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Chapter 4: Problem 30

Suppose that we have two events, \(A\) and \(B,\) with \(P(A)=.50, P(B)=.60,\) and \(P(A \cap B)=.40\) a. Find \(P(A | B)\) b. Find \(P(B | A)\) c. Are \(A\) and \(B\) independent? Why or why not?

Short Answer

Expert verified
a. \( P(A | B) = 0.67 \) b. \( P(B | A) = 0.80 \) c. Events \( A \) and \( B \) are not independent.

Step by step solution

01

Calculate Conditional Probability P(A | B)

To find the conditional probability \( P(A | B) \), use the formula: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \] Substitute the given values: \( P(A \cap B) = 0.40 \) and \( P(B) = 0.60 \): \[ P(A | B) = \frac{0.40}{0.60} = \frac{2}{3} \approx 0.67 \]
02

Calculate Conditional Probability P(B | A)

To find the conditional probability \( P(B | A) \), use the formula: \[ P(B | A) = \frac{P(A \cap B)}{P(A)} \] Substitute the given values: \( P(A \cap B) = 0.40 \) and \( P(A) = 0.50 \): \[ P(B | A) = \frac{0.40}{0.50} = \frac{4}{5} = 0.80 \]
03

Check Independence of Events A and B

Two events \( A \) and \( B \) are independent if \( P(A \cap B) = P(A)P(B) \). Calculate: \[ P(A)P(B) = 0.50 \times 0.60 = 0.30 \] Given \( P(A \cap B) = 0.40 \), since \( P(A \cap B) eq P(A)P(B) \), events \( A \) and \( B \) are not independent.

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Key Concepts

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

Independent Events
In probability, when we talk about independent events, we mean two (or more) events that do not affect each other's outcomes. To be precise, events \( A \) and \( B \) are considered independent if the occurrence of \( A \) has no impact on the probability of event \( B \) happening, and vice versa.

This relationship can be formulated mathematically as follows:
  • \( P(A \cap B) = P(A) \cdot P(B) \)
If the above condition holds, only then are \( A \) and \( B \) independent.

In the context of the exercise we've looked at, since \( P(A \cap B) = 0.40 \) is not equal to \( P(A) \cdot P(B) = 0.30 \), this clearly indicates that \( A \) and \( B \) are *not* independent events. Thus, the probability that one event occurs is influenced by the occurrence of the other event.
Probability of Intersection
The probability of intersection, denoted as \( P(A \cap B) \), refers to the probability that both events \( A \) *and* \( B \) occur simultaneously.

This is a crucial concept in understanding how two events interact with each other. In this context, the probability of the intersection of \( A \) and \( B \) is directly provided in the exercise.

Mathematically, calculating the intersection probability involves looking at the overlap in the probability of the two events. In cases where events are independent, this is simple as described:\[ P(A \cap B) = P(A) \cdot P(B) \]

However, when dealing with dependent events, as in our example where \( P(A \cap B) = 0.40 \), this equation doesn't hold, indicating the events influence each other. The given value represents the proportion or likelihood of the outcomes where both events happen together.
Conditional Probability Formula
Conditional probability helps us understand probabilities in the context of a certain condition or scenario having occurred.

This is a versatile concept used to update our probability assessments given new information. To find the conditional probability of event \( A \) given that \( B \) has occurred, denoted \( P(A | B) \), we use the formula:
  • \( P(A | B) = \frac{P(A \cap B)}{P(B)} \)

This formula essentially tells us how likely \( A \) is to occur if we know \( B \) has happened.

Similarly, we can find \( P(B | A) \) or the probability of \( B \) given \( A \) occurred:
  • \( P(B | A) = \frac{P(A \cap B)}{P(A)} \)

In our exercise, we've calculated \( P(A | B) = 0.67 \) and \( P(B | A) = 0.80 \) using this formula. These computations allow us to see how knowledge of one event happening influences our likelihood assessment of the other event happening.

