Probability
Compound Events and Conditional Probability
Grade 12
Question:
<p>For three events A, B and C, if 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 atleast one of the events occurs, is</p>
<p>(a) \(\frac{7}{32}\)</p>
<p>(b) \(\frac{7}{16}\)</p>
<p>(c) \(\frac{7}{8}\)</p>
<p>(d) \(\frac{3}{16}\)</p>
Step-by-Step Solution
Key Concept: Use the conditions on probabilities of exactly one event occurring and the intersection of all three events to set up a system of equations, then apply inclusion-exclusion principle.
<p><strong>Given:</strong></p><p>P(exactly one of A or B) = $\frac{1}{4}$</p><p>P(exactly one of B or C) = $\frac{1}{4}$</p><p>P(exactly one of C or A) = $\frac{1}{4}$</p><p>P(A ∩ B ∩ C) = $\frac{1}{16}$</p><p><strong>Using the condition for exactly one event occurring:</strong></p><p>P(exactly one of A or B) = P(A) + P(B) - 2P(A ∩ B) = $\frac{1}{4}$</p><p>Similarly for other pairs, we can set up equations and solve for P(A ∪ B ∪ C).</p><p><strong>By solving the system of equations with the given constraints:</strong></p><p>P(A ∪ B ∪ C) = $\frac{7}{16}$</p><p>∴ Answer is (b)</p>
Correct Answer: b