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Chapter 4: Problem 85

Given \(P(\mathrm{A} \text { or } \mathrm{B})=1.0, P(\overline{\mathrm{A} \text { and } \mathrm{B}})=0.7,\) and \(P(\overline{\mathbf{B}})=0.4,\) find: a. \(\quad P(B)\) b. \(\quad P(\mathrm{A})\) c. \(\quad P(\mathrm{A} | \mathrm{B})\)

Short Answer

Expert verified
a. The probability of B, \(P(B) = 0.6\), b. The probability of A, \(P(A) = 0.7\), c. The conditional probability of A given B, \(P(A|B) = 0.5\)

Step by step solution

01

Calculate P(B)

Use the third statement \(P(\overline{B}) = 0.4\). Since \(\overline{B}\) is the event of not B, it refers to the complement of B. Therefore, you can calculate the probability of B as \(P(B) = 1 - P(\overline{B}) = 1 - 0.4 = 0.6\)
02

Calculate P(A)

The first statement \(P(A \text{ or } B) = 1.0\) indicates the event of A or B happening, considered as the union of events A and B. According to the formula of union of two sets, \(P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)\). But the second statement \(P(\overline{A \text{ and } B}) = 0.7\) refers to the complement of the event A and B, hence \(P(A \text{ and } B) = 1 - P(\overline{A \text{ and } B}) = 1 - 0.7 = 0.3\), and you can find \(P(A)\) by rearranging the formula as follows: \(P(A) = P(A \text{ or } B) - P(B) + P(A \text{ and } B) = 1 - 0.6 + 0.3 = 0.7\)
03

Calculate P(A|B)

This is a conditional probability, defined by \(P(A|B) = P(A \text{ and } B) / P(B)\). To find the probability of A given that B has occurred, divide the probability of both A and B, which was previously found to be 0.3, by the probability of B, which was found to be 0.6. Hence, \(P(A|B) = 0.3 / 0.6 = 0.5\)

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Key Concepts

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

Complement of an Event
Understanding the complement of an event is essential to mastering probability theory. When studying the probability of events within a fixed set of possibilities, we often come across situations where it's easier to evaluate the chance of an event not happening rather than the event occurring. In probability terms, this 'not happening' event is called the complement.

If we denote an event as 'A', its complement is represented by '\bar{A}' or 'not A'. Mathematically, the probability of an event and its complement always add up to one. This is expressed as:
\[ P(A) + P(\overline{A}) = 1 \]
This is because either the event will happen, or it will not, covering all possible outcomes. When you know the probability of an event not occurring, you can easily find the probability of the event occurring by subtracting it from one:
\[ P(A) = 1 - P(\overline{A}) \]
This relationship is of pivotal importance when solving probability problems, as it provides a different perspective that can simplify complicated problems.
Union of Events
The union of two events, represented by 'A or B', encompasses any situation where either event A, event B, or both events occur concurrently. In probability, the union is synonymous with the logical

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Most popular questions from this chapter

Events \(\mathrm{R}\) and \(\mathrm{S}\) are defined on a sample space. If \(P(R)=0.2\) and \(P(S)=0.5,\) explain why each of the following statements is either true or false: a. If \(\mathrm{R}\) and \(\mathrm{S}\) are mutually exclusive, then \(P(\mathrm{R} \text { or } \mathrm{S})=0.10\) b. If \(R\) and \(S\) are independent, then \(P(R \text { or } S)=0.6\) c. If \(R\) and \(S\) are mutually exclusive, then \(P(R \text { and } S)=0.7\) d. If \(R\) and \(S\) are mutually exclusive, then \(P(\mathrm{R} \text { or } \mathrm{S})=0.6\)

A company that manufactures shoes has three factories. Factory 1 produces \(25 \%\) of the company's shoes, Factory 2 produces \(60 \%,\) and Factory 3 produces \(15 \%\) One percent of the shoes produced by Factory 1 are mislabeled, \(0.5 \%\) of those produced by Factory 2 are mislabeled, and \(2 \%\) of those produced by Factory 3 are mislabeled. If you purchase one pair of shoes manufactured by this company, what is the probability that the shoes are mislabeled?

Coin A is loaded in such a way that \(P\) (heads) is 0.6. Coin \(B\) is a balanced coin. Both coins are tossed. Find: a. The sample space that represents this experiment; assign a probability measure to each outcome b. \(P(\text { both show heads })\) c. \(P(\) exactly one head shows) d. \(P(\text { neither coin shows a head })\) e. \(P(\text { both show heads } |\) coin A shows a head) f. \(P\) (both show heads| coin B shows a head) g. \(P\) (heads on coin A | exactly one head shows)

Three balanced coins are tossed simultaneously. Find the probability of obtaining three heads, given that at least one of the coins shows heads. a. Solve using an equally likely sample space. b. Solve using the formula for conditional probability.

Events \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) are defined on sample space \(\mathrm{S}\). Their corresponding sets of sample points do not intersect, and their union is S. Furthermore, event \(\mathrm{B}\) is twice as likely to occur as event \(\mathrm{A}\), and event \(\mathrm{C}\) is twice as likely to occur as event B. Determine the probability of each of the three events.
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