Probability
Independent Events
Grade 12
Question:
<p>The probability of event \(A\) and \(B\) occurring together is \(\frac{3}{2}\) and that of \(A\) and \(B\) occurring together is \(\frac{3}{10}\) then \(P(A^c) + P(B^c)\) is equal to</p>
<p>(a) \(\frac{6}{5}\)</p>
<p>(b) \(\frac{11}{10}\)</p>
<p>(c) \(\frac{9}{10}\)</p>
<p>(d) \(\frac{23}{20}\)</p>
Step-by-Step Solution
Key Concept: Use the relationship P(A∩B) + P(A∩B^c) + P(A^c∩B) + P(A^c∩B^c) = 1 and the constraint that P(A∪B) = P(A) + P(B) - P(A∩B) to find P(A^c) + P(B^c) = 2 - P(A∪B).
<p><strong>Step 1:</strong> Interpret the given information. The probability of A or B occurring is P(A∪B) = 3/5, and probability of both A and B occurring is P(A∩B) = 3/10.</p><p><strong>Step 2:</strong> Use the formula: P(A∪B) = P(A) + P(B) - P(A∩B)</p><p>3/5 = P(A) + P(B) - 3/10</p><p>P(A) + P(B) = 3/5 + 3/10 = 6/10 + 3/10 = 9/10</p><p><strong>Step 3:</strong> Calculate P(A^c) + P(B^c)</p><p>P(A^c) + P(B^c) = [1 - P(A)] + [1 - P(B)]</p><p>= 2 - [P(A) + P(B)]</p><p>= 2 - 9/10</p><p>= 20/10 - 9/10</p><p>= 11/10</p><p>∴ Answer: B</p>
Correct Answer: B