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Chapter 2: Problem 120

Suppose that events \(A, B\), and \(C\) are independent. (a) Use a Venn diagram to find an expression for \(P(A \cup\) \(B \cup C\) ) that does not make use of a complement. (b) Find an expression for \(P(A \cup B \cup C)\) that does make use of a complement.

Short Answer

Expert verified
(a) \(P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A)P(B) - P(A)P(C) - P(B)P(C) + P(A)P(B)P(C)\) , (b) \(P(A \cup B \cup C) = 1 - (1 - P(A))(1 - P(B))(1 - P(C))\)

Step by step solution

01

Calculate the probability of the union of A, B, and C without using a complement

Using the formula for the probability of the union of three independent events, we have: \(P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)\). Since A, B and C are independent, the probability of their intersection is the product of their probabilities. This gives us: \(P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A)P(B) - P(A)P(C) - P(B)P(C) + P(A)P(B)P(C)\)
02

Calculate the probability of the union of A, B, and C using a complement

By applying De Morgan's law, we know that the complement of the union of events is the intersection of their complements. Therefore, we can write: \(P(A \cup B \cup C) = 1 - P(A' \cap B' \cap C')\). Because the events are independent, \(P(A' \cap B' \cap C') = P(A')P(B')P(C')\). Moreover, since the complement of an event A is 1 - P(A), we have: \(P(A \cup B \cup C) = 1 - (1 - P(A))(1 - P(B))(1 - P(C))\).

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Key Concepts

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

Independent Events
Independent events are fundamental to understanding probability. Two events are independent when the occurrence of one does not affect the probability of the other happening. Let's consider two such events, say event A and event B.
The concept follows:
  • If A happens, it doesn't change the probability of B happening.
  • Mathematically, we express this as: \( P(A \cap B) = P(A) \cdot P(B) \).
For example, rolling a die and flipping a coin are independent; the die's result doesn鈥檛 affect the coin flip. When we know events like A, B, and C are independent, it simplifies the calculations for probabilities involving these events, especially intersections and unions.
Union of Events
The union of events is about calculating the probability of at least one of several events happening. For events A, B, and C, the notation is \( A \cup B \cup C \).
To find this probability, we can use the formula:
\[ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C) \]
This formula adds the probabilities of each event occurring and subtracts the probabilities of multiple events occurring together to avoid double-counting.

It can be tricky without visual aids. A Venn diagram can help by showing the overlapping areas that represent multiple events occurring together. This visual tool can make the concept easier to grasp.
De Morgan's Law
De Morgan's Law is a powerful tool in probability and set theory. It provides a way to express complements of unions and intersections in terms of the other operation.
For events A, B, and C, De Morgan's Law regarding their union states that:
\[ (A \cup B \cup C)' = A' \cap B' \cap C' \]
This tells us the complement of the union is the intersection of the complements. But, why is this useful?
  • It simplifies calculating \( P(A \cup B \cup C) \) by shifting the focus to the complements.
  • The complement probability is usually simpler because of multiplication through independence: \( P(A' \cap B' \cap C') = P(A')P(B')P(C') \).
Through De Morgan's Law, we can find that the probability of the union is:
\[ P(A \cup B \cup C) = 1 - (1 - P(A))(1 - P(B))(1 - P(C)) \]. This is often more straightforward than the direct addition-subtraction formula.
Venn Diagram
A Venn diagram is an excellent visual tool for exploring relationships between events and understanding their unions and intersections.
Each event is represented as a circle, with overlaps indicating where events occur together. Visualizing events A, B, and C in a Venn diagram makes it easier to calculate the probability of their union or intersection.
  • Individual intersections (e.g., \( A \cap B \)) are shown as overlapping areas.
  • The union is seen as the entire area covered by any circle.
By using a Venn diagram, it's straightforward to see why and where we add or subtract probabilities, helping to prevent errors in complex calculations.
The diagram ensures that each part of the equation matches a real overlap, making abstract probability rules more tangible.

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Most popular questions from this chapter

Imagine that the test tube pictured contains \(2 n\) grains of sand, \(n\) white and \(n\) black. Suppose the tube is vigorously shaken. What is the probability that the two colors of sand will completely separate; that is, all of one color fall to the bottom, and all of the other color lie on top? (Hint: Consider the \(2 n\) grains to be aligned in a row. In how many ways can the \(n\) white and the \(n\) black grains be permuted?)

A man has \(n\) keys on a key ring, one of which opens the door to his apartment. Having celebrated a bit too much one evening, he returns home only to find himself unable to distinguish one key from another. Resourceful, he works out a fiendishly clever plan: He will choose a key at random and try it. If it fails to open the door, he will discard it and choose at random one of the remaining \(n-1\) keys, and so on. Clearly, the probability that he gains entrance with the first key he selects is \(1 / n\). Show that the probability the door opens with the third key he tries is also \(1 / n\). (Hint: What has to happen before he even gets to the third key?)

The final exam in History 101 consists of five essay questions that the professor chooses from a pool of seven that are given to the students a week in advance. For how many possible sets of questions does a student need to be prepared? In this situation, does order matter?

Suppose each of ten sticks is broken into a long part and a short part. The twenty parts are arranged into ten pairs and glued back together so that again there are ten sticks. What is the probability that each long part will be paired with a short part? (Note: This problem is a model for the effects of radiation on a living cell. Each chromosome, as a result of being struck by ionizing radiation, breaks into two parts, one part containing the centromere. The cell will die unless the fragment containing the centromere recombines with a fragment not containing a centromere.)

Recently the U.S. Senate Committee on Labor and Public Welfare investigated the feasibility of setting up a national screening program to detect child abuse. A team of consultants estimated the following probabilities: (1) one child in ninety is abused, \((2)\) a screening program can detect an abused child \(90 \%\) of the time, and (3) a screening program would incorrectly label \(3 \%\) of all nonabused children as abused. What is the probability that a child is actually abused given that the screening program makes that diagnosis? How does the probability change if the incidence of abuse is one in one thousand? Or one in fifty?
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