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Chapter 4: Problem 26

Suppose that \(A\) and \(B\) are events such that \(P(A \mid B)=P(B \mid A)\) and \(P(A \cup B)=\) 1 and \(P(A \cap B)>0\). Prove that \(P(A)>1 / 2\).

Short Answer

Expert verified
Under the given conditions, it is shown that \(P(A) > 1/2\).

Step by step solution

01

Understanding the Relationship of Conditional Probabilities

We know that the conditional probabilities of events \(A\) and \(B\) are equal, i.e., \(P(A \mid B) = P(B \mid A)\). Recall that \(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\) and \(P(B \mid A) = \frac{P(B \cap A)}{P(A)}\). Therefore, we can write:\[\frac{P(A \cap B)}{P(B)} = \frac{P(A \cap B)}{P(A)}.\]Since \(P(A \cap B) > 0\), we can cancel it from both sides, obtaining:\[P(B) = P(A).\]
02

Analyzing the Union of Events

It is given that \(P(A \cup B) = 1\). Using the formula for the probability of the union of two events, we have:\[P(A \cup B) = P(A) + P(B) - P(A \cap B).\]Substitute \(P(A \cup B) = 1\) and \(P(B) = P(A)\) into the equation:\[1 = P(A) + P(A) - P(A \cap B).\]
03

Simplifying the Equation

Simplify the equation from the previous step:\[1 = 2P(A) - P(A \cap B).\]Rearrange this to find \(P(A)\):\[2P(A) = 1 + P(A \cap B).\]
04

Solving for the Probability of Event A

Given \(P(A \cap B) > 0\), we substitute a small positive value such that:\[P(A \cap B) \Rightarrow 1 + P(A \cap B) > 1.\]Thus, dividing both sides by 2, we obtain:\[P(A) > \frac{1}{2}.\]
05

Conclusion

Thus, we have shown that given the conditions, the probability of event A is greater than 1/2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Union of Events
In probability theory, the union of events concept is a pivotal element in understanding how combined likelihoods work when dealing with multiple occurrences. The union of events, symbolized as \(A \cup B\), represents the scenario in which either event \(A\), event \(B\), or both occur. The formula for calculating the probability of a union of events is:
  • \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
This formula accounts for the potential overlap between the two events, which is subtracted to avoid double-counting. If you're given \(P(A \cup B) = 1\), it means that one or both of the events are certain to happen.
In our original problem, knowing the union equals one simplifies some calculations and provides insight into how these events interrelate. Clearly, having a union probability of 1 indicates a certainty where either of the two events or both occur in every possible outcome.
This sets the stage for finding individual probabilities, leveraging other known probabilities, such as the intersection.
Intersection of Events
The intersection of events is a fundamental concept used to quantify the probability of two events happening at the same time. Denoted by \(A \cap B\), it represents the instance where both event \(A\) and event \(B\) occur simultaneously. The probability of this intersection helps in understanding common occurrences between events and is crucial for further calculations in probability theory.
In our scenario, we know that \(P(A \cap B) > 0\), which is key for deriving further relationships. By knowing that the intersection of \(A\) and \(B\) exists (i.e., it is greater than zero), we can confidently simplify and solve equations that involve both these events.
The presence of an intersection helps us exploit the conditional probabilities we have (like \(P(A \mid B)\) and \(P(B \mid A)\)) to find other individual probabilities.
Ultimately, understanding the intersection allows us to manipulate equations more effectively and unlock insights that wouldn't be apparent if \(P(A \cap B)\) were zero.
Probability Equations
Probability equations form the backbone of solving problems where likelihoods need manipulation and derivation. Typically, they include expressions like:
  • \(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\)
  • \(P(B \mid A) = \frac{P(B \cap A)}{P(A)}\)
  • The union and intersection expressed as \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
These formulas allow us to break down and reconstruct the related probabilities systematically.
In exercises like the one given in the original solution, accurately re-arranging and simplifying these equations is crucial. Getting \(P(A \mid B) = P(B \mid A)\) tells us that the likelihood of \(A\) happening given \(B\) is the same as \(B\) happening given \(A\). This equality, paired with the given probability of the union and intersection, aids us in concluding that \(P(A) > \frac{1}{2}\).
Understanding and utilizing these probability equations ensures that we can find relationships and insights hidden within probability problems.

