<p>Suppose the probability for <em>A</em> to win a game against <em>B</em> is 0.4. If <em>A</em> has an option of playing either a "best of 3 games" or a "best of 5 games" match against <em>B</em>, which option should be chosen so that the probability of his winning the match is higher? (No game ends in a draw.)</p>
Step-by-Step Solution
Key Concept: A wins the match if he wins more games than B in the series. Calculate P(A wins match) for both scenarios using binomial probability, recognizing that with p=0.4<0.5, A is more likely to win with fewer games (lower variance favors the weaker player in shorter series).
<p><strong>Step 1: Best of 3 games</strong></p><p>A wins the match by winning 2 or 3 games out of 3.</p><p>P(A wins match) = P(exactly 2 wins) + P(exactly 3 wins)</p><p>= C(3,2)(0.4)²(0.6)¹ + C(3,3)(0.4)³(0.6)⁰</p><p>= 3(0.16)(0.6) + 1(0.064)</p><p>= 0.288 + 0.064 = <strong>0.352</strong></p><p><strong>Step 2: Best of 5 games</strong></p><p>A wins the match by winning 3, 4, or 5 games out of 5.</p><p>P(A wins match) = C(5,3)(0.4)³(0.6)² + C(5,4)(0.4)⁴(0.6)¹ + C(5,5)(0.4)⁵(0.6)⁰</p><p>= 10(0.064)(0.36) + 5(0.0256)(0.6) + 1(0.01024)</p><p>= 0.2304 + 0.0768 + 0.01024 = <strong>0.31744</strong></p><p><strong>Step 3: Compare</strong></p><p>0.352 > 0.31744</p><p>∴ A should choose the <strong>best of 3 games</strong> option since 0.352 > 0.317</p>
Correct Answer: A