For three events A, B and C, P (Exactly one of A or B occurs) = P (Exactly one of B or C occurs) = P(Exactly one of C or A occurs) = 14 and P (All the three events occur simultaneously) = 116. Then the probability that at least one of the events occurs, is

For three events A, B and C, P (Exactly one of A or B occurs) = P (Exactly one of B or C occurs) = P(Exactly one of C or A occurs) = \frac{1}{4} and P (All the three events occur simultaneously) = \frac{1}{16} . Then the probability that at least one of the events occurs, is

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Question

For three events A, B and C, P (Exactly one of A or B occurs) = P (Exactly one of B or C occurs) = P(Exactly one of C or A occurs) = $\frac{1}{4}$ and P (All the three events occur simultaneously) = $\frac{1}{16} .$

Then the probability that at least one of the events occurs, is

$\frac{7}{64}$

$\frac{3}{16}$

$\frac{7}{16}$

$\frac{7}{32}$

P (exactly one of A & B occur) = P(A) + P(B) – $2 P (A \cap B) = \frac{1}{4}$ …….(1)

P (exactly one of B & C occur) = P(B) + P(C) – $2 P (B \cap C) = \frac{1}{4}$ …….(2)

P (exactly one of C & A occur) = P(C) + P(A) – $2 P (C \cap A) = \frac{1}{4}$ …….(3)

Adding (1), (2) & (3) —————————————

2(P(A) + P(B) + P(C) – P(A $\cap$ B) – P(B $\cap$ C) – P(C $\cap$ A)) = $\frac{3}{4}$

⇒ P(A) + P(B) + P(C) – P(A $\cap$ B) – P(B $\cap$ C) – P(C $\cap$ A) = $\frac{3}{8}$

P(All three occurs simultaneously) = P(A $\cap$ B $\cap$ C) = $\frac{1}{16}$

$∴$ P (atleast one of events occur) = P(A $\cup$ B $\cup$ C) = $\frac{3}{8} + \frac{1}{16} = \frac{7}{16}$ .

Let's analyze the problem step by step. We are given three events A, B, and C with the following probabilities:

P(Exactly one of A or B occurs) = $\frac{1}{4}$
P(Exactly one of B or C occurs) = $\frac{1}{4}$
P(Exactly one of C or A occurs) = $\frac{1}{4}$
P(All three occur simultaneously) = $\frac{1}{16}$

We need to find P(at least one of A, B, or C occurs).

Step 1: Express "Exactly one of A or B occurs" in terms of set operations.

Exactly one of A or B occurs means A occurs without B, or B occurs without A. This is equivalent to (A ∩ B') ∪ (A' ∩ B). The probability is:

P(A) - P(A ∩ B) + P(B) - P(A ∩ B) = P(A) + P(B) - 2P(A ∩ B)

So, P(A) + P(B) - 2P(A ∩ B) = $\frac{1}{4}$ ...(1)

Similarly, for B and C: P(B) + P(C) - 2P(B ∩ C) = $\frac{1}{4}$ ...(2)

For C and A: P(C) + P(A) - 2P(C ∩ A) = $\frac{1}{4}$ ...(3)

Step 2: Add equations (1), (2), and (3):

2[P(A) + P(B) + P(C)] - 2[P(A∩B) + P(B∩C) + P(C∩A)] = $\frac{3}{4}$

Divide both sides by 2:

P(A) + P(B) + P(C) - [P(A∩B) + P(B∩C) + P(C∩A)] = $\frac{3}{8}$ ...(4)

Step 3: Use the inclusion-exclusion principle for three events.

P(at least one occurs) = P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A∩B) - P(B∩C) - P(C∩A) + P(A∩B∩C)

From equation (4), we have P(A) + P(B) + P(C) - [P(A∩B) + P(B∩C) + P(C∩A)] = $\frac{3}{8}$

And P(A∩B∩C) = $\frac{1}{16}$

So, P(A ∪ B ∪ C) = $\frac{3}{8}$ + $\frac{1}{16}$ = $\frac{6}{16}$ + $\frac{1}{16}$ = $\frac{7}{16}$

Final Answer: $\frac{7}{16}$

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