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The probability that event A occurs is 0.4, and the probabil
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01 Dec 2018, 08:27
01 Dec 2018, 08:27
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The probability that event A occurs is 0.4, and the probability that events A and B both occur is 0.25. If the probability that either event A or event B occurs is 0.6, what is the probability that event B will occur?
A. 0.05
B. 0.15
C. 0.45
D. 0.50
E. 0.55
A. 0.05
B. 0.15
C. 0.45
D. 0.50
E. 0.55
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Re: The probability that event A occurs is 0.4, and the probabil
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03 Dec 2018, 11:24
03 Dec 2018, 11:24
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Carcass wrote:
The probability that event A occurs is 0.4, and the probability that events A and B both occur is 0.25. If the probability that either event A or event B occurs is 0.6, what is the probability that event B will occur?
A. 0.05
B. 0.15
C. 0.45
D. 0.50
E. 0.55
A. 0.05
B. 0.15
C. 0.45
D. 0.50
E. 0.55
Given:
P(A) = 0.4
P(A and B) = 0.25
P(A or B) = 0.6
Let's apply the OR probability formula: P(A or B) = P(A) + P(B) - P(A and B)
Plug values into the formula to get: 0.6 = 0.4 + P(B) - 0.25
Simplify to get: 0.6 = 0.15 + P(B)
Solve: P(B) = 0.45
Answer: C
Re: The probability that event A occurs is 0.4, and the probabil
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30 Apr 2020, 02:17
30 Apr 2020, 02:17
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A small doubt here, it says the probability of EITHER A or B and not EITHER A or B or BOTH, so don't we have to subtract the intersection twice? Because we have to exclude the probability that both occur together completely.
If P(A) = 0.4 and P(A and B) is 0.25, then P(only A not B) is 0.4 - 0.25 = 0.15.
Now P(A or B and not both) is 0.6 = P(only A not B) + P(only B not A) = 0.15 + P(only B not A)
So P(only B not A) = 0.6 - 0.15 = 0.45.
Now it asks for P(B) so it includes both P(A and B) and P(only B not A) = 0.45 + 0.25 = 0.7.
It doesn't mention whether we want the probability of B with or without A here.
If P(A) = 0.4 and P(A and B) is 0.25, then P(only A not B) is 0.4 - 0.25 = 0.15.
Now P(A or B and not both) is 0.6 = P(only A not B) + P(only B not A) = 0.15 + P(only B not A)
So P(only B not A) = 0.6 - 0.15 = 0.45.
Now it asks for P(B) so it includes both P(A and B) and P(only B not A) = 0.45 + 0.25 = 0.7.
It doesn't mention whether we want the probability of B with or without A here.
