/*! 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 6 In a four-child family, what is ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 6

In a four-child family, what is the expected number of boys? (Assume that the probability of a boy being born is the same as the probability of a girl being born.)

Short Answer

Expert verified
The expected number of boys in a four-child family, assuming equal probability for boys and girls, is 2.

Step by step solution

01

List possible outcomes and their probabilities

Since there are 4 children in the family, the possible outcomes for the number of boys are 0, 1, 2, 3, and 4. The probability of each of them depends on the binomial distribution corresponding to the number of trials (n=4), the probability of success (number of boys, P(boy) = 0.5) and the probability of failure (number of girls, P(girl) = 0.5). We can use the equation for binomial probability: P(X = k) = \( \binom{n}{k} * p^k * (1 - p)^{n-k} \) Where: - P(X=k) is the probability of having k boys - n is the total number of children (4 in our case) - k is the number of boys in the family (from 0 to 4) - p is P(boy) = 0.5 - (1 - p) = P(girl) = 0.5
02

Calculate all binomial probabilities

Now, we will calculate the binomial probabilities for every possible number of boys in the family. P(0 boys) = \( \binom{4}{0} * 0.5^0 * (1 - 0.5)^{4-0} \) = \( \frac{0.5^0 * 0.5^4}{4!/(4! * 0!)} \) = 1 * 0.0625 = 0.0625 P(1 boy) = \( \binom{4}{1} * 0.5^1 * (1 - 0.5)^{4-1} \) = \( \frac{0.5^1 * 0.5^3}{4!/(3! * 1!)} \) = 4 * 0.0625 = 0.25 P(2 boys) = \( \binom{4}{2} * 0.5^2 * (1 - 0.5)^{4-2} \) = \( \frac{0.5^2 * 0.5^2}{4!/(2! * 2!)} \) = 6 * 0.0625 = 0.375 P(3 boys) = \( \binom{4}{3} * 0.5^3 * (1 - 0.5)^{4-3} \) = \( \frac{0.5^3 * 0.5^1}{4!/(3! * 1!)} \) = 4 * 0.0625 = 0.25 P(4 boys) = \( \binom{4}{4} * 0.5^4 * (1 - 0.5)^{4-4} \) = \( \frac{0.5^4 * 0.5^0}{4!/(4! * 0!)} \) = 1 * 0.0625 = 0.0625
03

Calculate expected value

The expected value is calculated by multiplying each outcome (number of boys) by its corresponding probability and adding all the results. Therefore, Expected value = 0 * P(0 boys) + 1 * P(1 boy) + 2 * P(2 boys) + 3 * P(3 boys) + 4 * P(4 boys) E = 0 * 0.0625 + 1 * 0.25 + 2 * 0.375 + 3 * 0.25 + 4 * 0.0625 = 0 + 0.25 + 0.75 + 0.75 + 0.25 = 2 Thus, the expected number of boys in a four-child family, assuming equal probability for boys and girls, is 2.

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!

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

FuPPING A CoIN An unbiased coin is tossed 1 million times. Show that the probability is at least. 99 that the proportion of times the coin will land heads is between \(.495\) and . 505 , inclusive. (Note: These results show that if an unbiased coin is tossed a very large number times, the proportion of times that the coin will land heads is very close to \(\frac{1}{2}\).) Hint: Use Chebychev's inequality.

Use the formula \(C(n, x) p^{x} q^{n-x}\) to determine the probability of the given event. The probability of no successful outcomes in six trials of a binomial experiment in which \(p=\frac{1}{3}\)

On average, a student takes 100 words/minute midway through an advanced court reporting course at the American Institute of Court Reporting. Assuming that the dictation speeds of the students are normally distributed and that the standard deviation is 20 words/minute, what is the probability that a student randomly selected from the course can take dictation at a speed a. Of more than 120 words/minute? b. Between 80 and 120 words/minute? c. Of less than 80 words/minute?

The probability that a fuse produced by a certain manufacturing process will be defective is \(\frac{1}{50}\). Is it correct to infer from this statement that there is at most 1 defective fuse in each lot of 50 produced by this process? Justify your answer.

The probability of an event \(E\) not occurring is .6. What are the odds in favor of \(E\) occurring? What are the odds against \(E\) occurring?
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Pure Maths

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.