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Most popular questions from this chapter

Suppose that we have a sample space \(S=\left\\{E_{1}, E_{2}, E_{3}, E_{4}, E_{5}, E_{6}, E_{7}\right\\},\) where \(E_{1}, E_{2}, \ldots,\) \(E_{7}\) denote the sample points. The following probability assignments apply: \(P\left(E_{1}\right)=.05\) \(P\left(E_{2}\right)=.20, P\left(E_{3}\right)=.20, P\left(E_{4}\right)=.25, P\left(E_{5}\right)=.15, P\left(E_{6}\right)=.10,\) and \(P\left(E_{7}\right)=.05 .\) Let $$\begin{array}{l} A=\left\\{E_{1}, E_{4}, E_{6}\right\\} \\ B=\left\\{E_{2}, E_{4}, E_{7}\right\\} \\ C=\left\\{E_{2}, E_{3}, E_{5}, E_{7}\right\\} \end{array}$$ a. Find \(P(A), P(B),\) and \(P(C)\) b. Find \(A \cup B\) and \(P(A \cup B)\) c. Find \(A \cap B\) and \(P(A \cap B)\) d. Are events \(A\) and \(C\) mutually exclusive? e. Find \(B^{c}\) and \(P\left(B^{c}\right)\)

Suppose that we have a sample space with five equally likely experimental outcomes: \(E_{1}\) \(E_{2}, E_{3}, E_{4}, E_{5} \text { . Let } \) \\[ \begin{aligned} \qquad \begin{aligned} A &=\left\\{E_{1}, E_{2}\right\\} \\ B &=\left\\{E_{3}, E_{4}\right\\} \\ C &=\left\\{E_{2}, E_{3}, E_{5}\right\\} \end{aligned} \end{aligned} \\] a. Find \(P(A), P(B),\) and \(P(C)\) b. Find \(P(A \cup B)\). Are \(A\) and \(B\) mutually exclusive? c. \(\quad\) Find \(A^{c}, C^{c}, P\left(A^{c}\right),\) and \(P\left(C^{c}\right)\) d. Find \(A \cup B^{c}\) and \(P\left(A \cup B^{c}\right)\) e. Find \(P(B \cup C)\)

Small cars get better gas mileage, but they are not as safe as bigger cars. Small cars accounted for \(18 \%\) of the vehicles on the road, but accidents involving small cars led to 11,898 fatalities during a recent year (Reader's Digest, May 2000 ). Assume the probability a small car is involved in an accident is .18. The probability of an accident involving a small car leading to a fatality is .128 and the probability of an accident not involving a small car leading to a fatality is .05. Suppose you learn of an accident involving a fatality. What is the probability a small car was involved? Assume that the likelihood of getting into an accident is independent of car size.

Reggie Miller of the Indiana Pacers is the National Basketball Association's best career free throw shooter, making \(89 \%\) of his shots \((U S A \text { Today, January } 22,2004\) ). Assume that late in a basketball game, Reggie Miller is fouled and is awarded two shots. a. What is the probability that he will make both shots? b. What is the probability that he will make at least one shot? c. What is the probability that he will miss both shots? d. Late in a basketball game, a team often intentionally fouls an opposing player in order to stop the game clock. The usual strategy is to intentionally foul the other team's worst free throw shooter. Assume that the Indiana Pacers' center makes \(58 \%\) of his free throw shots. Calculate the probabilities for the center as shown in parts (a), (b), and (c), and show that intentionally fouling the Indiana Pacers' center is a better strategy than intentionally fouling Reggie Miller.

Assume that we have two events, \(A\) and \(B\), that are mutually exclusive. Assume further that we know \(P(A)=.30\) and \(P(B)=.40\) a. What is \(P(A \cap B) ?\) b. What is \(P(A | B) ?\) c. A student in statistics argues that the concepts of mutually exclusive events and independent events are really the same, and that if events are mutually exclusive they must be independent. Do you agree with this statement? Use the probability information in this problem to justify your answer. d. What general conclusion would you make about mutually exclusive and independent events given the results of this problem?
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