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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 we have ten coins which are such that if the ith 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 nipped, it shows heads. What is the conditional probability that it was the fifth coin?

Use Exercise 15 to show that \(P(E \cup F)=P(E)+P(F)-P(E F)\).

Bill and George go target shooting together. Both shoot at a target at the same time. Suppose Bill hits the target with probability \(0.7\), whereas George, independently, hits the target with probability \(0.4\). (a) Given that exactly one shot hit the target, what is the probability that it was George's shot? (b) Given that the target is hit, what is the probability that George hit it?

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?

A fair coin is continually flipped. What is the probability that the first four flips are (a) \(H, H, H, H ?\) (b) \(T, H, H, H ?\) (c) What is the probability that the pattern \(T, H, H, H\) occurs before the pattern \(\boldsymbol{H}, \boldsymbol{H}, \boldsymbol{H}, \boldsymbol{H} ?\)
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