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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. 36✔ 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 must be 1, the probability of any one of the 36 equally likely outcomes is 1 (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3), (4,4), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) 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. P(A n B) To find the probability that "the red one is 1 given that the green one is 1," that is, to find P(AIB), we use the conditional probability formula P(AIB) = 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 1✔ the red die shows 1 and the green die shows 1

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. 36✔ 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 must be 1, the probability of any one of the 36 equally likely outcomes is 1 (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3), (4,4), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) 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. P(A n B) To find the probability that "the red one is 1 given that the green one is 1," that is, to find P(AIB), we use the conditional probability formula P(AIB) = 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 1✔ the red die shows 1 and the green die shows 1

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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. 36 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 must be 1, the probability of any one of the 36 equally likely outcomes is 1 (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 5), (3, 6), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (3, 4), (4, 4), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) 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(AIB), we use the conditional probability formula P(AIB) 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 1 = P(A n B) P(B) the red die shows 1 and the green die shows 1
Transcribed Image Text: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. 36 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 must be 1, the probability of any one of the 36 equally likely outcomes is 1 (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 5), (3, 6), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (3, 4), (4, 4), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) 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(AIB), we use the conditional probability formula P(AIB) 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 1 = P(A n B) P(B) the red die shows 1 and the green die shows 1
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 1 that one of the 36 outcomes occur. We previously determined that the probability of any one of the 36 equally likely outcomes is So, P(A n B) = 0.0278 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)} There are 6 outcomes in set B, each with probability 1 36 So, P(B) = 6 36
Transcribed Image Text: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 1 that one of the 36 outcomes occur. We previously determined that the probability of any one of the 36 equally likely outcomes is So, P(A n B) = 0.0278 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)} There are 6 outcomes in set B, each with probability 1 36 So, P(B) = 6 36
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