Let A, B and C be three events such that the probability that exactly one of A and B occurs is (1 - k), the probability that exactly one of B and C occurs is (1 - 2k), the probability that exactly one of C and A occurs is (1 - k) and the probability of all A, B and C occur simultaneously is k2, where 0

Question

Let A, B and C be three events such that the probability that exactly one of A and B occurs is (1 - k), the probability that exactly one of B and C occurs is (1 - 2k), the probability that exactly one of C and A occurs is (1 - k) and the probability of all A, B and C occur simultaneously is k2, where 0 < k < 1. Then the probability that at least one of A, B and C occurs isa)greater than 1/2b)greater than 1/4 but less than 1/2c)exactly equal to 1/2d)greater than 1/8 but less than 1/4Correct answer is option 'A'. Can you explain this answer?

Answer

P(A) + P(B) - 2P(A ∩ B) = 1 - k ... (i)
P(B) + P(C) - 2P(B ∩ C) = 1 - k ... (ii)
P(C) + P(A) - 2P(A ∩ C) = 1 - 2k ... (iii)
(1) + (2) + (3)
P(A) + P(B) + P(C) - P(A ∩ B) - P(B ∩ C) - P(C ∩ A) =

So,

Answer

Let's assign probabilities to the events A, B, and C:

P(A) = a
P(B) = b
P(C) = c

We are given the following information:

1. The probability that exactly one of A and B occurs is (1 - k):
P(A and not B) + P(not A and B) = 1 - k
ab + (1 - a)(1 - b) = 1 - k
ab + 1 - a - b + ab = 1 - k
2ab - a - b + 1 = 1 - k
2ab - a - b = -k

2. The probability that exactly one of B and C occurs is (1 - 2k):
P(B and not C) + P(not B and C) = 1 - 2k
bc + (1 - b)(1 - c) = 1 - 2k
bc + 1 - b - c + bc = 1 - 2k
2bc - b - c = -2k

3. The probability that exactly one of C and A occurs is (1 - k):
P(C and not A) + P(not C and A) = 1 - k
ca + (1 - c)(1 - a) = 1 - k
ca + 1 - c - a + ca = 1 - k
2ca - c - a = -k

4. The probability of all A, B, and C occurring simultaneously is k^2:
P(A and B and C) = k^2
abc = k^2

Now, let's solve these equations to find the values of a, b, and c in terms of k:

From equation 1: 2ab - a - b = -k
Adding k to both sides: 2ab - a - b + k = 0

From equation 2: 2bc - b - c = -2k
Adding 2k to both sides: 2bc - b - c + 2k = 0

From equation 3: 2ca - c - a = -k
Adding k to both sides: 2ca - c - a + k = 0

Now, let's multiply equation 1 by 2c, equation 2 by 2a, and equation 3 by 2b:

4abc - 2ac - 2bc + 2ck = 0
4abc - 2ab - 2ca + 2ak = 0
4abc - 2bc - 2ab + 2bk = 0

Adding these three equations together:

8abc - 2ac - 2bc - 2ab + 2ck + 2ak + 2bk = 0
8abc - 2(a + b + c)(a + b + c) + 2k(c + a + b) = 0
8abc - 2(a + b + c)^2 + 2k(c + a + b) = 0

Simplifying further:

4abc - (a + b + c)^2 + k(c + a + b) = 0
4k^2 - (a + b + c)^2 + k^2 = 0
3

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