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Chapter 1: Problem 28

If the occurrence of \(B\) makes \(A\) more likely, does the occurrence of \(A\) make \(B\) more likely?

Short Answer

Expert verified
In conclusion, just because the occurrence of B makes A more likely, it does not imply that the occurrence of A will make B more likely. The relationship between the occurrence of A making B more likely will depend on the values of probabilities of events A and B, and the probability of both events happening together, which can vary depending on the specific scenario.

Step by step solution

01

Understand conditional probability

Conditional probability is denoted by P(A|B), which means the probability of event A occurring given that event B has occurred. It is calculated using the formula: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\) Here, P(A 鈭� B) is the probability of both A and B happening, and P(B) is the probability of event B happening.
02

Analyse the given condition

We are given that the occurrence of B makes A more likely. Mathematically, this can be represented as: \(P(A|B) > P(A)\) Which means the probability of A happening, given that B has happened, is greater than the probability of A happening without any information about B.
03

Determine if the occurrence of A makes B more likely

Now, we need to find if the occurrence of A makes B more likely. In other words, we need to determine whether: \(P(B|A) > P(B)\) Using the formula of conditional probability, we can rewrite this expression as: \(\frac{P(A \cap B)}{P(A)} > P(B)\) Now, multiplying both sides by P(A): \(P(A \cap B) > P(A)P(B)\) Comparing this inequality to the initial condition, we cannot conclude a definitive relationship between the occurrence of A making B more likely or not. It is because the inequality depends on the values of the probabilities, which can vary.
04

Conclusion

In conclusion, just because the occurrence of B makes A more likely, it does not imply that the occurrence of A will make B more likely. The relationship between the occurrence of A making B more likely will depend on the values of probabilities of events A and B, and the probability of both events happening together, which can vary depending on the specific scenario.

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Most popular questions from this chapter

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?

If \(P(E)=0.9\) 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.

(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 a 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 a 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 a fair coin?

We say that \(E \subset F\) if every point in \(E\) is also in \(F\). Show that if \(E \subset F\), then $$ P(F)=P(E)+P\left(F E^{c}\right) \geqslant P(E) $$

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?
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