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Math Advanced Math ADV.ENG.MATH (LL) W/WILEYPLUS BUNDLE

To choose: The correct option from the given choices that gives the greater probability of hitting at least once. The correct option is a ) . Hitting with probability 1 2 and firing 1 shot _ . Rule used: Complementation Rule: P ( A c ) = 1 − P ( A ) Given: (a) Hitting with probability 1 2 and firing 1 shot, (b) Hitting with probability 1 4 and firing 2 shots, (c) Hitting with probability 1 8 and firing 4 shots Calculation: Consider the case of hitting with probability 1 2 and firing 1 shot. The probability here is 1 2 = 0.5 . Now, consider the case of hitting with probability 1 4 and firing 2 shots Let the event A i denote hitting a target in the i -th try. That is, P ( A i ) = 1 4 . Using rule mentioned, the value of P ( A i c ) is obtained as follows. P ( A i c ) = 1 − P ( A i ) = 1 − 1 4 = 4 − 1 4 = 3 4 Note that, P ( A 1 c ∩ A 2 c ) = P ( A 1 ∪ A 2 ) c . Therefore, P ( A 1 c ∩ A 2 c ) = P ( A 1 ∪ A 2 ) c = 1 − P ( A 1 ∪ A 2 ) Here, the events A 1 and A 2 are independent. Then, the require probability is obtained as follows. P ( A 1 ∪ A 2 ) = 1 − P ( A 1 c ∩ A 2 c ) = 1 − [ P ( A 1 c ) × P ( A 2 c ) ] = 1 − [ 3 4 × 3 4 ] Simplify further as follows. P ( A 1 ∪ A 2 ) = 1 − 9 16 = 16 − 9 16 = 7 16 = 0.4375 Thus, P ( hitting with probability 1 4 and firing 2 shots ) = 0.4375 . Consider the case of hitting with probability 1 8 and firing 4 shots The value of P ( A i c ) is obtained as follows. P ( A i c ) = 1 − P ( A i ) = 1 − 1 8 = 8 − 1 8 = 7 8 Therefore, the required probability is obtained as follows. P ( A 1 ∪ A 2 ∪ A 3 ∪ A 4 ) = 1 − P ( A 1 c ∩ A 2 c ∩ A 3 c ∩ A 4 c ) = 1 − [ P ( A 1 c ) × P ( A 2 c ) × P ( A 3 c ) × P ( A 4 c ) ] = 1 − [ 7 8 × 7 8 × 7 8 × 7 8 ] = 1 − 2,401 4,096 Simplify further as follows. P ( A 1 ∪ A 2 ∪ A 3 ∪ A 4 ) = 4,096 − 2,401 4,096 = 1,695 4,096 = 0.4138 Thus, P ( hitting with probability 1 8 and firing 4 shots ) = 0.4138 . Observe that, 0.5 > 0.4375 > 0.4138 . Hence, the correct option is a ) . Hitting with probability 1 2 and firing 1 shot _ .

To choose: The correct option from the given choices that gives the greater probability of hitting at least once. The correct option is a ) . Hitting with probability 1 2 and firing 1 shot _ . Rule used: Complementation Rule: P ( A c ) = 1 − P ( A ) Given: (a) Hitting with probability 1 2 and firing 1 shot, (b) Hitting with probability 1 4 and firing 2 shots, (c) Hitting with probability 1 8 and firing 4 shots Calculation: Consider the case of hitting with probability 1 2 and firing 1 shot. The probability here is 1 2 = 0.5 . Now, consider the case of hitting with probability 1 4 and firing 2 shots Let the event A i denote hitting a target in the i -th try. That is, P ( A i ) = 1 4 . Using rule mentioned, the value of P ( A i c ) is obtained as follows. P ( A i c ) = 1 − P ( A i ) = 1 − 1 4 = 4 − 1 4 = 3 4 Note that, P ( A 1 c ∩ A 2 c ) = P ( A 1 ∪ A 2 ) c . Therefore, P ( A 1 c ∩ A 2 c ) = P ( A 1 ∪ A 2 ) c = 1 − P ( A 1 ∪ A 2 ) Here, the events A 1 and A 2 are independent. Then, the require probability is obtained as follows. P ( A 1 ∪ A 2 ) = 1 − P ( A 1 c ∩ A 2 c ) = 1 − [ P ( A 1 c ) × P ( A 2 c ) ] = 1 − [ 3 4 × 3 4 ] Simplify further as follows. P ( A 1 ∪ A 2 ) = 1 − 9 16 = 16 − 9 16 = 7 16 = 0.4375 Thus, P ( hitting with probability 1 4 and firing 2 shots ) = 0.4375 . Consider the case of hitting with probability 1 8 and firing 4 shots The value of P ( A i c ) is obtained as follows. P ( A i c ) = 1 − P ( A i ) = 1 − 1 8 = 8 − 1 8 = 7 8 Therefore, the required probability is obtained as follows. P ( A 1 ∪ A 2 ∪ A 3 ∪ A 4 ) = 1 − P ( A 1 c ∩ A 2 c ∩ A 3 c ∩ A 4 c ) = 1 − [ P ( A 1 c ) × P ( A 2 c ) × P ( A 3 c ) × P ( A 4 c ) ] = 1 − [ 7 8 × 7 8 × 7 8 × 7 8 ] = 1 − 2,401 4,096 Simplify further as follows. P ( A 1 ∪ A 2 ∪ A 3 ∪ A 4 ) = 4,096 − 2,401 4,096 = 1,695 4,096 = 0.4138 Thus, P ( hitting with probability 1 8 and firing 4 shots ) = 0.4138 . Observe that, 0.5 > 0.4375 > 0.4138 . Hence, the correct option is a ) . Hitting with probability 1 2 and firing 1 shot _ .