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Most popular questions from this chapter

Luxco, a wholesale lightbulb manufacturer, has two factories. Factory A sells bulbs in lots that consists of 1000 regular and 2000 softglow bulbs each. Random sampling has shown that on the average there tend to be about 2 bad regular bulbs and 11 bad softglow bulbs per lot. At factory B the lot size is reversed - there are 2000 regular and 1000 softglow per lot- and there tend to be 5 bad regular and 6 bad softglow bulbs per lot. The manager of factory A asserts, "We're obviously the better producer; our bad bulb rates are 2 percent and .55 percent compared to B's .25 percent and .6 percent. We're better at both regular and softglow bulbs by half of a tenth of a percent each." "Au contraire," counters the manager of B, "each of our 3000 bulb lots contains only 11 bad bulbs, while A's 3000 bulb lots contain 13\. So our .37 percent bad bulb rate beats their .43 percent." Who is right?

Suppose that \(X\) and \(Y\) are continuous random variables with density functions \(f_{X}(x)\) and \(f_{Y}(y)\), respectively. Let \(f(x, y)\) denote the joint density function of \((X, Y)\). Show that $$ \int_{-\infty}^{\infty} f(x, y) d y=f_{X}(x) $$ and $$ \int_{-\infty}^{\infty} f(x, y) d x=f_{Y}(y) $$

A fair coin is tossed three times. Let \(X\) be the number of heads that turn up on the first two tosses and \(Y\) the number of heads that turn up on the third toss. Give the distribution of (a) the random variables \(X\) and \(Y\). (b) the random variable \(Z=X+Y\). (c) the random variable \(W=X-Y\).

It is desired to find the probability that in a bridge deal each player receives an ace. A student argues as follows. It does not matter where the first ace goes. The second ace must go to one of the other three players and this occurs with probability \(3 / 4 .\) Then the next must go to one of two, an event of probability \(1 / 2,\) and finally the last ace must go to the player who does not have an ace. This occurs with probability \(1 / 4 .\) The probability that all these events occur is the product \((3 / 4)(1 / 2)(1 / 4)=3 / 32 .\) Is this argument correct?

Probability theory was used in a famous court case: People v. Collins. \(^{10}\) In this case a purse was snatched from an elderly person in a Los Angeles suburb. A couple seen running from the scene were described as a black man with a beard and a mustache and a blond girl with hair in a ponytail. Witnesses said they drove off in a partly yellow car. Malcolm and Janet Collins were arrested. He was black and though clean shaven when arrested had evidence of recently having had a beard and a mustache. She was blond and usually wore her hair in a ponytail. They drove a partly yellow Lincoln. The prosecution called a professor of mathematics as a witness who suggested that a conservative set of probabilities for the characteristics noted by the witnesses would be as shown in Table 4.5 . The prosecution then argued that the probability that all of these characteristics are met by a randomly chosen couple is the product of the probabilities or \(1 / 12,000,000\), which is very small. He claimed this was proof beyond a reasonable doubt that the defendants were guilty. The jury agreed and handed down a verdict of guilty of second-degree robbery. $$ \begin{aligned} &\begin{array}{lc} \text { man with mustache } & 1 / 4 \\ \text { girl with blond hair } & 1 / 3 \\ \text { girl with ponytail } & 1 / 10 \\ \text { black man with beard } & 1 / 10 \\ \text { interracial couple in a car } & 1 / 1000 \\ \text { partly yellow car } & 1 / 10 \end{array}\\\ &\text { Table 4.5: Collins case probabilities. } \end{aligned} $$ If you were the lawyer for the Collins couple how would you have countered the above argument? (The appeal of this case is discussed in Exercise 5.1.34.)
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