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Suppose two events A and B are independent, with P(A) = 0 and P(B) = 0. By working through the following steps, you'll see why two independent events are not mutually exclusive. (a) What formula is used to compute P(A and B)? ⒸP(A) - P(B) O P(A) - P(B) OP(A)/ P(B) OP(A) + P(B) Is P(A and B) = 0? Explain. O Yes. Because both P(A) and P(B) are not equal to 0, P(A and B) * 0. O No. P(A and B) = 0. O No. Because both P(A) and P(B) are not equal to 0, P(A and B) = 0. O Yes. Even if P(A) - 0 or P(B) = 0, P(A and B) will always be non-zero. are not mutually exclusive? (b) Using the information from part (a), can you conclude that events A and O Yes. Two events being independent always implies they are mutually exclusive. O No. Because P(A and B) = 0, A and B are mutually exclusive. O Yes. Because P(A and B) = 0, A and B are not mutually exclusive. O No. Two events being independent always implies they are mutually exclusive.

Suppose two events A and B are independent, with P(A) = 0 and P(B) = 0. By working through the following steps, you'll see why two independent events are not mutually exclusive. (a) What formula is used to compute P(A and B)? ⒸP(A) - P(B) O P(A) - P(B) OP(A)/ P(B) OP(A) + P(B) Is P(A and B) = 0? Explain. O Yes. Because both P(A) and P(B) are not equal to 0, P(A and B) * 0. O No. P(A and B) = 0. O No. Because both P(A) and P(B) are not equal to 0, P(A and B) = 0. O Yes. Even if P(A) - 0 or P(B) = 0, P(A and B) will always be non-zero. are not mutually exclusive? (b) Using the information from part (a), can you conclude that events A and O Yes. Two events being independent always implies they are mutually exclusive. O No. Because P(A and B) = 0, A and B are mutually exclusive. O Yes. Because P(A and B) = 0, A and B are not mutually exclusive. O No. Two events being independent always implies they are mutually exclusive.

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Suppose two events A and B are independent, with P(A) = 0 and P(B) 0. By working through the following steps, you'll see why two independent events are not mutually exclusive. (a) What formula is used to compute P(A and B)? ⒸP(A) - P(B) OP(A) - P(B) O P(A) / P(B) O P(A) + P(B) Is P(A and B) = 0? Explain. O Yes. Because both P(A) and P(B) are not equal to 0, P(A and B) = 0. O No. P(A and B) = 0. O No. Because both P(A) and P(B) are not equal to 0, P(A and B) = 0. O Yes. Even if P(A) = 0 or P(B) = 0, P(A and B) will always be non-zero. (b) Using the information from part (a), can you conclude that events A and B are not mutually exclusive? O Yes. Two events being independent always implies they are mutually exclusive. O No. Because P(A and B) # 0, A and B are mutually exclusive. O Yes. Because P(A and B) = 0, A and B are not mutually exclusive. O No. Two events being independent always implies they are mutually exclusive.
Transcribed Image Text:Suppose two events A and B are independent, with P(A) = 0 and P(B) 0. By working through the following steps, you'll see why two independent events are not mutually exclusive. (a) What formula is used to compute P(A and B)? ⒸP(A) - P(B) OP(A) - P(B) O P(A) / P(B) O P(A) + P(B) Is P(A and B) = 0? Explain. O Yes. Because both P(A) and P(B) are not equal to 0, P(A and B) = 0. O No. P(A and B) = 0. O No. Because both P(A) and P(B) are not equal to 0, P(A and B) = 0. O Yes. Even if P(A) = 0 or P(B) = 0, P(A and B) will always be non-zero. (b) Using the information from part (a), can you conclude that events A and B are not mutually exclusive? O Yes. Two events being independent always implies they are mutually exclusive. O No. Because P(A and B) # 0, A and B are mutually exclusive. O Yes. Because P(A and B) = 0, A and B are not mutually exclusive. O No. Two events being independent always implies they are mutually exclusive.
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