Independence). (i) Recall that given events A, B,C, we have that P(AUBUC) = P(A)+ P(B)+ P(C) – P(An B) – P(AnC)- P(BnC)+ P(An BnC). that whenever events A, B, C are mutually independent, we have that P(AUBUC) =1- P(A°)P(B°)P(C®) = 1 – (1 – P(A)) · (1 – P(B))· (1 – P(C)) he probability in question can be calculated considerably faster than with the use of the formula (4.1)). The probability that A hits a target is 1/6, the probability that B hits it is 4/7. and the probability that C hits th
Independence). (i) Recall that given events A, B,C, we have that P(AUBUC) = P(A)+ P(B)+ P(C) – P(An B) – P(AnC)- P(BnC)+ P(An BnC). that whenever events A, B, C are mutually independent, we have that P(AUBUC) =1- P(A°)P(B°)P(C®) = 1 – (1 – P(A)) · (1 – P(B))· (1 – P(C)) he probability in question can be calculated considerably faster than with the use of the formula (4.1)). The probability that A hits a target is 1/6, the probability that B hits it is 4/7. and the probability that C hits th
A First Course in Probability (10th Edition)
10th Edition
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Chapter1: Combinatorial Analysis
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![1. (Independence). (i) Recall that given events A, B,C, we have that
P(AUBUC) = P(A)+ P(B) + P(C) – P(An B) – P(AnC) – P(BNC) + P(AN BnC).
(4.1)
Prove that whenever events A, B,C are mutually independent, we have that
P(AUBUC)= 1– P(A°)P(B°)P(C°) = 1- (1– P(A)) · (1 – P(B)) · (1 – P(C))
(and the probability in question can be calculated considerably faster than with the use of the formula (4.1)).
(ii) The probability that A hits a target is 1/6, the probability that B hits it is 4/7, and the probability that C hits the target is
2/5. Use (i) to find the probability that the target will be hit if A, B, and C each shoot at the target.
(iii) Suppose that the probability that a soldier firing his personal weapon hits an enemy warplane is p> 0. Show then, arguing as
in (i), that the probability that the warplane is hit at least once when n > 2 soldiers shoot at it is
1- (1– p)".
(4.2)
Evaluate the probability in (4.2) in the case when p = 0.0007 and n = 500, and then round your result to a 5-digit floating-point
number.
You may not include the proof of the result in (i) in your table with the answers (but do try to obtain the proof, and keep it for your
records).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F80470a4d-4352-4142-9842-1867e9f509a4%2F017b9892-876b-421b-89a2-dffaac9192e3%2Fysthtq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1. (Independence). (i) Recall that given events A, B,C, we have that
P(AUBUC) = P(A)+ P(B) + P(C) – P(An B) – P(AnC) – P(BNC) + P(AN BnC).
(4.1)
Prove that whenever events A, B,C are mutually independent, we have that
P(AUBUC)= 1– P(A°)P(B°)P(C°) = 1- (1– P(A)) · (1 – P(B)) · (1 – P(C))
(and the probability in question can be calculated considerably faster than with the use of the formula (4.1)).
(ii) The probability that A hits a target is 1/6, the probability that B hits it is 4/7, and the probability that C hits the target is
2/5. Use (i) to find the probability that the target will be hit if A, B, and C each shoot at the target.
(iii) Suppose that the probability that a soldier firing his personal weapon hits an enemy warplane is p> 0. Show then, arguing as
in (i), that the probability that the warplane is hit at least once when n > 2 soldiers shoot at it is
1- (1– p)".
(4.2)
Evaluate the probability in (4.2) in the case when p = 0.0007 and n = 500, and then round your result to a 5-digit floating-point
number.
You may not include the proof of the result in (i) in your table with the answers (but do try to obtain the proof, and keep it for your
records).
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