/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Suppose that parents are equally... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 11

Suppose that parents are equally likely to have (in total) one, two, or three offspring. A girl is selected at random; what is the probability that the family includes no older girl? (Assume that children are independent and equally likely to be male or female.)

Short Answer

Expert verified
The probability is \(\frac{1}{3}\).

Step by step solution

01

Understand the Problem

The problem involves calculating the probability that a randomly chosen girl is the first or only girl in the family when each family can have one, two, or three children. Each child has an equal chance of being male or female.
02

Analyze the Possible Family Configurations

Families can have 1, 2, or 3 children. For each child, the probability of being a girl is \(\frac{1}{2}\). We need to assess scenarios where the girl is the first girl or the only girl.
03

List Out Possible Outcomes

For 1 child, the family must have 1 girl. For 2 children (GG, GB, BG, BB), we consider BG. For 3 children (GGG, GGB, GBG, BGG, GBB, BGB, BBG, BBB), BGX, and BGB are possible, where X is any gender.
04

Calculate Probability for Each Scenario

Calculate the probability for each scenario where the first girl appears. For 1 child, the family can only be G. For 2 children, only the sequence BG fits. For 3 children, sequences like BGX and BGB fit.
05

Find Total Probability for Each Family Size

For 1 child: Probability = \(\frac{1}{2}\). For 2 children: Probability = \(\frac{1}{4}\) (for BG). For 3 children: Probability = \(\frac{1}{8}\) (for BGX) + \(\frac{1}{8}\) (for BGB) = \(\frac{1}{4}\).
06

Compute Overall Probability

The overall probability is the average of these probabilities, weighted by family size probabilities (\(\frac{1}{3}\) for each family size). Total probability = \(\frac{1}{3}(\frac{1}{2}) + \frac{1}{3}(\frac{1}{4}) + \frac{1}{3}(\frac{1}{4}) = \frac{1}{3}\).
07

Conclusion

The probability that a randomly selected girl is the first girl in her family is \(\frac{1}{3}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Conditional Probability
Conditional probability is a concept used to calculate the probability of an event occurring, given that another event has already occurred. It helps us refine our guesses based on new information. In probability theory, the probability of event A happening given event B is denoted as \( P(A|B) \).

For our exercise, suppose we want to find the probability that a family composed of 2 children has the sequence BG (Boy then Girl) when we know at least one child is a girl. By specifying that the family includes a girl, we are narrowing the sample space, only focusing on scenarios where the family has at least one girl. Conditional probability, in this case, allows us to consider only relevant sequences instead of all possible child sequences.

The real power of conditional probability lies in its ability to focus calculations on current conditions or known facts, thereby simplifying complex probability issues.
Independent Events
In probability theory, events are considered independent if the occurrence of one does not affect the likelihood of the other. This is crucial because it simplifies calculations significantly.

We utilize independent events in this exercise when assuming that the gender of one child does not influence the gender of another. For instance, the probability of the first child being a girl is \( \frac{1}{2} \), and the probability of the second child being a girl is also \( \frac{1}{2} \). These probabilities are independent of each other, and hence, multiply when calculating the joint probabilities of multiple independent events.

This assumption allows us to straightforwardly compute scenarios like GG (Girl, Girl) or BG (Boy, Girl) by multiplying individual probabilities, making our task manageable without complex dependencies.
Random Selection
Random selection is a fundamental part of probability theory, ensuring that every outcome within a set has an equal chance of being chosen. This principle is essential when dealing with scenarios like picking a child at random from a family.

In our problem, we randomly select a girl from a family. This means that each child has an equal probability of being selected, regardless of birth order or other family specifics. Therefore, each child having a probability of \( \frac{1}{2} \) for being a girl ensures unbiased selection for our analysis.

Random selection supports fairness and impartiality, ensuring that the conclusions drawn from probability calculations are as unbiased and representative of reality as possible.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two roads join Ayton to Beaton, and two further roads join Beaton to the City. Ayton is directly connected to the City by a railway. All four roads and the railway are each independently blocked by snow with probability \(p\). I am at Ayton. (a) Find the probability that I can drive to the City. (b) Find the probability that I can travel to the City. (c) Given that I can travel to the City, what is the probability that the railway is blocked?

Suppose that for events \(S, A\), and \(B\), $$ \begin{aligned} \mathbf{P}(S \mid A) & \geq \mathbf{P}(S) \\ \mathbf{P}(A \mid S \cap B) & \geq \mathbf{P}(A \mid S) \\ \mathbf{P}\left(A \mid S^{c}\right) & \geq \mathbf{P}\left(A \mid S^{c} \cap B\right) \end{aligned} $$ (a) Show that, except in trivial cases, \(\mathbf{P}(S \mid A \cap B) \geq \mathbf{P}(S \mid B)\). (b) Show that \(\mathbf{P}(S \mid A) \geq \mathbf{P}(A)\). (c) Show that if \((*)\) is replaced by \(\mathbf{P}(S \mid B) \geq \mathbf{P}(S)\), then \(\mathbf{P}(S \mid A \cap B) \geq \mathbf{P}(S \mid A)\).

Simpson's Paradox Two drugs are being tested. Of 200 patients given drug \(A, 60\) are cured; and of 1100 given drug \(B, 170\) are cured. If we assume a homogeneous group of patients, find the probabilities of successful treatment with \(A\) or \(B\). Now closer investigation reveals that the 200 patients given drug \(A\) were in fact 100 men, of whom 50 were cured, and 100 women of whom 10 were cured. Further, of the 1100 given drug \(B, 100\) were men of whom 60 were cured, and 1000 were women of whom 110 were cured. Calculate the probability of cure for men and women receiving each drug; note that \(B\) now seems better than \(A\). (Results of this kind indicate how much care is needed in the design of experiments. Note that the paradox was described by Yule in 1903 , and is also called the Yule-Simpson paradox.)

Box \(A\) contains three red balls and two white balls; box \(B\) contains two red balls and two white balls. A fair die is thrown. If the upper face of the die shows 1 or 2 , a ball is drawn at random from box \(A\) and put in box \(B\) and then a ball is drawn at random from box \(B\). If the upper face of the die shows \(3,4,5\) or 6 , a ball is drawn at random from box \(B\) and put in box \(A\), and then a ball is drawn at random from box \(A\). What are the probabilities (a) That the second ball drawn is white? (b) That both balls drawn are red? (c) That the upper face of the red die showed 3, given that one ball drawn is white and the other red?

You roll a fair die \(n\) times. What is the probability that (a) You have rolled an odd number of sixes? (b) You have not rolled a six on two successive rolls? (c) You rolled a one before you rolled a six, given that you have rolled at least one of each?
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.