ADV.ENG.MATH (LL) W/WILEYPLUS BUNDLE

ADV.ENG.MATH (LL) W/WILEYPLUS BUNDLE

10th Edition

ISBN: 9781119809210

Author: Kreyszig

Publisher: WILEY

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Chapter 24 Solutions

ADV.ENG.MATH (LL) W/WILEYPLUS BUNDLE

Ch. 24.1 - Prob. 1P Ch. 24.1 - Prob. 2P Ch. 24.1 - Prob. 3P Ch. 24.1 - Prob. 4P Ch. 24.1 - Prob. 5P Ch. 24.1 - Prob. 6P Ch. 24.1 - Prob. 7P Ch. 24.1 - Prob. 8P Ch. 24.1 - Prob. 9P Ch. 24.1 - Prob. 10P

Ch. 24.1 - Prob. 11P Ch. 24.1 - Prob. 12P Ch. 24.1 - Prob. 13P Ch. 24.1 - Prob. 14P Ch. 24.1 - Prob. 15P Ch. 24.1 - Prob. 16P Ch. 24.1 - Prob. 17P Ch. 24.1 - Prob. 18P Ch. 24.1 - Prob. 19P Ch. 24.1 - Prob. 20P Ch. 24.2 - Prob. 1P Ch. 24.2 - Prob. 2P Ch. 24.2 - Prob. 3P Ch. 24.2 - Prob. 4P Ch. 24.2 - Prob. 5P Ch. 24.2 - Prob. 6P Ch. 24.2 - Prob. 7P Ch. 24.2 - Prob. 8P Ch. 24.2 - Prob. 9P Ch. 24.2 - Prob. 10P Ch. 24.2 - Prob. 11P Ch. 24.2 - Prob. 12P Ch. 24.2 - Prob. 13P Ch. 24.2 - Prob. 14P Ch. 24.2 - Prob. 15P Ch. 24.2 - Prob. 16P Ch. 24.2 - Prob. 17P Ch. 24.2 - Prob. 18P Ch. 24.2 - Prob. 19P Ch. 24.2 - Prob. 20P Ch. 24.3 - Prob. 1P Ch. 24.3 - Prob. 2P Ch. 24.3 - Prob. 3P Ch. 24.3 - Prob. 4P Ch. 24.3 - Prob. 5P Ch. 24.3 - Prob. 6P Ch. 24.3 - Prob. 7P Ch. 24.3 - Prob. 8P Ch. 24.3 - Prob. 9P Ch. 24.3 - Prob. 10P Ch. 24.3 - Prob. 11P Ch. 24.3 - Prob. 12P Ch. 24.3 - Prob. 13P Ch. 24.3 - Prob. 14P Ch. 24.3 - Prob. 15P Ch. 24.3 - Prob. 16P Ch. 24.3 - Prob. 17P Ch. 24.3 - Prob. 18P Ch. 24.3 - Prob. 19P Ch. 24.4 - Prob. 1P Ch. 24.4 - Prob. 2P Ch. 24.4 - Prob. 3P Ch. 24.4 - Prob. 4P Ch. 24.4 - Prob. 5P Ch. 24.4 - Prob. 6P Ch. 24.4 - Prob. 7P Ch. 24.4 - Prob. 8P Ch. 24.4 - Prob. 9P Ch. 24.4 - Prob. 10P Ch. 24.4 - Prob. 11P Ch. 24.4 - Prob. 12P Ch. 24.4 - Prob. 15P Ch. 24.5 - Prob. 1P Ch. 24.5 - Prob. 2P Ch. 24.5 - Prob. 3P Ch. 24.5 - Prob. 4P Ch. 24.5 - Prob. 5P Ch. 24.5 - Prob. 6P Ch. 24.5 - Prob. 7P Ch. 24.5 - Prob. 8P Ch. 24.5 - Prob. 9P Ch. 24.5 - Prob. 10P Ch. 24.5 - Prob. 11P Ch. 24.5 - Prob. 12P Ch. 24.5 - Prob. 14P Ch. 24.5 - Prob. 15P Ch. 24.6 - Prob. 1P Ch. 24.6 - Prob. 2P Ch. 24.6 - Prob. 3P Ch. 24.6 - Prob. 4P Ch. 24.6 - Prob. 5P Ch. 24.6 - Prob. 6P Ch. 24.6 - Prob. 7P Ch. 24.6 - Prob. 8P Ch. 24.6 - Prob. 9P Ch. 24.6 - Prob. 10P Ch. 24.6 - Prob. 11P Ch. 24.6 - Prob. 12P Ch. 24.6 - Prob. 13P Ch. 24.6 - Prob. 14P Ch. 24.6 - Prob. 15P Ch. 24.6 - Prob. 16P Ch. 24.6 - Prob. 17P Ch. 24.6 - Prob. 18P Ch. 24.6 - Prob. 19P Ch. 24.7 - Prob. 1P Ch. 24.7 - Prob. 3P Ch. 24.7 - Prob. 4P Ch. 24.7 - Prob. 5P Ch. 24.7 - Prob. 6P Ch. 24.7 - Prob. 7P Ch. 24.7 - Prob. 8P Ch. 24.7 - Prob. 9P Ch. 24.7 - Prob. 10P Ch. 24.7 - Prob. 11P Ch. 24.7 - Prob. 12P Ch. 24.7 - Prob. 13P Ch. 24.7 - Prob. 14P Ch. 24.7 - Prob. 15P Ch. 24.7 - Prob. 17P Ch. 24.7 - Prob. 18P Ch. 24.8 - Prob. 1P Ch. 24.8 - Prob. 2P Ch. 24.8 - Prob. 3P Ch. 24.8 - Prob. 4P Ch. 24.8 - Prob. 5P Ch. 24.8 - Prob. 6P Ch. 24.8 - Prob. 7P Ch. 24.8 - Prob. 8P Ch. 24.8 - Prob. 9P Ch. 24.8 - Prob. 10P Ch. 24.8 - Prob. 11P Ch. 24.8 - Prob. 12P Ch. 24.8 - Prob. 13P Ch. 24.9 - Prob. 1P Ch. 24.9 - Prob. 2P Ch. 24.9 - Prob. 3P Ch. 24.9 - Prob. 4P Ch. 24.9 - Prob. 5P Ch. 24.9 - Prob. 6P Ch. 24.9 - Prob. 7P Ch. 24.9 - Prob. 8P Ch. 24.9 - Prob. 9P Ch. 24.9 - Prob. 10P Ch. 24.9 - Prob. 11P Ch. 24.9 - Prob. 12P Ch. 24.9 - Prob. 13P Ch. 24.9 - Prob. 14P Ch. 24.9 - Prob. 15P Ch. 24.9 - Prob. 16P Ch. 24.9 - Prob. 17P Ch. 24.9 - Prob. 18P Ch. 24.9 - Prob. 19P Ch. 24.9 - Prob. 20P Ch. 24 - Prob. 1RQ Ch. 24 - Prob. 2RQ Ch. 24 - Prob. 3RQ Ch. 24 - Prob. 4RQ Ch. 24 - Prob. 5RQ Ch. 24 - Prob. 6RQ Ch. 24 - Prob. 7RQ Ch. 24 - Prob. 9RQ Ch. 24 - Prob. 10RQ Ch. 24 - Prob. 11RQ Ch. 24 - Prob. 12RQ Ch. 24 - Prob. 13RQ Ch. 24 - Prob. 14RQ Ch. 24 - Prob. 15RQ Ch. 24 - Prob. 16RQ Ch. 24 - Prob. 17RQ Ch. 24 - Prob. 18RQ Ch. 24 - Prob. 19RQ Ch. 24 - Prob. 20RQ Ch. 24 - Prob. 21RQ Ch. 24 - Prob. 22RQ Ch. 24 - Prob. 23RQ Ch. 24 - Prob. 24RQ Ch. 24 - Prob. 25RQ

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