/*! 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 A ball is drawn from an urn cont... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 11

A ball is drawn from an urn containing three white and three black balls. After the ball is drawn, it is then replaced and another ball is drawn. This goes on indefinitely. What is the probability that of the first four balls drawn, exactly two are white?

Short Answer

Expert verified
The probability of drawing exactly 2 white balls in the first four draws is \(\frac{3}{8}\).

Step by step solution

01

Define the probabilities for drawing one white or black ball

Let's define the probabilities of drawing one white or one black ball. There's a total of 6 balls in the urn, with 3 white balls and 3 black balls. The probability of drawing one white ball is the number of white balls divided by the total number of balls: \(P(W) = \frac{3}{6} = \frac{1}{2}\) Similarly, the probability of drawing one black ball is the number of black balls divided by the total number of balls: \(P(B) = \frac{3}{6} = \frac{1}{2}\)
02

Using the combination formula

Now, let's use the combination formula to find the possible ways we can draw exactly 2 white balls and 2 black balls in 4 draws. The combination formula is given by \(C(n, r) = \frac{n!}{r!(n-r)!}\), where n is the total number of items and r is the number of items to be chosen. Here, we choose 2 white balls from 4 draws, so n=4 and r=2. We calculate \(C(4, 2) = \frac{4!}{2!(4-2)!} = \frac{4\cdot3\cdot2\cdot1}{2\cdot1\cdot2\cdot1} = 6\)
03

Calculate the probability

To calculate the probability of exactly two white balls and two black balls in four draws, we need to multiply the probabilities of drawing white and black balls with the number of possible ways. So, \(P(\text{exactly 2 white balls}) = C(4, 2) \times P(W)^2 \times P(B)^2\) Substituting the values, we get: \(P(\text{exactly 2 white}) = 6 \times (\frac{1}{2})^2 \times (\frac{1}{2})^2 = 6 \times \frac{1}{16} = \frac{3}{8}\) Thus, the probability of drawing exactly 2 white balls in the first four draws is \(\frac{3}{8}\).

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

A coin, having probability \(p\) of landing heads, is flipped until a head appears for the \(r\) th time. Let \(N\) denote the number of flips required. Calculate \(E[N]\). Hint: There is an easy way of doing this. It involves writing \(N\) as the sum of \(r\) geometric random variables.

Suppose that an experiment can result in one of \(r\) possible outcomes, the ith outcome having probability \(p_{i}, i=1, \ldots, r, \sum_{i=1}^{r} p_{i}=1 .\) If \(n\) of these experiments are performed, and if the outcome of any one of the \(n\) does not affect the outcome of the other \(n-1\) experiments, then show that the probability that the first outcome appears \(x_{1}\) times, the second \(x_{2}\) times, and the \(r\) th \(x_{r}\) times is $$ \frac{n !}{x_{1} ! x_{2} ! \ldots x_{r} !} p_{1}^{x_{1}} p_{2}^{x_{2}} \cdots p_{r}^{x_{r}} \quad \text { when } x_{1}+x_{2}+\cdots+x_{r}=n $$ This is known as the multinomial distribution.

If the distribution function of \(F\) is given by $$ F(b)=\left\\{\begin{array}{ll} 0, & b<0 \\ \frac{1}{2}, & 0 \leq b<1 \\ \frac{3}{3}, & 1 \leq b<2 \\ \frac{4}{3}, & 2 \leq b<3 \\ \frac{9}{10}, & 3 \leq b<3.5 \\ 1, & b \geq 3.5 \end{array}\right. $$ calculate the probability mass function of \(X\).

A television store owner figures that 50 percent of the customers entering his store will purchase an ordinary television set, 20 percent will purchase a color television set, and 30 percent will just be browsing. If five customers enter his store on a certain day, what is the probability that two customers purchase color sets, one customer purchases an ordinary set, and two customers purchase nothing?

Let \(X\) and \(Y\) be independent random variables with means \(\mu_{x}\) and \(\mu_{y}\) and variances \(\sigma_{x}^{2}\) and \(\sigma_{y}^{2}\). Show that $$ \operatorname{Var}(X Y)=\sigma_{x}^{2} \sigma_{y}^{2}+\mu_{y}^{2} \sigma_{x}^{2}+\mu_{x}^{2} \sigma_{y}^{2} $$
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Decision 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.