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Chapter 6: Problem 96

Suppose that a box contains 25 bulbs, of which 20 are good and the other 5 are defective. Consider randoml 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. \(P(F | E) = 0.792\) \n c. \(P(G | E 鈭� F) = 0.783\) \n d. The probability that all three selected bulbs are good is 0.496.

Step by step solution

01

Compute P(E)

This is the probability of picking a good bulb on the first try. There are 20 good bulbs out of the total of 25, so \( P(E) = \frac{20}{25} = 0.8 \)
02

Compute P(F | E)

This is the probability of picking a good bulb on the second try given that a good bulb was selected first. After picking the first bulb which was good, there are 19 good bulbs left and 24 bulbs in total. Therefore, \( P(F | E) = \frac{19}{24} = 0.792 \)
03

Compute P(G | E 鈭� F)

This is the probability of picking a good bulb on the third try given that the first two selections were good bulbs. After picking two bulbs which were good, there are 18 good bulbs left and 23 bulbs in total. Therefore, \( P(G | E 鈭� F) = \frac{18}{23} = 0.783 \).
04

Compute the probability that all three selected bulbs are good

The probability that all three bulbs are good can be calculated by multiplying the probabilities of E, F and G. Therefore, \( P(E 鈭� F 鈭� G) = P(E) 脳 P(F | E) 脳 P(G | E 鈭� F) = 0.8 脳 0.792 脳 0.783 = 0.496 \).

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Key Concepts

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

Random Selection
Random selection is a fundamental concept in probability, where items are chosen randomly from a group. Imagine a box with several items, in this case, bulbs, from which you randomly pick without replacement. This means once a bulb is selected, it is not put back into the box for further selections.

In the bulb example, you first pick from a total of 25 bulbs. The randomness dictates that each bulb has an equal chance of being selected in the first draw.
  • If you pick a good bulb first, the total number of bulbs decreases to 24 for your second pick.
  • The same decrement applies as you pick the next one, and so on.
Here, each selection impacts the subsequent ones, showcasing the dynamic nature of random selection without replacement.
Conditional Probability
Conditional probability is a way of finding the probability of an event occurring given that another event has already occurred. It is denoted as \( P(A | B)\), meaning the probability of event A happening given that B has happened.

In the bulb scenario:
  • When asking for \( P(F | E)\), you're calculating the probability of getting a good bulb on the second try given that a good bulb was picked first.
  • This probability changes depending on the outcomes of prior events in a sequence.
To calculate conditional probability effectively:
  • Understand that the total number of outcomes decreases as events progress.
  • Formulas are used based on remaining possibilities.
This concept is crucial as it helps to narrow down possibilities by taking into account prior outcomes.
Permutations and Combinations
Permutations and combinations are essential tools in understanding probability, dealing with the arrangement of items.

* **Permutations** count the number of possible ways to order items. Here, order matters. * **Combinations** focus on the selection of items, where order doesn鈥檛 matter. In our case of bulb selection:
  • You're dealing more with combinations because the order in which bulbs are picked isn鈥檛 as crucial as selecting three good ones.
  • This affects how you compute the entire probability that all bulbs picked are good.
Knowing when to use permutations and combinations can clarify complex probability calculations and simplify real-world problem-solving approaches.

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Most popular questions from this chapter

Consider the chance experiment in which both tennis racket head size and grip size are noted for a randomly selected customer at a particular store. The six possible outcomes (simple events) and their probabilities are displayed in the following table: a. The probability that grip size is \(4 \frac{1}{2}\) in. (event \(\mathrm{A}\) ) is $$ P(A)=P\left(O_{2} \text { or } O_{5}\right)=.20+.15=.35 $$ How would you interpret this probability? b. Use the result of Part (a) to calculate the probability that grip size is not \(4 \frac{1}{2}\) in. c. What is the probability that the racket purchased has an oversize head (event \(B\) ), and how would you interpret this probability? d. What is the probability that grip size is at least \(4 \frac{1}{2}\) in.?

Two different airlines have a flight from Los Angeles to New York that departs each weekday morning at a certain time. Let \(E\) denote the event that the first airline's flight is fully booked on a particular day, and let \(F\) denote the event that the second airline's flight is fully booked on that same day. Suppose that \(P(E)=.7, P(F)=.6\), and \(P(E \cap F)=.54\). a. Calculate \(P(E \mid F)\) the probability that the first airline's flight is fully booked given that the second airline's flight is fully booked. b. Calculate \(P(F \mid E)\).

A mutual fund company offers its customers several different funds: a money market fund, three different bond funds, two stock funds, and a balanced fund. Among customers who own shares in just one fund, the percentages of customers in the different funds are as follows: \(\begin{array}{lr}\text { Money market } & 20 \% \\ \text { Short-term bond } & 15 \% \\ \text { Intermediate-term bond } & 10 \% \\ \text { Long-term bond } & 5 \% \\ \text { High-risk stock } & 18 \% \\ \text { Moderate-risk stock } & 25 \% \\ \text { Balanced fund } & 7 \%\end{array}\) A customer who owns shares in just one fund is to be selected at random. a. What is the probability that the selected individual owns shares in the balanced fund? b. What is the probability that the individual owns shares in a bond fund? c. What is the probability that the selected individual does not own shares in a stock fund?

A certain company sends \(40 \%\) of its overnight mail parcels by means of express mail service \(A_{1}\). Of these parcels, \(2 \%\) arrive after the guaranteed delivery time (use \(L\) to denote the event late delivery). If a record of an overnight mailing is randomly selected from the company's files, what is the probability that the parcel went by means of \(A_{1}\) and was late?

A company that manufactures video cameras produces a basic model and a deluxe model. Over the past year, \(40 \%\) of the cameras sold have been the basic model. Of those buying the basic model, \(30 \%\) purchase an extended warranty, whereas \(50 \%\) of all purchasers of the deluxe model buy an extended warranty. If you learn that a randomly selected purchaser bought an extended warranty, what is the probability that he or she has a basic model?
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