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Chapter 2: Problem 75

Let \(A\) and \(B\) be two events defined on a sample space \(S\) such that \(P\left(A \cap B^{C}\right)=0.1, P\left(A^{C} \cap B\right)=0.3\), and \(P\left((A \cup B)^{C}\right)=0.2\). Find the probability that at least one of the two events occurs given that at most one occurs.

Short Answer

Expert verified
The probability that at least one of the two events occurs given that at most one occurs is 0.5.

Step by step solution

01

Understand The Problem Statement

Here we have three given probabilities: \(P\left(A \cap B^{C}\right)=0.1, P\left(A^{C} \cap B\right)=0.3\), and \(P\left((A \cup B)^{C}\right)=0.2\). Since \(A \cap B^{C}\) and \(A^{C} \cap B\) are mutually exclusive, their union equals \(A \oplus B\), which is the event that either A occurs or B occurs, but not both. And \((A \cup B)^{C}\) is the event that neither A nor B occurs.
02

Calculate The Probabilities of \(A \oplus B\) and \(S \backslash (A \cap B)\)

From given conditions, we can figure out \(P(A \oplus B) = P(A \cap B^{C}) + P(A^{C} \cap B) = 0.1 + 0.3 = 0.4\). Also, \(P((A \cup B)^{C}) = 0.2\) tells us that the probability of either A or B or both occurring is \(1 - 0.2 = 0.8\). This is equivalent to \(P(A)+P(B)-P(A \cap B) = 0.8\). But since \(P(A \cap B) = P(A) + P(B) - P(A \cup B) = P(A \oplus B) - P((A \cup B)^{C}) = 0.4 - 0.2 = 0.2\), the desired probability \(P(S \backslash (A \cap B)) = 1 - P(A \cap B) = 1 - 0.2 = 0.8\).
03

Apply The Formula of Conditional Probability

Now we apply the formula of conditional probability, \(P(A|B) = \frac{P(A \cap B)}{P(B)}\), to calculate the wanted probability. Here the event A is 'at least one of the events occurs' which is \(A \cup B\), and event B is 'at most one occurs' which is \(S \backslash (A \cap B)\). Hence, the desired probability \(P(A \cup B|S \backslash (A \cap B)) = \frac{P((A \cup B) \cap (S \backslash (A \cap B)))}{P(S \backslash (A \cap B))} = \frac{P(A \oplus B)}{P(S \backslash (A \cap B))} = \frac{0.4}{0.8} = 0.5\).

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Key Concepts

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

Conditional Probability
Understanding conditional probability is crucial when dealing with events that are dependent on one another. It refers to the likelihood of an event occurring given that another event has already occurred. The formula to compute conditional probability is denoted as \(P(A|B) = \frac{P(A \cap B)}{P(B)}\), where \(P(A|B)\) is the probability of event \(A\) occurring given that \(B\) has occurred, \(P(A \cap B)\) is the probability of both events \(A\) and \(B\) occurring, and \(P(B)\) is the probability of event \(B\) occurring.

In our exercise example, we dealt with finding the probability of at least one event happening, given that at most one can occur. This is a classic example of conditional probability because the occurrence of at least one event is dependent on the restriction that no more than one event can occur simultaneously.
Mutually Exclusive Events
Mutually exclusive events are those that cannot happen at the same time. When dealing with probability, if you鈥檙e told events are mutually exclusive, like in our exercise with \(A \cap B^{C}\) and \(A^{C} \cap B\), you can immediately know that the occurrence of one event eliminates the possibility of the other occurring.

In probability terms, this means that if \(A\) and \(B\) are mutually exclusive events, then \(P(A \cap B) = 0\). When calculating probabilities involving mutually exclusive events, you can simply add their individual probabilities to find the probability of at least one occurring, as in: \(P(A \oplus B) = P(A) + P(B)\), assuming \(A\) and \(B\) cannot both occur.
Probability Formulas
Several key probability formulas underpin the study of chance and likelihood. Beyond the basics of individual event probabilities, some essential formulas include the addition rule for mutually exclusive events, \(P(A \cup B) = P(A) + P(B)\) when \(A\) and \(B\) are mutually exclusive, and \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) when they're not.

Other important formulas include the complement rule, \(P(A^{C}) = 1 - P(A)\), and the aforementioned conditional probability formula. These formulas were used to tackle the three-part exercise provided, enabling the calculation of the overall likelihood of at least one event occurring under given conditions. The ability to correctly apply these formulas allows for the exploration of more complex probability scenarios.

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Most popular questions from this chapter

Suppose that two cards are drawn simultaneously from a standard fifty-two-card poker deck. Let \(A\) be the event that both are either a jack, queen, king, or ace of hearts, and let \(B\) be the event that both are aces Are \(A\) and \(B\) independent? (Note: There are 1326 equally likely ways to draw two cards from a poker deck.)

Two events, \(A\) and \(B\), are defined on a sample space \(S\) such that \(P(A \mid B)=0.6, P(\) At least one of the events occurs \()=0.8\), and \(P(\) Exactly one of the events occurs \()=0.6 .\) Find \(P(A)\) and \(P(B)\).

School board officials are debating whether to require all high school seniors to take a proficiency exam before graduating. A student passing all three parts (mathematics, language skills, and general knowledge) would be awarded a diploma; otherwise, he or she would receive only a certificate of attendance. A practice test given to this year's ninety-five hundred seniors resulted in the following numbers of failures: $$ \begin{array}{lc} \hline \text { Subject Area } & \text { Number of Students Failing } \\ \hline \text { Mathematics } & 3325 \\ \text { Language skills } & 1900 \\ \text { Ceneral knowledge } & 1425 \\ \hline \end{array} $$ If "Student fails mathematics," "Student fails language skills," and "Student fails general knowledge" are independent events, what proportion of next year's seniors can be expected to fail to qualify for a diploma? Does independence seem a reasonable assumption in this situation?

A telephone solicitor is responsible for canvassing three suburbs. In the past, \(60 \%\) of the completed calls to Belle Meade have resulted in contributions, compared to \(55 \%\) for Oak Hill and \(35 \%\) for Antioch. Her list of telephone numbers includes one thousand households from Belle Meade, one thousand from Oak Hill, and two thousand from Antioch. Suppose that she picks a number at random from the list and places the call. What is the probability that she gets a donation?

Two fair dice are rolled. What is the probability that the number on the first die was at least as large as 4 given that the sum of the two dice was \(8 ?\)
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