(2019, Suppose Airport A handles 80% of the airline traffic in a region, Airport

Question

(2019, Suppose Airport A handles 80% of the airline traffic in a region, Airport B handles l 5% of the traflic, and Airport C handles only 5% of the airline traffic. The detection rates for weapons at these airports are 3, .5, and.02. .Let W indicate a person is carrying a weapon. Let W" indicate a person is not carrying a weapon. indicate the person is at Airport A, and A" indicates the person is not at Airport A. B indicate the person is at Airport B, and B° indicates the person is not at Airport B. C indicate the person is at Airport C, and C" indicates the person is not at Airport C . Let A Let (1o) a. Complete the table below. Show your work separately below the table. P(WIA)- is found to be carrying a weapon, what is the probability that the person is at Airport A? )c. Ifa person is found to be carrying a weapon, what is the probability that the person is at Airport B? (2) d. Ifa person is found to be carrying a weapon, what is the probability that the person is at Airport A? (2)e. If a person is found to be carrying a weapon, which airport is the person most likely to use? Explain. (2) f If a person is not found to be carrying a weapon, which airport is the person most likely to use? Explain. (20) 10. Let x be a hypergeometric random variable with N- 11, n 5, and M-3. (1) a. Calculate P(0). (1) b. Calculate P(1). (1) c. Calculate P(2). (I) d. Calculate P(3). (2) e. Show how to calculate the mean. The mean (2) f. Show how to calculate the variance. The variance (1) g. What is the interval of? ?? (1) h. What is the interval of? 20? (l) i. What is the interval of 3?? (2) j. What proportion of the population falls in the interval of ? ± ?? (2) k. What propo (2) I. What proportion of the population falls in the interval of ? (1) m. Do these results agree with the Empirical Rule? (1) n. Do these results agree with Tchebysheff's Theorem? rtion of the population falls in the interval of u + 20? 2a? (1) o. Explain the results in one or two sentences.

Explanation / Answer

a) Here, we are given that:

P(A) = 0.8, P(B) = 0.15 and P(C) = 0.05

Also, using the given detection rates, we get:

P(W | A) = 0.3
P(W | B) = 0.5
P(W | C) = 0.02

Using law of total probability, we get:

P(W) = P(W | A)P(A) + P(W | B)P(B) + P(W | C)P(C) = 0.3*0.8 + 0.5*0.15 + 0.02*0.05 = 0.316

Therefore, now using bayes theorem, we get here:
P(A | W) = 0.3*0.8 / 0.316 = 0.7595
P(B | W) = 0.5*0.15 / 0.316 = 0.2373
P(C | W) = 0.02*0.05 / 0.316 = 0.0032

Now, we have here: P(Wc) = 1 - P(W) = 1 - 0.316 = 0.684

Therefore,
P(A | Wc) = 0.8*(1 - 0.3) / 0.684 = 0.8187
P(B | Wc) = 0.15*(1 - 0.5) / 0.684 = 0.1096
P(C | Wc) = 0.05*(1 - 0.02) / 0.684 = 0.0716

b) P(A | W) = 0.7595 ( from the above part )

c) P(B | W) = 0.2373 ( from above part )

d) P(A | W) = 0.7595 ( same problem as part b) )

e) We know that:

P(A | W) = 0.7595
P(B | W) = 0.2373
P(C | W) = 0.0032

Therefore given that the person is carrying weapon, he is most likely to be at airport A

f) Here, we have from part a)

P(A | Wc) = 0.8187 which is greater than P(B | Wc) and P(C | Wc), therefore Again the person is most likely to be at airport A.

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