/*! 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 24 3.24. Each of 2 balls is painted... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 24

3.24. Each of 2 balls is painted either black or gold and then placed in an urn. Suppose that each ball is colored black with probability \(\frac{1}{2}\) and that these events are independent. (a) Suppose that you obtain information that the gold paint has been used (and thus at least one of the balls is painted gold). Compute the conditional probability that both balls are painted gold. (b) Suppose now that the urn tips over and 1 ball falls out. It is painted gold. What is the probability that both balls are gold in this case? Explain.

Short Answer

Expert verified
The short answer for this problem is: (a) The conditional probability that both balls are painted gold, given that at least one ball is painted gold, is \(\frac{1}{3}\). (b) The conditional probability that both balls are painted gold, given that the first ball observed is gold, is \(\frac{1}{2}\).

Step by step solution

01

Calculate P(A)

To start, we need to find the probability of both balls being painted gold (event A). Since each ball has a \(\frac{1}{2}\) probability of being painted black or gold, and the events are independent, to find P(A), we simply square the probability of painting one ball gold: \[P(A) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\]
02

Calculate P(B1)

We want the probability that at least one ball is painted gold. We can find this by calculating the probability that both balls are painted black and then subtract that probability from 1, as they are complementary events: \[P(B1) = 1 - P(\text{both balls black})\] \[P(\text{both balls black}) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\] \[P(B1) = 1 - \frac{1}{4} = \frac{3}{4}\]
03

Calculate P(A 鈭� B1)

Since A is a subset of B1 (if both balls are gold, it's certain that at least one ball is gold), we have: \[P(A \cap B1) = P(A) = \frac{1}{4}\]
04

Calculate the conditional probability P(A|B1)

Now, we plug the probabilities into the formula for conditional probability: \[P(A|B1) = \frac{P(A \cap B1)}{P(B1)} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}\] Thus, the conditional probability that both balls are painted gold, given that at least one ball is painted gold, is \(\frac{1}{3}\). **(b) Calculate P(A|B2)** Now, we need to compute the conditional probability P(A|B2). We will use the same formula: \[P(A|B2) = \frac{P(A \cap B2)}{P(B2)}\]
05

Calculate P(B2)

First, we need to find the probability that the first ball observed is gold. Since there's a \(\frac{1}{2}\) chance that either ball is gold, we have: \[P(B2) = \frac{1}{2}\]
06

Calculate P(A 鈭� B2)

Now we need to find the probability of both events A and B2 occurring. If both balls are gold, observing one ball as gold is certain: \[P(A \cap B2) = P(A) = \frac{1}{4}\]
07

Calculate the conditional probability P(A|B2)

Now, we plug the probabilities into the formula for conditional probability: \[P(A|B2) = \frac{P(A \cap B2)}{P(B2)} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}\] Thus, the conditional probability that both balls are painted gold, given that the first ball observed is gold, is \(\frac{1}{2}\).

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!

Key Concepts

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

Independent Events
Independent events are a fundamental concept in probability theory. If you are dealing with independent events, the outcome of one event does not affect the outcome of another. When it comes to the exercise, each ball being either black or gold is independent of the other, meaning the color of one ball doesn't influence the color of the other.

In mathematical terms, two events A and B are said to be independent if the probability of both occurring is the product of their individual probabilities:
  • If event A occurs with probability \( P(A) \) and event B with probability \( P(B) \), then the probability of both A and B occurring (\( P(A \cap B) \)) is \( P(A) \times P(B) \).
  • For example, if both balls being gold is event A and the probability of one ball being gold is \( \frac{1}{2} \), then \( P(A) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \).

Understanding independent events simplifies complex probability problems as it allows for straightforward calculation of joint probabilities.
Probability Calculation
Probability is the measure of how likely an event is to occur. To calculate probabilities, especially in the context of independent events, we use simple mathematical operations. In our exercise, we need to determine the probability of different combinations of two balls either being black or gold.

