(A\) has 4 red and 2 white faces, whereas die \(B\) has 2 red and 4 white faces. A fair coin is flipped once. If it lands on heads, the game continues with die \(A\); if it lands tails, then die \(B\) is to be used. (a) Show that the probability of red at any throw is \(\frac{1}{2}\).

A friend randomly chooses two cards, without replacement, from an ordinary deck of 52 playing cards. In each of the following situations, determine the conditional probability that both cards are aces. (a) You ask your friend if one of the cards is the ace of spades and your friend answers in the affirmative. (b) You ask your friend if the first card selected is an ace and your friend answers in the affirmative. (c) You ask your friend if the second card selected is an ace and your friend answers in the affirmative. (d) You ask your friend if either of the cards selected is an ace and your friend answers in the affirmative.

If you had to construct a mathematical model for events \(E\) and \(F\), as described in parts (a) through (e), would you assume that they were independent events? Explain your reasoning. (a) \(E\) is the event that a businesswoman has blue eyes, and \(F\) is the event that her"secretary has blue eyes. (b) \(E\) is the event that a professor owns a car, and \(F\) is the event that he is listed in the telephone book. (c) \(E\) is the event that a man is under 6 feet tall, and \(F\) is the event that he weighs over 200 pounds. (d) \(E\) is the event that a woman lives in the United States, and \(F\) is the event that she lives in the western hemisphere. (e) \(E\) is the event that it will rain tomorrow, and \(F\) is the event that it will rain the day after tomorrow.

The color of a person's eyes is determined by a single pair of genes. If they are both blue-eyed genes, then the person will have blue eyes; if they are both brown-eyed genes, then the person will. have brown eyes; and if one of them is a blue-eyed gene and the other a brown-eyed gene, then the person. will have brown eyes. (Because of the latter fact we say that the brown-eyed gene is dominant over the blue-eyed one.) A newborn child independently receives one eye gene from each of its parents and the gene it receives from a parent is equally likely to be either of the two eye genes of that parent. Suppose that Smith and both of his parents have brown eyes, but Smith's sister has blue eyes. (a) What is the probability that Smith possesses a blue-eyed gene? Suppose that Smith's wife has blue eyes. (b) What is the probability that their first child will have blue eyes? (c) If their first child has brown eyes, what is the probability that their next child will also have brown eyes?

The following method was proposed to estimate the number of people over the age of 50 that reside in a town of known population 100,000 . "As you walk along the streets, keep a running count of the percentage of people that you encounter who are over \(50 .\) Do this for a few days; then multiply the obtained percentage by 100,000 to obtain the estimate." Comment on this method. HINT: Let \(p\) denote the proportion of people in this town who are over \(50 .\) Furthermore, let \(\alpha_{1}\) denote the proportion of time that a person under the age of 50 spends in the streets, and let \(\alpha_{2}\) be the corresponding value for those over \(50 .\) What quantity does the method suggested estimate? When is it approximately equal to \(p ?\)

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