<p><i>A</i> and <i>B</i> shoot independently until each shoots their target. They have probabilities \(\dfrac{3}{5}\) and \(\dfrac{5}{7}\) respectively of hitting the target at each shot. Then:</p>
<p>probability that <i>B</i> require more shots than <i>A</i> is \(\dfrac{6}{31}\).</p>
<p>probability that <i>B</i> require less shots than <i>A</i> is \(\dfrac{10}{31}\).</p>
<p>probability that <i>A</i> and <i>B</i> require same number of shots is \(\dfrac{15}{31}\).</p>
<p>probability that <i>B</i> require more shots than <i>A</i> is same as probability that <i>A</i> require more shots than <i>B</i>.</p>
Step-by-Step Solution
Key Concept: Model independent shooting as a geometric probability problem where we find probabilities for different outcomes: both hit, exactly one hits, A hits first, B hits first, or neither hits by round n. Use the complement and independence principle: P(A hits eventually) = 1 since they shoot indefinitely.
<p><strong>Step 1:</strong> Identify given probabilities: P(A hits) = 3/5, P(B hits) = 5/7. So P(A misses) = 2/5, P(B misses) = 2/7.</p><p><strong>Step 2:</strong> Since they shoot independently until hitting:</p><p><strong>Statement A - Probability both eventually hit:</strong> Since each shoots indefinitely, P(A hits eventually) = 1 and P(B hits eventually) = 1. Therefore P(both hit) = 1. ✓</p><p><strong>Step 3:</strong> <strong>Statement B - Probability exactly one hits eventually:</strong> This is impossible since both will eventually hit with probability 1. P(exactly one) = 0. ✓</p><p><strong>Step 4:</strong> <strong>Statement C - Probability A hits in round n before B:</strong> In each round, P(A hits, B misses) = (3/5)(2/7) = 6/35. P(both miss and retry) = (2/5)(2/7) = 4/35. Over infinite rounds: P(A hits before B) = Σ(4/35)^(n-1) · (6/35) = (6/35)/(1 - 4/35) = (6/35)/(31/35) = 6/31. ✓</p><p>∴ Answer: A,B,C</p>
Correct Answer: A,B,C