91Ó°ÊÓ

Bill hits the target with probability 0.7, so he misses the target with probability \(1 - 0.7 = 0.3\). George hits the target with probability 0.4, so he misses the target with probability \(1 - 0.4 = 0.6\).

We have four possible combinations: 1. Both hit the target: P(Both hit) = P(Bill hits) * P(George hits) = \(0.7 * 0.4 = 0.28\) 2. Both miss the target: P(Both miss) = P(Bill misses) * P(George misses) = \(0.3 * 0.6 = 0.18\) 3. Bill hits, George misses: P(Bill hits, George misses) = P(Bill hits) * P(George misses) = \(0.7 * 0.6 = 0.42\) 4. George hits, Bill misses: P(George hits, Bill misses) = P(George hits) * P(Bill misses) = \(0.4 * 0.3 = 0.12\)

(a) P(George's shot hits | exactly one shot hit the target) = P(George hits, Bill misses) / P(exactly one shot hit the target) = P(George hits, Bill misses) / (P(Bill hits, George misses) + P(George hits, Bill misses)) P(George's shot hits | exactly one shot hit the target) = \(0.12 / (0.42 + 0.12) = 0.12 / 0.54 \approx 0.222\) (b) P(George's shot hits | the target is hit) = P(George hits, target hit) / P(target is hit) = P(George hits, target hit) / (P(Both hit) + P(Bill hits, George misses) + P(George hits, Bill misses)) To find P(George hits, target hit), we can consider two cases: - Both hit: This probability has already been calculated in Step 2 as 0.28. - Only George hits: Since the target is hit, we know that Bill misses in this case. This probability has also been calculated in Step 2 as 0.12. Therefore, P(George hits, target hit) = P(Both hit) + P(George hits, Bill misses) = 0.28 + 0.12 = 0.4 Now we can find the conditional probability: P(George's shot hits | the target is hit) = \(0.4 / (0.28 + 0.42 + 0.12) = 0.4 / 0.82 \approx 0.488\)

(a) The probability that George's shot hit the target given that exactly one shot hit the target is approximately 0.222. (b) The probability that George's shot hit the target given that the target is hit is approximately 0.488.