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The events A and B are independent, the probability that
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24 Mar 2020, 08:58
24 Mar 2020, 08:58
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Question Stats:
62% (02:14) correct
37% (03:22) wrong based on 61 sessions
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The events A and B are independent, the probability that event A occurs is greater than 0, and the probability that event A occurs is twice the probability that event B occurs. The probability that at least one of events A and B occurs is 8 times the probability that both events A and B occur. What is the probability that event A occurs?
A. 1/12
B. 1/8
C. 1/6
D. 1/3
E. 2/3
A. 1/12
B. 1/8
C. 1/6
D. 1/3
E. 2/3
Re: The events A and B are independent, the probability that
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24 Mar 2020, 09:39
24 Mar 2020, 09:39
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a = 2b
Probability (either A or B or both) = 8 times Probability (A and B)
a*(1-b) + b*(1-a) + ab = 8*ab
Substituting a=2b in the second equation:
2b*(1-b) + b*(1-2b) + 2b*b = 8*2b*b
3b-2b^2 = 16b^2
3b = 18b^2
b = 3/18 = 1/6
So, a = 2b = 1/3
Probability (either A or B or both) = 8 times Probability (A and B)
a*(1-b) + b*(1-a) + ab = 8*ab
Substituting a=2b in the second equation:
2b*(1-b) + b*(1-2b) + 2b*b = 8*2b*b
3b-2b^2 = 16b^2
3b = 18b^2
b = 3/18 = 1/6
So, a = 2b = 1/3
General Discussion
Re: The events A and B are independent, the probability that
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25 Mar 2020, 05:44
25 Mar 2020, 05:44
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P(A or B) = P(A) + P(B) - P(A and B).....(1)
since A and B are independent, then P(A and B)= P(A) * P(B)
Given,
P(A or B) = 8*P(A and B)
P(A) = 2*P(B) ; P(B) = P(A)/2
therefore,
8* P(A and B) = 8 (P(A) *P(A)/2) = 4 (P(A)^2)
substituting the given values into (1),
4(P(A)^2) = P(A) + P(A)/2 -P(A)^2)/2
4(P(A)^2) + P(A)^2/2 =3P(A)/2
9(P(A)^2)/2 = 3P(A)/2
Hence,
P(A) =1/3
since A and B are independent, then P(A and B)= P(A) * P(B)
Given,
P(A or B) = 8*P(A and B)
P(A) = 2*P(B) ; P(B) = P(A)/2
therefore,
8* P(A and B) = 8 (P(A) *P(A)/2) = 4 (P(A)^2)
substituting the given values into (1),
4(P(A)^2) = P(A) + P(A)/2 -P(A)^2)/2
4(P(A)^2) + P(A)^2/2 =3P(A)/2
9(P(A)^2)/2 = 3P(A)/2
Hence,
P(A) =1/3
