CA Foundation Exam > CA Foundation Questions > For any two events A1, A2letP(A1) = 2/3,P(A2)...
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For any two events A1, A2 letP(A1) = 2/3,P(A2) = 3/8 and P(A1 ∩ A2) =1/4 then A1, A2 are:
- a)Mutually exclusive but not independent events
- b)Mutually exclusive and independent events
- c)Independent but not mutually exclusive
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
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For any two events A1, A2letP(A1) = 2/3,P(A2) = 3/8 andP(A1∩ A2) =...
To use Bayes' theorem, we need to first find P(A1 and B) and P(A2 and B). We can use the formula:
P(A and B) = P(A) * P(B|A)
where P(B|A) is the probability of event B given that event A has occurred.
For A1 and B:
P(A1 and B) = P(B|A1) * P(A1)
P(B|A1) is given as 1/2, since if A1 has occurred, then only half of the remaining cards can be black.
P(A1) is given as 2/3.
So, P(A1 and B) = (1/2) * (2/3) = 1/3.
For A2 and B:
P(A2 and B) = P(B|A2) * P(A2)
P(B|A2) is given as 3/7, since if A2 has occurred, then 3 of the remaining 7 cards are black.
P(A2) is given as 3/8.
So, P(A2 and B) = (3/7) * (3/8) = 9/56.
Now we can apply Bayes' theorem:
P(A1|B) = P(A1 and B) / P(B)
P(A2|B) = P(A2 and B) / P(B)
where P(B) is the probability of event B occurring, which is the sum of the probabilities of A1 and B and A2 and B:
P(B) = P(A1 and B) + P(A2 and B) = 1/3 + 9/56 = 25/56.
So,
P(A1|B) = (1/3) / (25/56) = 56/225
P(A2|B) = (9/56) / (25/56) = 9/25
Therefore, the probability that the card drawn is from deck 1 given that it is a black card is 56/225, and the probability that it is from deck 2 given that it is a black card is 9/25.
P(A and B) = P(A) * P(B|A)
where P(B|A) is the probability of event B given that event A has occurred.
For A1 and B:
P(A1 and B) = P(B|A1) * P(A1)
P(B|A1) is given as 1/2, since if A1 has occurred, then only half of the remaining cards can be black.
P(A1) is given as 2/3.
So, P(A1 and B) = (1/2) * (2/3) = 1/3.
For A2 and B:
P(A2 and B) = P(B|A2) * P(A2)
P(B|A2) is given as 3/7, since if A2 has occurred, then 3 of the remaining 7 cards are black.
P(A2) is given as 3/8.
So, P(A2 and B) = (3/7) * (3/8) = 9/56.
Now we can apply Bayes' theorem:
P(A1|B) = P(A1 and B) / P(B)
P(A2|B) = P(A2 and B) / P(B)
where P(B) is the probability of event B occurring, which is the sum of the probabilities of A1 and B and A2 and B:
P(B) = P(A1 and B) + P(A2 and B) = 1/3 + 9/56 = 25/56.
So,
P(A1|B) = (1/3) / (25/56) = 56/225
P(A2|B) = (9/56) / (25/56) = 9/25
Therefore, the probability that the card drawn is from deck 1 given that it is a black card is 56/225, and the probability that it is from deck 2 given that it is a black card is 9/25.
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For any two events A1, A2letP(A1) = 2/3,P(A2) = 3/8 andP(A1∩ A2) =...
They are Independent because the product of their individual probabilities is equal to the probability of their intersection and are not Mutually exclusive because the sum of their individual probabilities is not equal to unity(1).
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For any two events A1, A2letP(A1) = 2/3,P(A2) = 3/8 andP(A1∩ A2) =1/4then A1, A2are: a)Mutually exclusive but not independent eventsb)Mutually exclusive and independent eventsc)Independent but not mutually exclusived)None of theseCorrect answer is option 'C'. Can you explain this answer?
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