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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

According to a study conducted by a risk assessment firm (Associated Press, December 8,2005 ), drivers residing within one mile of a restaurant are \(30 \%\) more likely to be in an accident in a given policy year. Consider the following two events: \(A=\) event that a driver has an accident during a policy year \(R=\) event that a driver lives within one mile of a restaurant Which of the following four probability statements is consistent with the findings of this survey? Justify your choice. i. \(P(A \mid R)=.3\) iii. \(\frac{P(A \mid R)}{P\left(A \mid R^{C}\right)}=.3\) ii. \(P\left(A \mid R^{C}\right)=.3 \quad\) iv. \(\frac{P(A \mid R)-P\left(A \mid R^{C}\right)}{P\left(A \mid R^{C}\right)}=.3\)

The article "Anxiety Increases for Airline Passengers After Plane Crash" (San Luis Obispo Tribune, November 13,2001 ) reported that air passengers have a 1 in 11 million chance of dying in an airplane crash. This probability was then interpreted as "You could fly every day for 26,000 years before your number was up." Comment on why this probability interprctation is mislcading.

Suppose that a new Internet company Mumble.com requires all employees to take a drug test. Mumble.com can afford only the inexpensive drug test - the one with a \(5 \%\) false-positive rate and a \(10 \%\) false-negative rate. (That means that \(5 \%\) of those who are not using drugs will incorrectly test positive and that \(10 \%\) of those who are actually using drugs will test negative.) Suppose that \(10 \%\) of those who work for Mumble.com are using the drugs for which Mumble is checking. (Hint: It may be helpful to draw a tree diagram to answer the questions that follow.) a. If one employee is chosen at random, what is the probability that the employee both uses drugs and tests positive? b. If one employee is chosen at random, what is the probability that the employee does not use drugs but tests positive anyway? c. If one employee is chosen at random, what is the probability that the employee tests positive? d. If we know that a randomly chosen employee has tested positive, what is the probability that he or she uses drugs?

A construction firm bids on two different contracts. Let \(E_{1}\) be the event that the bid on the first contract is successful, and define \(E_{2}\) analogously for the second contract. Suppose that \(P\left(E_{1}\right)=.4\) and \(P\left(E_{2}\right)=.2\) and that \(E_{1}\) and \(E_{2}\) are independent events. a. Calculate the probability that both bids are successful (the probability of the event \(E_{1}\) and \(E_{2}\) ). b. Calculate the probability that neither bid is successful (the probability of the event \(\left(\right.\) not \(\left.E_{1}\right)\) and \(\left(\right.\) not \(\left.E_{2}\right)\) ). c. What is the probability that the firm is successful in at least one of the two bids?

A medical research team wishes to evaluate two different treatments for a disease. Subjects are selected two at a time, and then one of the pair is assigned to each of the two treatments. The treatments are applied, and each is either a success (S) or a failure (F). The researchers keep track of the total number of successes for each treatment. They plan to continue the chance experiment until the number of successes for one treatment exceeds the number of successes for the other treatment by \(2 .\) For example, they might observe the results in the table below. The chance experiment would stop after the sixth pair, because Treatment 1 has 2 more successes than Treatment \(2 .\) The researchers would conclude that Treatment 1 is preferable to Treatment \(2 .\) Suppose that Treatment 1 has a success rate of \(.7\) (i.e., \(P(\) success \()=.7\) for Treatment 1 ) and that Treatment 2 has a success rate of \(.4\). Use simulation to estimate the probabilities in Parts (a) and (b). (Hint: Use a pair of random digits to simulate one pair of subjects. Let the first digit represent Treatment 1 and use \(1-7\) as an indication of a
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