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Chapter 2: Problem 12

Consider randomly selecting a student at a large university, and let \(A\) be the event that the selected student has a Visa card and \(B\) be the analogous event for MasterCard. Suppose that \(P(A)=.6\) and \(P(B)=.4\). a. Could it be the case that \(P(A \cap B)=.5\) ? Why or why not? b. From now on, suppose that \(P(A \cap B)=.3\). What is the probability that the selected student has at least one of these two types of cards? c. What is the probability that the selected student has neither type of card? d. Describe, in terms of \(A\) and \(B\), the event that the selected student has a Visa card but not a MasterCard, and then calculate the probability of this event. e. Calculate the probability that the selected student has exactly one of the two types of cards.

Short Answer

Expert verified
a. No, because 0.5 > max(0.6, 0.4). b. 0.7 c. 0.3 d. 0.3 e. 0.4

Step by step solution

01

Check if P(A 鈭� B) = 0.5 is possible

Using the principle of probability, we know that the probability of an intersection cannot exceed the smaller individual probabilities of the events. Since \( P(A) = 0.6 \) and \( P(B) = 0.4 \), the maximum possible value for \( P(A \cap B) \) is 0.4. Hence, \( P(A \cap B) = 0.5 \) is not possible because it exceeds \( P(B) \).
02

Calculate P(A 鈭� B) with P(A 鈭� B) = 0.3

Using the formula for the union of two events \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \), substitute the given values: \( P(A \cup B) = 0.6 + 0.4 - 0.3 = 0.7 \). Thus, the probability that a student has at least one of these two types of cards is 0.7.
03

Calculate probability of having neither card

The probability that a student has neither type of card is the complement of the probability of having at least one type. Thus, \( P((A \cup B)^c) = 1 - P(A \cup B) = 1 - 0.7 = 0.3 \).
04

Describe and calculate P(A and not B)

The event of having a Visa card but not a MasterCard is represented by \( A \cap B^c \). To find this probability, use \( P(A \cap B^c) = P(A) - P(A \cap B) = 0.6 - 0.3 = 0.3 \).
05

Calculate probability of having exactly one type of card

To determine the probability of having exactly one type of card, add the probabilities of having only a Visa (calculated as \( P(A \cap B^c) = 0.3 \)) and only a MasterCard (\( P(B \cap A^c) = P(B) - P(A \cap B) = 0.4 - 0.3 = 0.1 \)). Therefore, \( P((A \cap B^c) \cup (B \cap A^c)) = 0.3 + 0.1 = 0.4 \).

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Key Concepts

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

Probability of Events
Probability helps us measure how likely an event is to happen. In our exercise, we consider two specific events: a student having a Visa card or a MasterCard. We assign these events some probabilities: 0.6 for a Visa card (\( P(A) = 0.6 \)) and 0.4 for a MasterCard (\( P(B) = 0.4 \)). These probabilities give us a numerical way to discuss the likelihood of these events occurring when a student is randomly selected.
Two main rules in probability are that the probability of an event is always between 0 and 1, and the sum of all possible probabilities in a sample space equals 1. If you add up the probabilities of having a Visa, a MasterCard, and not having any card, it should make a total of 1, considering all possibilities for this selection.
Intersection and Union of Events
The concepts of intersection (\( A \cap B \)) and union (\( A \cup B \)) of events are essential in probability theory. When we talk about intersection (\( A \cap B \)), we mean all outcomes that are common to both events. It's akin to asking what the probability is that the selected student possesses both types of cards simultaneously.
In the given exercise, it's requested to assess if \( P(A \cap B) = 0.5 \) could be plausible. Since an intersection's value can't exceed the smallest individual probability, and here, the smallest is 0.4 (\( P(B) \)), this isn't feasible.
The union of events (\( A \cup B \)) refers to all possible outcomes included in either event. To find this, we use:
  • \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
In the exercise, if \( P(A \cap B) = 0.3 \), then \( P(A \cup B) = 0.7 \). This states there's a 0.7 probability of a student having at least one of the cards.
Complementary Events
In probability, events often have counterparts called complementary events. These express every possible outcome that is not covered by the original event.
For example, if \( A \) is the event "a student has a Visa card," the complement \( A^c \) is "a student does not have a Visa card." The complement rule notes that \( P(A^c) = 1 - P(A) \), which lets us compute any missing probability.
In the steps provided, we calculate the probability of a student having neither card, represented as \(( A \cup B )^c\). By the complement rule, \( P(( A \cup B )^c) = 1 - P(A \cup B) \), yielding 0.3. So, a student is 30% likely not to own any card.
Cardinality of Sets
The cardinality of a set considers the count of distinct elements within the set. In probability, when dealing with events represented as sets, it's key to understand Set Math to determine the probability of combined events.
In our exercise, when the probability of a student possessing only a Visa card (\( A \cap B^c \)) or only a MasterCard (\( B \cap A^c \)) is considered, the cardinality analogy helps clarify how many unique setups lead to one card possession over the other.
  • \( P(A \cap B^c) = P(A) - P(A \cap B) \), helping us find that 30% of students have only Visa.
  • \( P(B \cap A^c) = P(B) - P(A \cap B) \), covering another 10% only with MasterCard.
Thus, for students with exactly one card, we unite these separate probabilities (\( P(( A \cap B^c ) \cup ( B \cap A^c )) = 0.3 + 0.1 = 0.4 \)), covering that 40% scope, stressing the importance of understanding distinct count via cardinality.

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Most popular questions from this chapter

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