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Chapter 6: Problem 101

Suppose that a box contains 25 light bulbs, of which 20 are good and the other 5 are defective. Consider randomly selecting three bulbs without replacement. Let \(E\) denote the event that the first bulb selected is good, \(F\) be the event that the second bulb is good, and \(G\) represent the event that the third bulb selected is good. a. What is \(P(E)\) ? b. What is \(P(F \mid E)\) ? c. What is \(P(G \mid E \cap F)\) ? d. What is the probability that all three selected bulbs are good?

Short Answer

Expert verified
a) \(P(E) = 0.8\) \n b) \(P(F | E) = 0.79167\) \n c) \(P(G | E \cap F) = 0.78261\) \n d) The probability that all three selected bulbs are good is \(0.49688\).

Step by step solution

01

Calculate \(P(E)\)

The event \(E\) is that the first bulb selected is good. Since there are 20 good bulbs out of a total of 25, the probability that the first bulb selected is good, \(P(E)\), is simply the ratio of good bulbs to total bulbs, written mathematically as \(P(E) = \frac{20}{25} = 0.8\).
02

Calculate \(P(F | E)\)

The event \(F\) is that the second bulb is good and it is conditional on event \(E\), the first bulb being good. If the first bulb was good, we have 19 good bulbs left of a total of 24 bulbs, thus \(P(F|E) = \frac{19}{24} = 0.79167.\)
03

Calculate \(P(G | E \cap F)\)

The event \(G\) is that the third bulb is good and it is conditional on both the first bulb and the second bulb being good. If the first two bulbs were good, we have 18 good bulbs left out of a total of 23 bulbs, thus \(P(G|E \cap F) = \frac{18}{23} = 0.78261\) .
04

Calculate the probability that all three selected bulbs are good

The probability that all three selected bulbs are good is simply the multiplication of the probabilities of each individual event, since they are dependent events: \(\Pi P (good) = P(E) \times P(F|E) \times P(G|E \cap F) = 0.8 \times 0.79167 \times 0.78261 = 0.49688.\)

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Key Concepts

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

Conditional Probability
Conditional probability is a fundamental concept in statistics that deals with the likelihood of an event occurring given that another event has already occurred. Think of it as updating your expectations based on new information.

Regarding the light bulb example, once we know the first bulb (\(E\)) is good, the proportion of good bulbs out of the remaining total changes. The conditional probability of selecting a good second bulb (\(P(F | E)\)) is different from the initial unconditional probability because our sample space has been updated: we're now choosing from 24 bulbs instead of the original 25, with one less good bulb to pick.
Dependent Events
Dependent events in probability are those whose outcomes are linked. The likelihood of a subsequent event changes based on the outcomes of the preceding ones.

In the context of our non-replacement light bulb scenario, each selection affects the next. If the first bulb selected is good (\(E\)), this affects the probability of the subsequent bulb (\(F\)) also being good. The events are dependent because the selection of each bulb alters the composition of what remains in the box, influencing the odds for the next draw.
Probability Theory
Probability theory is the mathematical framework that describes the nature of random events. It helps to predict the likeliness of outcomes in situations where there is some form of uncertainty.

Learning probability theory involves understanding concepts like randomness, events, sample spaces, and probabilities themselves, as seen in our exercise. For each bulb picked, probability theory gives us the tools to determine how likely it is to be good or defective, establishing a systematic way to make informed predictions based on the evidence at hand.
Combinatorics
Combinatorics is the branch of mathematics dealing with counting and arrangements of objects. It's closely related to probability, as it provides methods to enumerate possibilities, which is critical for calculating probabilities in more complex scenarios.

While not directly involved in our step-by-step light bulb example, understanding combinatorics would be essential if, for instance, we wanted to know how many ways we could select three bulbs in total. It enables us to calculate the total number of outcomes and thus the likelihood of any specific outcome occurring.

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Most popular questions from this chapter

Consider the chance experiment in which the type of transmission-automatic (A) or manual (M) - is recorded for each of the next two cars purchased from a certain dealer. a. What is the set of all possible outcomes (the sample space)? b. Display the possible outcomes in a tree diagram. c. List the outcomes in each of the following events. Which of these events are simple events? i. \(B\) the event that at least one car has an automatic transmission ii. \(C\) the event that exactly one car has an automatic transmission iii. \(D\) the event that neither car has an automatic transmission d. What outcomes are in the event \(B\) and \(C ?\) In the event \(B\) or \(C\) ?

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Define the term chance experiment, and give an example of a chance experiment with four possible outcomes.

A new model of laptop computer can be ordered with one of three screen sizes ( 10 inches, 12 inches, 15 inches) and one of four hard drive sizes \((50 \mathrm{~GB}, 100 \mathrm{~GB}\), \(150 \mathrm{~GB}\), and \(200 \mathrm{~GB}\) ). Consider the chance experiment in which a laptop order is selected and the screen size and hard drive size are recorded. a. Display possible outcomes using a tree diagram. b. Let \(A\) be the event that the order is for a laptop with a screen size of 12 inches or smaller. Let \(B\) be the event that the order is for a laptop with a hard drive size of at most \(100 \mathrm{~GB}\). What outcomes are in \(A^{C}\) ? In \(A \cup B ?\) In \(A \cap B\) ? c. Let \(C\) denote the event that the order is for a laptop with a \(200 \mathrm{~GB}\) hard drive. Are \(A\) and \(C\) disjoint events? Are \(B\) and \(C\) disjoint?

After all students have left the classroom, a statistics professor notices that four copies of the text were left under desks. At the beginning of the next lecture, the professor distributes the four books at random to the four students \((1,2,3\), and 4\()\) who claim to have left books. One possible outcome is that 1 receives 2's book, 2 receives 4's book, 3 receives his or her own book, and 4 receives l's book. This outcome can be abbreviated \((2,4,3,1)\). a. List the 23 other possible outcomes. b. Which outcomes are contained in the event that exactly two of the books are returned to their correct owners? Assuming equally likely outcomes, what is the probability of this event? c. What is the probability that exactly one of the four students receives his or her own book? d. What is the probability that exactly three receive their own books? e. What is the probability that at least two of the four students receive their own books?
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