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A couple plans on having three children. Suppose that the probability of any given child being female is \(0.5,\) and also suppose that the genders of each child are independent events. a. Write out all outcomes in the sample space for the genders of the three children. b. What should be the probability associated with each outcome? Using the sample space constructed in part a, find the probability that the couple will have c. two girls and one boy. d. at least one child of each gender.

Short Answer

Expert verified
a) {GGG, GGB, GBG, BGG, BGB, BBG, BBB, GBG}; b) 0.125 each; c) 0.375; d) 0.75.

Step by step solution

01

Identifying the Sample Space

When a couple plans to have three children, each child can either be a girl (G) or a boy (B). Therefore, the sample space will include all combinations of G and B for the three children. The sample space is: {GGG, GGB, GBG, BGG, BGB, BBG, BBB, GBG}, representing all possible outcomes of the genders of the three children. Each outcome is an equally likely event.
02

Determining Probability of Each Outcome

Since the probability of each child being either a girl (G) or a boy (B) is independent and equal (0.5), we can multiply the probabilities for each child. Thus, the probability of any specific outcome, like GGG or BGB, is \( (0.5) \times (0.5) \times (0.5) = 0.125 \). This applies to each of the eight outcomes.
03

Finding Probability of Two Girls and One Boy

Next, count the number of outcomes with two girls (G) and one boy (B). These are: GGB, GBG, BGG. There are three such outcomes. Since each has a probability of 0.125, the total probability for having two girls and one boy is \( 3 \times 0.125 = 0.375 \).
04

Finding Probability of At Least One Child of Each Gender

To find the probability of having at least one child of each gender, we exclude cases of all children being the same gender: GGG or BBB. The outcomes that include both genders are GGB, GBG, BGG, BGB, BBG, and BGB. There are six such outcomes. Therefore, the probability is \( 6 \times 0.125 = 0.75 \).

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Key Concepts

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

Sample Space
In probability theory, the concept of a sample space is fundamental because it represents the set of all possible outcomes of a probabilistic experiment. In this exercise, the sample space is the list of all possible gender combinations for three children.
Each child can either be a girl (G) or a boy (B), which yields multiple combinations. For three children, the possible combinations are: {GGG, GGB, GBG, BGG, BGB, BBG, BBB, GBG}.
When dealing with probability, establishing a clear sample space ensures that all potential outcomes are considered. This step is crucial for calculating other probabilities. It is like setting the stage before the main drama of probability calculation begins.
Independent Events
Independent events are events where the outcome of one event does not affect the outcome of another. In the context of this exercise, each child can independently be a boy or a girl, and the gender of one child does not influence the gender of the others.
Think about it this way: flipping a fair coin three times and getting heads or tails does not depend on the result of any of the previous flips. Just as each child in this scenario being a boy or girl remains uninfluenced by the outcome of the others. Recognizing that events are independent is crucial as it simplifies calculations and assumptions during probability assessments, making modeling real-world situations easier.
Probability of Independent Events
Probability for independent events is calculated using the multiplication rule. If you want to know the likelihood of multiple independent events happening together, you multiply the probability of each event. In this exercise, the probability of having a girl or a boy is 0.5 per child, and each child is an independent event.
To find the probability of each possible combination, like GGB or BGB, you multiply the probability of a girl or boy consecutively for all three children: \[(0.5) \times (0.5) \times (0.5) = 0.125\]This means each different combination of children has an equal probability of 0.125. This mathematics is straightforward when dealing with independent events because there are no contingent dependencies to consider.
Combinations and Permutations
Understanding combinations and permutations is important when dealing with probability. A permutation is an arrangement of objects in a specific order, whereas a combination is a selection of objects without regard to the order.
In this problem, we are interested in the combination of results – meaning which results occur, not the order in which they happen. For instance, having two girls and one boy can occur in different arrangements such as GGB, GBG, and BGG, but for the combination calculation, the order doesn't matter. Understanding the difference between these concepts can greatly impact how probabilities are calculated and understood, especially as exercises increase in complexity and the need for different calculation approaches arises.

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Most popular questions from this chapter

A teacher announces a pop quiz for which the student is completely unprepared. The quiz consists of 100 true-false questions. The student has no choice but to guess the answer randomly for all 100 questions. a. Simulate taking this quiz by random guessing. Number a sheet of paper 1 to 100 to represent the 100 questions. Write a \(\mathrm{T}\) (true) or \(\mathrm{F}\) (false) for each question, by predicting what you think would happen if you repeatedly flipped a coin and let a tail represent a T guess and a head represent an F guess. (Don't actually flip a coin, but merely write down what you think a random series of guesses would look like.) b. How many questions would you expect to answer correctly simply by guessing? c. The table shows the 100 correct answers. The answers should be read across rows. How many questions did you answer correctly? d. The above answers were actually randomly generated by the Simulating the Probability of Head With a Fair Coin applet on the text CD. What percentage were true, and what percentage would you expect? Why are they not necessarily identical? e. Are there groups of answers within the sequence of 100 answers that appear nonrandom? For instance, what is the longest run of Ts or Fs? By comparison, which is the longest run of Ts or Fs within your sequence of 100 answers? (There is a tendency in guessing what randomness looks like to identify too few long runs in which the same outcome occurs several times in a row.)

In the opening scene of Tom Stoppard's play Rosencrantz and Guildenstern Are Dead, about two Elizabethan contemporaries of Hamlet, Guildenstern flips a coin 91 times and gets a head each time. Suppose the coin was balanced. a. Specify the sample space for 91 coin flips, such that each outcome in the sample space is equally likely. How many outcomes are in the sample space? b. Show Guildenstern's outcome for this sample space. Show the outcome in which only the second flip is a tail. c. What's the probability of the event of getting a head 91 times in a row? d. What's the probability of at least one tail, in the 91 flips? e. State the probability model on which your solutions in parts \(\mathrm{c}\) and \(\mathrm{d}\) are based.

Serena Williams won the 2010 Wimbledon Ladies' Singles Championship. For the seven matches she played in the tournament, her total number of first serves was \(379,\) total number of good first serves was 256 , and total number of double faults was 15 . a. Find the probability that her first serve is good. b. Find the conditional probability of double faulting, given that her first serve resulted in a fault. c. On what percentage of her service points does she double fault?

Since events of low probability eventually happen if you observe enough trials, a monkey randomly pecking on a typewriter could eventually write a Shakespeare play just by chance. Let's see how hard it would be to type the title of Macbeth properly. Assume 50 keys for letters and numbers and punctuation. Find the probability that the first seven letters that a monkey types are macbeth. (Even if the monkey can type 60 strokes a minute and never sleeps, if we consider each sequence of seven keystrokes as a trial, we would wait on the average over 100,000 years before seeing this happen!)

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