For three events A, B and C, P(Exactly one of A or B occurs) = P(Exactly one of B or C occurs) = P(E

Q: Wait, why do we subtract 2P(A \cap B) instead of just P(A \cap B)?

That is a fantastic question! Think about the Venn diagram. When you add P(A) and P(B), the region P(A \cap B) is counted twice—once inside A and once inside B. To represent the 'exactly one' region, we need to remove that intersection entirely. Since it was counted twice, we must subtract it twice to clear it out. If we only subtracted it once, we would still have one 'layer' of the intersection left!

Q: Do I need to find the individual probabilities P(A), P(B), and P(C) to solve this?

Not at all! This is the beauty of the problem. By using the symmetry provided, we can group the terms into the exact structure required by the Principle of Inclusion-Exclusion. We never need to isolate A, B, or C individually; we only need the aggregate values.

Q: Is there a trick to remembering the Inclusion-Exclusion formula?

Yes! Think of it as an alternating series: Add the singles, subtract the pairs, add the triples, subtract the quadruples, and so on. It's a beautiful, rhythmic pattern that holds for any number of sets. Just remember: + (singles), - (pairs), + (triples).

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:

Options:

(A) $\frac{7}{32}$

(B) $\frac{7}{16}$

(C) $\frac{7}{64}$

(D) $\frac{3}{16}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are dealing with three events, $A$, $B$, and $C$. Our goal is to find the probability that at least one of them occurs, which is represented by the union $P(A \cup B \cup C)$.

### Decoding the 'Exactly One' Condition

The problem states that the probability of 'exactly one of $A$ or $B$' is $\frac{1}{4}$. In set theory, this region is the symmetric difference, denoted as $A \Delta B$.

This region includes everything in $A$ and everything in $B$, but we must exclude the overlap where both occur. Mathematically, this is expressed as:

$$P(A) + P(B) - 2P(A \cap B) = \frac{1}{4}$$

The factor of $2$ is necessary because when we add $P(A)$ and $P(B)$, the intersection $P(A \cap B)$ is counted twice. To remove it completely from the union, we subtract it twice.

### The Power of Symmetry

The problem specifies that this condition holds for all pairs: $(A, B)$, $(B, C)$, and $(C, A)$. We therefore have three identical equations:

$$P(A) + P(B) - 2P(A \cap B) = \frac{1}{4}$$

$$P(B) + P(C) - 2P(B \cap C) = \frac{1}{4}$$

$$P(C) + P(A) - 2P(C \cap A) = \frac{1}{4}$$

Summing these three equations, we observe that each individual probability and each pairwise intersection appears twice. This yields:

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

Dividing by $2$, we isolate the core components of the Inclusion-Exclusion Principle:

$$\sum P(A) - \sum P(A \cap B) = \frac{3}{8}$$

### The Final Bridge

The Principle of Inclusion-Exclusion states that:

$$P(A \cup B \cup C) = \sum P(A) - \sum P(A \cap B) + P(A \cap B \cap C)$$

We have already determined that $\sum P(A) - \sum P(A \cap B) = \frac{3}{8}$. The problem also provides the triple intersection value, $P(A \cap B \cap C) = \frac{1}{16}$.

Substituting these values into the formula:

$$P(A \cup B \cup C) = \frac{3}{8} + \frac{1}{16}$$

Finding a common denominator, we get:

$$P(A \cup B \cup C) = \frac{6}{16} + \frac{1}{16} = \frac{7}{16}$$

The probability that at least one event occurs is **$\frac{7}{16}$**. By embracing symmetry and the structure of the Inclusion-Exclusion Principle, we have arrived at the solution efficiently.

Select Answer:

Visualized Solution

Visualizing the Three Events

Defining 'Exactly One of A or B'

Equations for All Pairs

Summing the Equations

Simplifying the Sum

The Triple Intersection

The Union Formula

Substituting the Values

Final Calculation

Conclusion

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

Decoding the 'Exactly One' Condition

The Power of Symmetry

The Final Bridge

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