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Chapter 4: Problem 67

Given that \(A\) and \(B\) are two independent events, find their joint probability for the following. a. \(P(A)=.29 \quad\) and \(\quad P(B)=.65\) b. \(P(A)=.03\) and \(P(B)=.28\)

Short Answer

Expert verified
The joint probabilities are .1885 and .0084 respectively.

Step by step solution

01

Calculate joint probability for the first part

Use the joint probability formula \(P(A \cap B) = P(A) \times P(B)\). Insert the given probabilities into the formula: \(P(A \cap B) = .29 \times .65 = .1885\).
02

Calculate joint probability for the second part

Again, use the joint probability formula \(P(A \cap B) = P(A) \times P(B)\). Insert the given probabilities into the formula: \(P(A \cap B) = .03 \times .28 = .0084\).
03

Present the results

The joint probability of events A and B when given \(P(A)=.29\) and \(P(B)=.65\) equals to .1885. Whereas, when given \(P(A)=.03\) and \(P(B)=.28\), the joint probability equals to .0084.

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Most popular questions from this chapter

Twenty percent of a town's voters favor letting a major discount store move into their neighborhood, \(63 \%\) are against it, and \(17 \%\) are indifferent. What is the probability that a randomly selected voter from this town will either be against it or be indifferent? Explain why this probability is not equal to \(1.0\).

The probability that a student graduating from Suburban State University has student loans to pay off after graduation is .60. The probability that a student graduating from this university has student loans to pay off after graduation and is a male is \(.24\). Find the conditional probability that a randomly selected student from this university is a male given that this student has student loans to pay off after graduation.

Given that \(P(A \mid B)=.44\) and \(P(A\) and \(B)=.33\), find \(P(B)\).

A gambler has four cards - two diamonds and two clubs. The gambler proposes the following game to you: You will leave the room and the gambler will put the cards face down on a table. When you return to the room, you will pick two cards at random. You will win $$\$ 10$$ if both cards are diamonds, you will win $$\$ 10$$ if both are clubs, and for any other outcome you will lose $$\$ 10$$. Assuming that there is no cheating, should you accept this proposition? Support your answer by calculating your probability of winning $$\$ 10$$.

How is the multiplication rule of probability for two dependent events different from the rule for two independent events?
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