Chapter 1: Problem 20
Three dice are thrown. What is the probability the same number appears on exactly two of the three dice?
/*! 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}
Learning Materials
Features
Discover
Chapter 1: Problem 20
Three dice are thrown. What is the probability the same number appears on exactly two of the three dice?
All the tools & learning materials you need for study success - in one app.
Get started for free
If the occurrence of \(B\) makes \(A\) more likely, does the occurrence of \(A\) make \(B\) more likely?
An urn contains \(b\) black balls and \(r\) red balls. One of the balls is drawn at random, but when it is put back in the urn \(c\) additional balls of the same color are put in with it. Now suppose that we draw another ball. Show that the probability that the first ball drawn was black given that the second ball drawn was red is \(b /(b+r+c)\).
For events \(E_{1}, E_{2}, \ldots, E_{n}\) show that $$ P\left(E_{1} E_{2} \cdots E_{n}\right)=P\left(E_{1}\right) P\left(E_{2} \mid E_{1}\right) P\left(E_{3} \mid E_{1} E_{2}\right) \cdots P\left(E_{n} \mid E_{1} \cdots E_{n-1}\right) $$
Suppose that \(P(E)=0.6\). What can you say about \(P(E \mid F)\) when (a) \(E\) and \(F\) are mutually exclusive? (b) \(E \subset F ?\) (c) \(F \subset E ?\)
Three prisoners are informed by their jailer that one of them has been chosen at random to be executed, and the other two are to be freed. Prisoner \(A\) asks the jailer to tell him privately which of his fellow prisoners will be set free, claiming that there would be no harm in divulging this information, since he already knows that at least one will go free. The jailer refuses to answer this question, pointing out that if \(A\) knew which of his fellows were to be set free, then his own probability of being executed would rise from \(\frac{1}{3}\) to \(\frac{1}{2}\), since he would then be one of two prisoners. What do you think of the jailer's reasoning?
What do you think about this solution?
We value your feedback to improve our textbook solutions.