Probability
Geometric Probability / Independent Events
Grade 12
Question:
<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 repeated trials as geometric distributions where success probability remains constant each attempt; calculate probabilities of hitting on specific attempts or within given attempts using the formula P(first success on nth trial) = (1-p)^(n-1)·p.
<p><strong>Step 1:</strong> Identify given probabilities</p><p>P(A hits target) = 3/5, so P(A misses) = 2/5</p><p>P(B hits target) = 5/7, so P(B misses) = 2/7</p><p><strong>Step 2:</strong> For geometric distribution, probability of first success on nth trial:</p><p>P(A hits on nth shot) = (2/5)^(n-1) · (3/5)</p><p>P(B hits on nth shot) = (2/7)^(n-1) · (5/7)</p><p><strong>Step 3:</strong> Calculate key probabilities</p><p>P(both hit on first shot) = (3/5) · (5/7) = 15/35 = 3/7</p><p>P(A hits on 2nd shot) = (2/5) · (3/5) = 6/25</p><p>P(B hits on 2nd shot) = (2/7) · (5/7) = 10/49</p><p><strong>Step 4:</strong> For compound events with independence:</p><p>P(A hits within 2 shots) = 3/5 + (2/5)·(3/5) = 3/5 + 6/25 = 21/25</p><p>P(B hits within 2 shots) = 5/7 + (2/7)·(5/7) = 5/7 + 10/49 = 45/49</p><p><strong>Step 5:</strong> For both hitting within n shots (independence):</p><p>P(both hit within 2 shots) = (21/25) · (45/49) = 945/1225 = 189/245 = 27/35</p><p>∴ Answer: A,B,C</p>
Correct Answer: A,B,C