Chapter 1: Problem 28
If the occurrence of \(B\) makes \(A\) more likely, does the occurrence of \(A\) make \(B\) more likely?
Chapter 1: Problem 28
If the occurrence of \(B\) makes \(A\) more likely, does the occurrence of \(A\) make \(B\) more likely?
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Get started for freeIf \(P(E)=0.9\) and \(P(F)=0.8\), show that \(P(E F) \geqslant 0.7\). In general, show that $$ P(E F) \geqslant P(E)+P(F)-1 $$ This is known as Bonferroni's inequality.
Stores \(A, B\), and \(C\) have 50,75, and 100 employees, and, respectively, 50,60, and 70 percent of these are women. Resignations are equally likely among all employees, regardless of sex. One employee resigns and this is a woman. What is the probability that she works in-store \(C\)?
Suppose all \(n\) men at a party throw their hats in the center of the room. Each man then randomly selects a hat. Show that the probability that none of the \(n\) men selects his own hat is $$ \frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-+\cdots \frac{(-1)^{n}}{n !} $$ Note that as \(n \rightarrow \infty\) this converges to \(e^{-1}\). Is this surprising?
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?
Let \(E\) and \(F\) be mutually exclusive events in the sample space of an experiment. Suppose that the experiment is repeated until either event \(E\) or event \(F\) occurs. What does the sample space of this new super experiment look like? Show that the probability that event \(E\) occurs before event \(F\) is \(P(E) /[P(E)+P(F)]\). Hint: Argue that the probability that the original experiment is performed \(n\) times and \(E\) appears on the \(n\) th time is \(P(E) \times(1-p)^{n-1}, n=1,2, \ldots\), where \(p=P(E)+\) \(P(F)\). Add these probabilities to get the desired answer.
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