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Chapter 1: Problem 4

Let \(E, F, G\) be three events. Find expressions for the events that of \(E, F, G\) (a) only \(F\) occurs, (b) both \(E\) and \(F\) but not \(G\) occur, (c) at least one event occurs, (d) at least two events occur, (e) all three events occur, (f) none occurs, (g) at most one occurs, (h) at most two occur.

Short Answer

Expert verified
\( (a) F \cap E^{c} \cap G^{c}; \newline (b) E \cap F \cap G^{c}; \newline (c) E \cup F \cup G; \newline (d) (E \cap F) \cup (E \cap G) \cup (F \cap G); \newline (e) E \cap F \cap G; \newline (f) E^{c} \cap F^{c} \cap G^{c}; \newline (g) (E \cap F^{c} \cap G^{c}) \cup (E^{c} \cap F \cap G^{c}) \cup (E^{c} \cap F^{c} \cap G) \cup (E^{c} \cap F^{c} \cap G^{c}); \newline (h) (E \cap F \cap G^{c}) \cup (E \cap F^{c} \cap G) \cup (E^{c} \cap F \cap G) \cup (E \cap F^{c} \cap G^{c}) \cup (E^{c} \cap F \cap G^{c}) \cup (E^{c} \cap F^{c} \cap G) \)

Step by step solution

01

Exercise (a) - only F occurs

An expression for the event that only F occurs (and neither E nor G) is: \[ F \cap E^{c} \cap G^{c} \]
02

Exercise (b) - both E and F but not G occur

An expression where both events E and F occur, and G does not occur, is: \[ E \cap F \cap G^{c} \]
03

Exercise (c) - at least one event occurs

An expression for the event that at least one of the events E, F, or G occurs is: \[ E \cup F \cup G \]
04

Exercise (d) - at least two events occur

An expression for the event that at least two of the events E, F, or G occur is: \[ (E \cap F) \cup (E \cap G) \cup (F \cap G) \]
05

Exercise (e) - all three events occur

An expression for the event that all three events E, F, and G occur is: \[ E \cap F \cap G \]
06

Exercise (f) - none occurs

An expression for the event that none of the events E, F, or G occurs is: \[ E^{c} \cap F^{c} \cap G^{c} \]
07

Exercise (g) - at most one occurs

An expression for the event that "at most one" of the events E, F, or G occurs is: \[ (E \cap F^{c} \cap G^{c}) \cup (E^{c} \cap F \cap G^{c}) \cup (E^{c} \cap F^{c} \cap G) \cup (E^{c} \cap F^{c} \cap G^{c}) \]
08

Exercise (h) - at most two occur

An expression for the event that "at most two" of the events E, F, or G occur is: \[ (E \cap F \cap G^{c}) \cup (E \cap F^{c} \cap G) \cup (E^{c} \cap F \cap G) \cup (E \cap F^{c} \cap G^{c}) \cup (E^{c} \cap F \cap G^{c}) \cup (E^{c} \cap F^{c} \cap G)\]

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Most popular questions from this chapter

Suppose we have ten coins which are such that if the \(i\) th one is flipped then heads will appear with probability \(i / 10, i=1,2, \ldots, 10\). When one of the coins is randomly selected and flipped, it shows heads. What is the conditional probability that it was the fifth coin?

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.

For events \(E_{1}, E_{2}, \ldots, E_{n}\) show that $$ P\left(E_{1} E_{2} \cdots E_{n}\right)=P\left(E_{1}\right) P\left(E_{2} \mid E_{1}\right) P\left(E_{3} \mid E_{1} E_{2}\right) \cdots P\left(E_{n} \mid E_{1} \cdots E_{n-1}\right) $$

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?

(a) A gambler has in his pocket a fair coin and a two-headed coin. He selects one of the coins at random, and when he flips it, it shows heads. What is the probability that it is the fair coin? (b) Suppose that he flips the same coin a second time and again it shows heads. Now what is the probability that it is the fair coin? (c) Suppose that he flips the same coin a third time and it shows tails. Now what is the probability that it is the fair coin?
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