Probability
Addition Theorem of Probability
Grade 12
Question:
<p>For three events \(A\), \(B\) and \(C\),<br>\(P(\text{Exactly one of } A \text{ or } B \text{ occurs})\)<br>\(= P(\text{Exactly one of } B \text{ or } C \text{ occurs})\)<br>\(= P(\text{Exactly one of } C \text{ or } A \text{ occurs}) = \dfrac{1}{4}\) and<br>\(P(\text{All the three events occur simultaneously}) = \dfrac{1}{16}\).<br>Then the probability that at least one of the events occurs, is</p>
<p>\(\dfrac{3}{16}\)</p>
<p>\(\dfrac{7}{32}\)</p>
<p>\(\dfrac{7}{16}\)</p>
<p>\(\dfrac{7}{64}\)</p>
Step-by-Step Solution
Key Concept: Use the symmetric condition that exactly one event occurs in each pair equals 1/4, combined with P(A∩B∩C) = 1/16, to set up equations using inclusion-exclusion principle and solve for P(A∪B∪C).
<p><strong>Step 1:</strong> Translate the given conditions.</p><p>P(Exactly one of A or B) = P(A∩B' ∪ A'∩B) = P(A) + P(B) - 2P(A∩B) = 1/4</p><p>P(Exactly one of B or C) = P(B) + P(C) - 2P(B∩C) = 1/4</p><p>P(Exactly one of C or A) = P(C) + P(A) - 2P(C∩A) = 1/4</p><p><strong>Step 2:</strong> Add all three equations.</p><p>2[P(A) + P(B) + P(C)] - 2[P(A∩B) + P(B∩C) + P(C∩A)] = 3/4</p><p>∴ P(A) + P(B) + P(C) - [P(A∩B) + P(B∩C) + P(C∩A)] = 3/8</p><p><strong>Step 3:</strong> Apply inclusion-exclusion principle.</p><p>P(A∪B∪C) = P(A) + P(B) + P(C) - P(A∩B) - P(B∩C) - P(C∩A) + P(A∩B∩C)</p><p>P(A∪B∪C) = 3/8 + 1/16 = 6/16 + 1/16 = 7/16</p><p>∴ Answer: C (7/16)</p>
Correct Answer: C