Probability
Addition Theorem of Probability
Grade 12
Question:
<p>For three events \(A\), \(B\) and \(C\),<br>\(P(\text{Exactly one of } A \text{ or } B \text{ occurs}) = P(\text{Exactly one of } B \text{ or } C \text{ occurs})\)<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>\(\dfrac{7}{16}\)</p>
<p>\(\dfrac{7}{64}\)</p>
<p>\(\dfrac{3}{16}\)</p>
<p>\(\dfrac{7}{32}\)</p>
Step-by-Step Solution
Key Concept: Express 'exactly one of X or Y' as (P(X)−P(X∩Y))+(P(Y)−P(X∩Y)) = P(X)+P(Y)−2P(X∩Y), then use the given equalities to establish relationships between individual probabilities and intersections.
<p><strong>Step 1:</strong> Express the given conditions using the formula for exactly one event occurring.</p><p>P(Exactly one of A or B) = P(A)+P(B)−2P(A∩B) = p₁</p><p>P(Exactly one of B or C) = P(B)+P(C)−2P(B∩C) = p₁ (equal by given)</p><p>P(Exactly one of C or A) = P(C)+P(A)−2P(C∩A) = 1/4</p><p><strong>Step 2:</strong> Let P(A)=a, P(B)=b, P(C)=c, and P(A∩B∩C)=1/16.</p><p>From conditions 1 and 2: a+b−2P(A∩B) = b+c−2P(B∩C)</p><p>This gives: a−2P(A∩B) = c−2P(B∩C)</p><p><strong>Step 3:</strong> From condition 3: a+c−2P(A∩C) = 1/4</p><p><strong>Step 4:</strong> By symmetry of the equal conditions and the constraint P(A∩B∩C)=1/16, assume a=b=c and P(A∩B)=P(B∩C)=P(A∩C)=p (say).</p><p>Then: 2a−2p = 1/4, so a−p = 1/8</p><p><strong>Step 5:</strong> Using inclusion-exclusion for at least one event:</p><p>P(A∪B∪C) = a+a+a−p−p−p+1/16 = 3a−3p+1/16</p><p>= 3(a−p)+1/16 = 3(1/8)+1/16 = 3/8+1/16 = 6/16+1/16 = 7/16</p><p><strong>∴ Answer: D (7/16)</strong></p>
Correct Answer: D