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Section 5.4 Lecture Problems

# Section 5.4 Lecture Problems - ASW ESBE(5e Chapter 5...

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ASW ESBE (5e) Chapter 5 Solutions to Selected Problems Section 5.4 for the binomial distribution. Problems 32, 33, 34, 35, 36. 32. Let X be the r.v. that counts the number of radars that detect an incoming missile of the n radars that are deployed. This is a binomial situation, because either the deployed radars detect an incoming missile attack (= “success”) or they don’t (= “failure”). Given: 0.90% probability that a radar will detect a missile attack. Probabilities will be calculated with the binomial probability distribution, ( 29 ( 29 ( 29 x n x p p ! x n x! n! x|n, p prob - - × × - = 1 In all questions, p = 0.90 (the success probability), as given. (Because each radar individually, by itself, has a 0.90 probability of detecting an incoming missile attack, as stated in the third sentence of the problem text). (Note: the failure probability = 1- p = 1-0.9 = 0.1) a. Want prob (single radar will detect an attack) Given: n = 1 p = 0.90 x = 1 Want prob( x =1| n =1, p =0.90) Solution: Some Section 5.4 Problem Solutions 1

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( 29 ( 29 9 0 1 9 0 1 1 0 9 0 0 1 1 1 0 9 0 1 1 1 1 90 0 1 1 0 1 1 1 1 . . . . ! ! ! . . ! ! ! . , p |n x prob = × × = × × = × × - = = = = - Some Section 5.4 Problem Solutions 2
b. Want prob (at least one of two deployed radars will detect an attack) “At least one of the two” means any one of the radars alone or both. In terms of the integer values of the random variable, x ≥ 1 so that x = 0 or 1 or 2. Given: n = 2 p = 0.90 x ≥ 1 Want prob ( x ≥1| n =2, p =0.90) Note: For n = 2, x ≥ 1 includes the values x = 1 or 2. Solution: ( 29 ( 29 ( 29 ( 29 ( 29 2 90 0 2 2 1 90 0 2 1 90 0 2 1 . , p |n x prob . , p |n x prob . , p |n x prob = = = + = = = = = = For term (1): ( 29 ( 29 18 0 1 0 9 0 2 1 0 9 0 1 1 2 1 0 9 0 1 2 1 2 1 1 1 1 2 1 . . . . . ! ! ! . . ! ! ! = × × = × × = × × - = - For term (2): ( 29 ( 29 81 0 1 9 0 1 1 0 9 0 0 2 2 1 0 9 0 2 2 2 2 2 2 0 2 2 2 2 . . . . ! ! ! . . ! ! ! = × × = × × = × × - = - Then Some Section 5.4 Problem Solutions 3

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( 29 ( 29 ( 29 99 0 81 0 18 0 2 1 90 0 2 1 . . . . , p |n x prob = + = + = = = ALTERNATIVE: For n = 2, x ≥ 1 are the values x = 1 or 2; this list does not include x = 0. By complements, prob ( x ≥1| n =2, p =0.90) = 1 – prob ( x =0| n =2, p =0.90) But prob ( x =0| n =2, p =0.90) = 1 × 0.90 0 × 0.10 2 = 0.01 So prob ( x ≥1| n =2, p =0.90) = 1 – 0.01 = 0.99 as before c. Want prob (at least one of three deployed radars will detect an attack) “At least one of the three” means any one of the radars alone, any pair of radars (of which there are three pairs), or all three. In terms of the integer values of the random variable, x ≥ 1 so that x = 0 or 1 or 2 or 3. Given: n = 3 p = 0.90 x ≥ 1 Want prob( x =1| n =3, p =0.90) Note: For n = 3, x ≥ 1 includes the values x = 1, 2, or 3. Solution: ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 3 90 0 3 3 2 90 0 3 2 1 90 0 3 1 90 0 3 1 . , p |n x prob . , p |n x prob . , p |n x prob . , p |n x prob = = = + = = = + = = = = = = For term (1): ( 29 ( 29 027 0 01 0 9 0 3 1 0 9 0 1 1 3 1 0 9 0 1 3 1 3 1 2 1 1 3 1 . . . . . ! ! ! . . ! ! ! = × × = × × = × × - = - Some Section 5.4 Problem Solutions 4
For term (2): ( 29 ( 29 243 0 1 0 81 0 3 1 0 9 0 1 2 3 1 0 9 0 2 3 2 3 2 1 2 2 3 2 . . . . . ! ! ! . . ! ! ! = × × = × × = × × - = - For term (3): ( 29 ( 29 729 0 1 9 0 1 1 0 9 0 0 3 3 1 0 9 0 3 2 3 3 3 3 3 0 3 3 3 3 . . . . ! ! ! . . ! ! ! = × × = × × = × × - = - Then ( 29 ( 29 ( 29 ( 29 999 0 729 0 243 0 027 0 3 2 1 90 0 3 1 . . . . . , p |n x prob = + + = + + = = = ALTERNATIVE: For n = 3, x ≥ 1 are the values x = 1 or 2 or; this list does not include x = 0. By complements, prob ( x ≥1| n =3, p =0.90) = 1 – prob ( x =0| n =2, p =0.90) But prob ( x =0| n =3, p =0.90) = 1 × 0.90 0 × 0.10 3 = 0.001 So prob ( x ≥1| n =3, p =0.90) = 1 – 0.001 = 0.999 as before You can probably guess that the chance of at least one of four deployed radars is 0.9999. You’d be correct.

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