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Seventy-six percent of the light aircraft that disappear while in flight in a ce

ID: 3242181 • Letter: S

Question

Seventy-six percent of the light aircraft that disappear while in flight in a certain country are subsequently discovered. Of the aircraft that are discovered, 64% have an emergency locator, whereas 88% of the aircraft not discovered do not have such a locator. Suppose a light aircraft has disappeared. (Round your answers to three decimal places.) (a) If it has an emergency locator, what is the probability that it will not be discovered? (b) If it does not have an emergency locator, what is the probability that it will be discovered?

Explanation / Answer

Let Probability of Discovery = P(D)

Probability of not being discovered = P(D')

Probability of having an Emergency Locator = P(L) and

Probability of not having an emergency Locator = P(L')

Given:

(1) P(D) = 0.76, Therefore P(D') = 1-0.76 = 0.24

(2) P(L/D) i.e Probability of having a locator given that the aircraft was discovered = 0.64

Therefore P(L'/D) i.e Probability of not having a locator given that the aircraft was discovered = 1 - 0.64 = 0.36

(3) P(L'/D') Probability of not having a locator given that the aircraft was not discovered = 0.88

Therefore P(L/D') i.e Probability of having a locator given that the aircraft was not discovered = 1 - 0.88 = 0.12

We need to find the probability that the aircraft had a locator. By the probability assembly rule

P(L) = P(L/D) * P(D) + P(L/D') * P(D') = 0.64*0.76 + 0.12 * 0.24 = 0.5152

Therefore P(L') (Probability the aircraft does not have a locator) = 1-0.5152 = 0.4848

(a) If it has an emergency locator, what is the probability that it will not be discovered i.e P(D'/L).

By Bayes Theorem P(D'/L) = P(L/D') * P(D')/P(L) = (0.12 * 0.24)/0.5152 = 0.056

(b) If it has does not have an emergency locator, what is the probability that it will be discovered i.e P(D/L').

By Bayes Theorem P(D/L') = P(L'/D) * P(D)/P(L') = (0.36 * 0.76)/0.4848 = 0.564

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