So, P(A n B) is the probability that both dice show a 1. Of the 36 possible outcomes, there is only one outcome where this occurs, (1, 1). Therefore, P(A n B) will be equal to the probability that one of the 36 outcomes occur. We previously determined that the probability of any one of the 36 equally likely outcomes is 1 so, P(A n B) = | 36 We now find P(B) the probability that the green (second) die shows a 1. The set B is as follows. B = {(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1)}

College Algebra
10th Edition
ISBN:9781337282291
Author:Ron Larson
Publisher:Ron Larson
Chapter8: Sequences, Series,and Probability
Section8.7: Probability
Problem 11ECP: A manufacturer has determined that a machine averages one faulty unit for every 500 it produces....
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This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any
points for the skipped part, and you will not be able to come back to the skipped part.
Tutorial Exercise
Find the conditional probability of the given event when two fair dice (one red and one green) are rolled.
The red one is 1, given that the green one is 1.
Step 1
Begin by considering the sample spaces and the probabilities of the outcomes.
Recall that for the experiment of rolling two dice (the first one red and the second one green) and observing
the face-up number on each die, the sample space S is the following 36-element set.
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6).
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
|(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
l(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
S =
Recall also that since both dice are fair, the outcomes are equally likely; that is, the probability of any one
outcome is the same as that of any other.
Because probabilities are never negative and the sum of the probabilities of all outcomes in a sample space
1
must be 1, the probability of any one of the 36 equally likely outcomes is
36
36
Step 2
Recall now that the probability of an event is the sum of the probabilities of the outcomes in the event.
The events under consideration are as follows.
A: The face-up number on the red (first) die is 1.
B: The face-up number on the green (second) die is 1.
To find the probability that "the red one is 1 given that the green one is 1," that is, to find P(A|B), we use the
P(A n B)
conditional probability formula P(A|B) =
P(B)
To do so, we need both P(A n B) and P(B).
To find the first of these, P(A n B) note that the event (A n B) can be described by
the red die shows 1 and the green die shows
the red die shows 1 and the green die shows 1
Step 3
So, P(A n B) is the probability that both dice show a 1. Of the 36 possible outcomes, there is only one
outcome where this occurs, (1, 1). Therefore, P(A n B) will be equal to the probability that one of the 36
outcomes occur. We previously determined that the probability of any one of the 36 equally likely outcomes is
1 so, P(A n B) = (
36
We now find P(B) the probability that the green (second) die shows a 1. The set B is as follows.
В 3 (1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1)}
Transcribed Image Text:This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise Find the conditional probability of the given event when two fair dice (one red and one green) are rolled. The red one is 1, given that the green one is 1. Step 1 Begin by considering the sample spaces and the probabilities of the outcomes. Recall that for the experiment of rolling two dice (the first one red and the second one green) and observing the face-up number on each die, the sample space S is the following 36-element set. (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6). (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), |(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), l(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) S = Recall also that since both dice are fair, the outcomes are equally likely; that is, the probability of any one outcome is the same as that of any other. Because probabilities are never negative and the sum of the probabilities of all outcomes in a sample space 1 must be 1, the probability of any one of the 36 equally likely outcomes is 36 36 Step 2 Recall now that the probability of an event is the sum of the probabilities of the outcomes in the event. The events under consideration are as follows. A: The face-up number on the red (first) die is 1. B: The face-up number on the green (second) die is 1. To find the probability that "the red one is 1 given that the green one is 1," that is, to find P(A|B), we use the P(A n B) conditional probability formula P(A|B) = P(B) To do so, we need both P(A n B) and P(B). To find the first of these, P(A n B) note that the event (A n B) can be described by the red die shows 1 and the green die shows the red die shows 1 and the green die shows 1 Step 3 So, P(A n B) is the probability that both dice show a 1. Of the 36 possible outcomes, there is only one outcome where this occurs, (1, 1). Therefore, P(A n B) will be equal to the probability that one of the 36 outcomes occur. We previously determined that the probability of any one of the 36 equally likely outcomes is 1 so, P(A n B) = ( 36 We now find P(B) the probability that the green (second) die shows a 1. The set B is as follows. В 3 (1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1)}
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