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dee6991
20.09.2020 •
Mathematics
Suppose that an experiment has five possible outcomes, which are denoted {1,2,3,4,5}. Let A be the event {1,3,4} and let B be the event {2,4,5}. (Notice that we did not say that the five outcomes are equally likely: the probability distributions could be anything.) For each of the following relations, tell whether it could possibly hold. If it could, give a numerical example using a probability distribution of your own choice: if it could not, explain why not (what rule is violated)
a. P(A) = P(B)
b. P(A) = 2P(B)
c. P(A) = 1 - P(B)
d. P(A) + P(B) > 1
e. P(A) - P(B) < 0
f. P(A) - P(B) > 1
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Ответ:
a. P(A) = P(B)
c. P(A) = 1 - P(B)
a and c are true . The rest are false.
Step-by-step explanation:
Two events A and B are said to be equally likely when one event is as likely to occur as the other. In other words each event should occur in equal number in repeated trials. For example when a fair coin is tossed the head is likely to appear as the tail, and the proportion of times each side is expected to appear is 1/2.
So when the events A= {1,3,4} B = {2,4,5} are equally likely then suppose their probability is 1/2.
a. P(A) = P(B) True
1/2= 1/2
b. P(A) = 2P(B) False
1/2 is not equal to 1
c. P(A) = 1 - P(B) True
1/2= 1-1/2= 1/2
d. P(A) + P(B) > 1 False
1/2 + 1/2 is not greater than 1
e. P(A) - P(B) < 0 False
1/2-1/2= 0 is not less than 0
f. P(A) - P(B) > 1 False
1/2-1/2= 0 is not greater than 1
Ответ:
no. but if you give me brainliest ill be happy to help
Step-by-step explanation: