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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 \(P(\mathrm{R} \text { or } \mathrm{S})=0.6\)

Short Answer

Expert verified
The statements in parts a and c of the exercise are false while the statements in parts b and d are true.

Step by step solution

01

Part A Analysis and Solution

Since \( R \) and \( S \) are said to be mutually exclusive, they cannot both occur at the same time. Hence the probability of either \( R \) or \( S \) occurring is the sum of their probabilities i.e. \( P(R \text { or } S)=P(R)+P(S) = 0.2 + 0.5 = 0.7 \). Therefore, the statement that \( P(R \text { or } S)=0.10 \) is false.
02

Part B Analysis and Solution

When \( R \) and \( S \) are independent, the probability that either \( R \) or \( S \) occurs is given by the formula \( P(R \text { or } S) = P(R) + P(S) - P(R)P(S) = 0.2 + 0.5 - 0.2*0.5 = 0.6 \). Therefore, the claim that \( P(R \text { or } S) = 0.6 \) is true.
03

Part C Analysis and Solution

Mutually exclusive events cannot occur simultaneously. Therefore, the probability of \( R \) and \( S \) occurring together is zero if they are mutually exclusive. Thus, the statement \( P(R \text { and } S) = 0.7 \) is false.
04

Part D Analysis and Solution

As in part a, the probability of either \( R \) or \( S \) occurring, when \( R \) and \( S \) are mutually exclusive, is the sum of their individual probabilities. Hence, the statement \( P(R \text { or } S) = 0.6) \) is true.

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Key Concepts

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

Mutually Exclusive Events
Understanding mutually exclusive events is fundamental to grasping the basics of probability theory. Two events are considered mutually exclusive when they cannot occur at the same time. Take, for example, flipping a coin; it cannot land both heads and tails simultaneously. The outcome must be one or the other, which makes these events mutually exclusive.

When calculating the probability of either of two mutually exclusive events occurring, we simply add their individual probabilities:
\[ P(A \text{ or } B) = P(A) + P(B) \]
This formula applies because there is no overlap between the events, which rules out the possibility of double-counting any shared outcomes. This principle was exemplified in our exercise, where the probability of either event R or S occurring was the sum of their probabilities, not 0.10 as one might incorrectly assume if not considering the definition of mutually exclusive events. If events R and S were mutually exclusive, the correct calculation would have yielded a combined probability of 0.7, not 0.6 or 0.10.
Independent Events
Independent events represent another crucial concept in the study of probability. Two events are independent if the occurrence of one does not affect the likelihood of the occurrence of the other. Again, considering the flip of a coin – whether it lands on heads or tails in one toss does not influence the outcome of the next toss.

For independent events, unlike mutually exclusive ones, we have a specific formula to calculate the probability of at least one event occurring:
\[ P(A \text{ or } B) = P(A) + P(B) - P(A) \times P(B) \]
This subtraction of \( P(A) \times P(B) \) accounts for the overcounting that happens when adding probabilities of independent events. In our original problem, the correct probability of R or S occurring, if they are independent, is 0.6, as derived from the provided solution, demonstrating the effectiveness of this formula.
Sample Space
The concept of a sample space is fundamental to all of probability theory. It is the set of all possible outcomes of a probability experiment. If you were rolling a standard six-sided die, the sample space would be \( S = \{1, 2, 3, 4, 5, 6\} \), consisting of all possible outcomes that could occur.

When we refer to events occurring within a certain sample space, we're essentially highlighting subsets of this space. An event is a specific outcome or a set of outcomes that we’re focusing on. For instance, having an event where the die rolls an even number is a subset of the sample space, \( \{2, 4, 6\} \), which contains all of the even numbers from the full range of possible outcomes. Understanding the sample space allows us to visualize all potential outcomes and calculate probabilities accurately by considering all relevant events within this space.

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Most popular questions from this chapter

If \(P(\mathrm{A})=0.5, P(\mathrm{B})=0.3,\) and \(P(\mathrm{A} \text { and } \mathrm{B})=0.2\) find \(P(\mathrm{A} \text { or } \mathrm{B})\)

a. Explain what is meant by the statement: "When a single die is rolled, the probability of a 1 is \(\frac{1}{6} "\) b. Explain what is meant by the statement: "When one coin is tossed one time, there is a \(50-50\) chance of getting a tail."

Suppose that \(P(\mathrm{A})=0.3, \quad P(\mathrm{B})=0.4,\) and a \(P(\mathrm{A} \text { and } \mathrm{B})=0.12\) a. What is \(P(A | B) ?\) b. What is \(P(B | A) ?\) c. Are \(A\) and \(B\) independent?

In sports, championships are often decided by two teams playing in a championship series Often the fans of the losing team claim they were unlucky and their team is actually the better team. Suppose Team \(A\) is the better team, and the probability it will defeat Team \(B\) in any one game is 0.6 a. What is the probability that the better team, Team \(A\), will win the series if it is a one-game series? b. What is the probability that the better team, Team A, will win the series if it is a best out of three series? c. What is the probability that the better team, Team A, will win the series if it is a best out of seven series? d. Suppose the probability that A would beat B in any given game were actually 0.7 Recompute parts \(a-c\) c. Suppose the probability that A would beat B in any given game were actually \(0.9 .\) Recompute parts a-c. f. What is the relationship between the "best" team winning and the number of games played? The best team winning and the probabilities that each will win?

A box contains four red and three blue poker chips. Three poker chips are to be randomly selected, one at a time. a. What is the probability that all three chips will be red if the selection is done with replacement? b. What is the probability that all three chips will be red if the selection is done without replacement? c. Are the drawings independent in either part a or b? Justify your answer.

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