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Chapter 4: Problem 103

When is the following addition rule used to find the probability of the union of two events \(A\) and \(B\) ? $$P(A \text { or } B)=P(A)+P(B)$$ Give one example where you might use this formula.

Short Answer

Expert verified
The addition rule \(P(A or B) = P(A) + P(B)\) is used when two events \(A\) and \(B\) are mutually exclusive, i.e., they cannot occur at the same time. An example of its use would be in a single throw of a die, where events such as 'getting a 2' and 'getting a 3' are mutually exclusive. The probability of getting a 2 or a 3 is determined using this rule and equals 1/3.

Step by step solution

01

Mutually Exclusive Principle

The first step is to understand the principle of mutually exclusive events. In probability theory, two events are mutually exclusive if they can't occur at the same time. In other words, it's impossible for both events \(A\) and \(B\) to happen simultaneously.
02

Addition Rule for Probability

The addition rule for mutually exclusive events states that the probability of either event \(A\) or event \(B\) taking place is equal to the sum of their individual probabilities.
03

Application and Example

This principle is applied when we have two mutually exclusive events and we need to find the probability of either event happening. For example, when tossing a fair dice, the events \(A: 'getting a 2'\) and \(B: 'getting a 3'\) are mutually exclusive. The probability of getting a 2 or a 3 is \(P(A) = 1/6\) and \(P(B) = 1/6\). So, \(P(A or B) = P(A) + P(B) = 1/6 + 1/6 = 1/3\).

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Key Concepts

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

Addition Rule for Probability
The addition rule for probability is a method used to find the likelihood of either one of two events occurring. It is particularly useful when dealing with events that cannot happen at the same time, known as mutually exclusive events. When events are mutually exclusive, the probability of either event occurring is simply the sum of their individual probabilities.
The formula for the addition rule when dealing with mutually exclusive events is:\[ P(A \text{ or } B) = P(A) + P(B) \]
This rule is only applicable when the events do not overlap. In other words, the occurrence of one event rules out the possibility of the other happening at the same time.
Probability of the Union of Events
The union of events in probability indicates that either event occurs. If Events \(A\) and \(B\) are being considered, the union refers to any situation where either \(A\), \(B\), or both occur.
In the context of mutually exclusive events, the union (denoted usually as \(A \cup B\)) simplifies to a straightforward calculation thanks to the addition rule:
  • For mutually exclusive events, since there is no overlap, there isn鈥檛 a need to subtract the probability of the intersection (\(P(A \cap B)\)) because it equals zero.
  • The formula thus remains as \(P(A \text{ or } B) = P(A) + P(B)\).
Knowing when to use this can save significant calculation time and ensures accuracy in probability assessments.
Mutually Exclusive Principle
The concept of mutually exclusive events is a fundamental idea in probability. Two events are mutually exclusive if they cannot happen at the same time. A classic example is the rolling of a die: when you roll a single die, getting a '4' and getting a '6' are mutually exclusive events, as the die cannot show both numbers simultaneously in one roll.
Understanding this principle is vital because it dictates whether or not the addition rule for probability can be applied directly without any modifications.
  • Only mutually exclusive events fit perfectly into the simplest form of the addition rule, where \(P(A \text{ or } B) = P(A) + P(B)\).
  • For non-mutually exclusive events, one would have to adjust the calculation by subtracting the intersection probability \(P(A \cap B)\).
Thus, identifying mutually exclusive events ensures employing the correct method for calculating probabilities.

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Most popular questions from this chapter

A box contains 10 red marbles and 10 green marbles. a. Sampling at random from the box five times with replacement, you have drawn a red marble all five times. What is the probability of drawing a red marble the sixth time? b. Sampling at random from the box five times without replacement, you have drawn a red marble all five times. Without replacing any of the marbles, what is the probability of drawing a red marble the sixth time? c. You have tossed a fair coin five times and have obtained heads all five times. A friend argues that according to the law of averages, a tail is due to occur and, hence, the probability of obtaining a head on the sixth toss is less than \(.50 .\) Is he right? Is coin tossing mathematically equivalent to the procedure mentioned in part a or the procedure mentioned in part b? Explain.

The probability of a student getting an A grade in an economics class is \(.24\) and that of getting a B grade is \(.28\). What is the probability that a randomly selected student from this class will get an \(\mathrm{A}\) or a \(\mathrm{B}\) in this class? Explain why this probability is not equal to \(1.0\)

A random sample of 400 college students was asked if college athletes should be paid. The following table gives a two-way classification of the responses. $$\begin{array}{lcc} \hline & \text { Should Be Paid } & \text { Should Not Be Paid } \\ \hline \text { Student athlete } & 90 & 10 \\ \text { Student nonathlete } & 210 & 90 \end{array}$$ a. If one student is randomly selected from these 400 students, find the probability that this student i. is in favor of paying college athletes ii. favors paying college athletes given that the student selected is a nonathlete iii. is an athlete and favors paying student athletes iv. is a nonathlete or is against paying student athletes b. Are the events "student athlete" and "should be paid" independent? Are they mutually exclusive? Explain why or why not.

In a sample survey, 1800 senior citizens were asked whether or not they have ever been victimized by a dishonest telemarketer. The following table gives the responses by age group. $$\begin{array}{l|llcc} & & & \begin{array}{c} \text { Have Been } \\ \text { Victimized } \end{array} & \begin{array}{c} \text { Have Never } \\ \text { Been Victimized } \end{array} \\ \hline & 60-69 & \text { (A) } & 106 & 698 \\ \text { Age } & 70-79 & \text { (B) } & 145 & 447 \\ & 80 \text { or over (C) } & 61 & 343 \\ \hline \end{array}$$ Suppose one person is randomly selected from these senior citizens. Find the following probabilities a. \(P(\) have been victimized or \(\mathrm{B}\) ) b. \(P(\) have never been victimized or \(\mathrm{C}\) )

Briefly explain the three approaches to probability. Give one example of each approach.
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