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Chapter 3: Problem 83

Die \(A\) has 4 red and 2 white faces, whereas die \(B\) has 2 red and 4 white faces. A fair coin is flipped once. If it lands on heads, the game continues with dic \(A ;\) if it lands on tails, then dic \(B\) is to be used. (a) Show that the probability of red at any throw is \(\frac{1}{2}.\) (b) If the first two throws result in red, what is the probability of red at the third throw? (c) If red turns up at the first two throws, what is the probability that it is die \(A\) that is being used?

Short Answer

Expert verified
The short answers for the given parts of the problem are as follows: (a) The probability of red at any throw is \(\frac{1}{2}\). (b) If the first two throws result in red, the probability of red at the third throw is \(\frac{1}{2}\). (c) If red turns up at the first two throws, the probability that it is die \(A\) that is being used is \(\frac{2}{3}\).

Step by step solution

01

Compute the Probability of Red at Any Single Throw

The total probability of red at any single throw, \(P(R)\), may be calculated by summing the probability of drawing red from die \(A\) times the probability of choosing die \(A\), and the probability of drawing red from die \(B\) times the probability of choosing die \(B\). This gives: \[P(R) = P(R|A)P(A) + P(R|B)P(B) \ =>\ =>\frac{4}{6}\frac{1}{2} + \frac{2}{6}\frac{1}{2} = \frac{1}{2}.\]
02

Compute the Probability of Red at the Third Throw Given Two Reds

Given that the first two throws landed on red, the probability of drawing red at the third throw is still \(P(R) = \frac{1}{2}\), because the third throw is independent of the results of the previous throws. This is because each die is re-selected by a coin flip before each throw, and each die throw is independent of previous throws.
03

Compute the Probability of Using Die A Given Two Reds

This is an application of Bayes' theorem. The probability of using die \(A\) given that red turned up in the first two throws, \(P(A|R,R)\), is given by the probability of drawing red from die \(A\) times the probability of choosing die \(A\), divided by the total probability of drawing red. This is computed as follows: \[P(A|R,R) = \frac{P(R|R,A) P(A)}{P(R|R)} => \frac{\frac{4}{6}\frac{1}{2}}{\frac{1}{2}} = \frac{4}{6} = \frac{2}{3}\].

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Key Concepts

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

Independent Events
Independent events are situations where the outcome of one event does not affect the outcome of another. In our exercise, each die is chosen before every roll using a fair coin flip. This means each roll is independent.
For example, whether the previous rolls showed a red or a white face, the next roll has the same chance of being red or white when the coin decides the die to use.
  • Each die selection is a separate event.
  • The probability of the outcome for one roll is unaffected by previous outcomes.
Because of this, even if the first two rolls result in red, it doesn't increase the likelihood of red appearing in the third roll.
Conditional Probability
Conditional probability is the likelihood of an event occurring given that another event has already occurred. It is central to figuring out complex probability situations.
In the exercise, we want the probability of the third throw being red, given that the first two throws were red. However, due to independent events, the probability remains the same,
because each roll's result is influenced only by the die chosen, or the previous throws. Thus, knowing previous throws does not change the odds of the next result.
  • This concept shows how probability can still be determined despite conditions being placed on events.
  • Realizing independence helps clarify why certain conditions do not affect probability.
Bayes' Theorem
Bayes' theorem is a way to calculate conditional probabilities. It uses prior knowledge to calculate the probability of an event.
For example, using Bayes' theorem, we determine the likelihood that die A is used when the first two throws were red. It involves comparing how likely each die is to cause red outcomes.
The formula is \[ P(A|R,R) = \frac{P(R|R,A) \cdot P(A)}{P(R|R)} \]
  • It uses prior probability of choosing a die and the probability of getting red outcomes.
  • Helps us revise previous probability beliefs.
Bayes鈥� theorem enhances understanding by relating one probability to another through known variables.
Fair Coin
A fair coin means it has an equal chance of landing on heads or tails. This is crucial in our exercise because each side results in using different dice with different probabilities of landing red.
It ensures that the probability calculations for the choice of die are unbiased. With a fair coin, each die has an equal chance of being selected before each throw.
  • Each die selection is random and equal.
  • Equal distribution is essential for fair probability calculations.
This provides a balanced and unpredictable element to the selection process, ensuring each roll is determined independently. The fairness of the coin flip means neither die is favored.

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Most popular questions from this chapter

\(A\) and \(B\) flip coins. \(A\) starts and continues flipping until a tail occurs, at which point \(B\) starts flipping and continues until there is a tail. Then \(A\) takes over, and so on. Let \(P_{1}\) be the probability of the coin's landing on heads when \(A\) flips and \(P_{2}\) when \(B\) flips. The winner of the game is the first one to get (a) 2 heads in a row; (b) a total of 2 heads; (c) 3 heads in a row; (d) a total of 3 heads. In each case, find the probability that \(A\) wins.

Suppose that each child born to a couple is equally likely to be a boy or a girl, independently of the sex distribution of the other children in the family. For a couple having 5 children, compute the probabilities of the following events: (a) All children are of the same sex. (b) The 3 eldest are boys and the others girls. (c) Exactly 3 are boys. (d) The 2 oldest are girls. (e) There is at least 1 girl.

An engineering system consisting of \(n\) components is said to be a \(k\) -out- of- \(n\) system \((k \leq n)\) if the system functions if and only if at least \(k\) of the \(n\) components function. Suppose that all components function independently of one another. (a) If the \(i\) th component functions with probability \(P_{i}, i=\) \(1,2,3,4,\) compute the probability that a 2 -out-of- 4 system functions. (b) Repeat part (a) for a 3 -out-of- 5 system. (c) Repeat for a \(k\) -out-of- \(n\) system when all the \(P_{i}\) equal \(p\) (that is, \(\left.P_{i}=p, i=1,2, \ldots, n\right)\)

Consider a sample of size 3 drawn in the following manner: We start with an urn containing 5 white and 7 red balls. At each stage, a ball is drawn and its color is noted. The ball is then returned to the urn, along with an additional ball of the same color. Find the probability that the sample will contain exactly. (a) 0 white balls; (b) 1 white ball; (c) 3 white balls; (d) 2 white balls.

The following method was proposed to estimate the number of people over the age of 50 who reside in a town of known population 100,000: "As you walk along the streets, keep a running count of the percentage of people you encounter who are over 50\. Do this for a few days; then multiply the percentage you obtain by 100,000 to obtain the estimate." Comment on this method. Hint: Let \(p\) denote the proportion of people in the town who are over \(50 .\) Furthermore, let \(\alpha_{1}\) denote the proportion of time that a person under the age of 50 spends in the streets, and let \(\alpha_{2}\) be the corresponding value for those over \(50 .\) What quantity does the method suggested estimate? When is the estimate approximately equal to \(p ?\)
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