Here is a quick guide on calculating these probabilities:
  • Start by identifying the probability of each event separately. The probability that one ball is gold is \( \frac{1}{2} \).
  • For independent events, calculate the combined probability by multiplying the probabilities of the individual events. For example, for both balls to be gold, the probability is \( \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) = \frac{1}{4} \).
  • For more complex scenarios like determining the probability that at least one ball is gold, consider complementary events, which we will discuss in the next section.
Being able to correctly calculate these probabilities allows you to assess different outcomes and better understand the likelihood of each scenario.
Complementary Events
Complementary events are events that cover all possible outcomes of a random experiment. If one event occurs, the other cannot, and vice versa. When considering the likelihood of at least one ball being gold, thinking about complementary events helps simplify the calculation.

  • The probability of both balls being black can be found by squaring the probability of one ball being black: \( \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) = \frac{1}{4} \).
  • The complementary event to both balls being black is having at least one ball gold, which can be found using: \( P(\text{at least one gold}) = 1 - P(\text{both black}) = 1 - \frac{1}{4} = \frac{3}{4} \).
To use complementary events effectively, remember that the probability of an event and its complement always add up to one. This method can often hasten your calculation and serve as a check for your final answer.

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 certain organism possesses a pair of each of 5 different genes (which we will designate by the first 5 letters of the English alphabet). Each gene appears in 2 forms (which we designate by lowercase and capital letters). The capital letter will be assumed to be the dominant gene, in the sense that if an organism possesses the gene pair \(x X,\) then it will outwardly have the appearance of the \(X\) gene. For instance, if \(X\) stands for brown eyes and \(x\) for blue eyes, then an individual having either gene pair \(X X\) or \(x X\) will have brown eyes, whereas one having gene pair \(x x\) will have blue eyes. The characteristic appearance of an organism is called its phenotype, whereas its genetic constitution is called its genotype. (Thus, 2 organisms with respective genotypes \(a A, b B, c c, d D, e e\) and \(A A, B B, c c,\) \(D D, e e\) would have different genotypes but the same phenotype. In a mating between 2 organisms, each one contributes, at random, one of its gene pairs of each type. The 5 contributions of an organism (one of each of the 5 types) are assumed to be independent and are also independent of the contributions of the organism's mate. In a mating between organisms having genotypes \(a A, b B, c C\) \(d D, e E\) and \(a a, b B, c c, D d, e e\) what is the probability that the progeny will (i) phenotypically and (ii) genotypically resemble (a) the first parent? (b) the second parent? (c) either parent? (d) neither parent?

There are 3 coins in a box. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75 percent of the time. When one of the 3 coins is selected at random and flipped, it shows heads. What is the probability that it was the two-headed coin?

A person tried by a 3 -judge panel is declared guilty if at least 2 judges cast votes of guilty. Suppose that when the defendant is in fact guilty, each judge will independently vote guilty with probability \(.7,\) whereas when the defendant is in fact innocent, this probability drops to \(.2 .\) If 70 percent of defendants are guilty, compute the conditional probability that judge number 3 votes guilty given that (a) judges 1 and 2 vote guilty; (b) judges 1 and 2 cast 1 guilty and 1 not guilty vote; (c) judges 1 and 2 both cast not guilty votes. Let \(E_{i}, i=1,2,3\) denote the event that judge \(i\) casts a guilty vote. Are these events independent? Are they conditionally independent? Explain.

Prostate cancer is the most common type of cancer found in males. As an indicator of whether a male has prostate cancer, doctors often perform a test that measures the level of the prostate-specific antigen (PSA) that is produced only by the prostate gland. Although PSA levels are indicative of cancer, the test is notoriously unreliable. Indeed, the probability that a noncancerous man will have an elevated PSA level is approximately. \(135,\) increasing to approximately.268 if the man does have cancer. If, on the basis of other factors, a physician is 70 percent certain that a male has prostate cancer, what is the conditional probability that he has the cancer given that (a) the test indicated an elevated PSA level? (b) the test did not indicate an elevated PSA level? Repeat the preceding calculation, this time assuming that the physician initially believes that there is a 30 percent chance that the man has prostate cancer.

Suppose that you continually collect coupons and that there are \(m\) different types. Suppose also that each time a new coupon is obtained, it is a type \(i\) coupon with probability \(p_{i}, i=1, \ldots, m .\) Suppose that you have just collected your \(n\)th coupon. What is the probability that it is a new type? Hint: Condition on the type of this coupon.
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Mechanics 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.