MATH SOLVE

3 months ago

Q:
# Seventy percent of light aircraft that disappear while in flight in a certain country are subsequently discovered. Of the aircraft that are discovered, 60% have an emergency locator, whereas 90% of the aircraft not discovered do not have an emergency locator. Suppose that a light aircraft has disappeared.If it has an emergency locator, what is the probability that it will not be discovered?

Accepted Solution

A:

Answer:Figure out the various probabilities first, that will make the rest of the questions easier:P(discovered) = .7P(not discovered) = 1 - .7 = .3P(locator|discovered) = .6P(no locator|discovered) = 1 - .6 = .4P(locator|not discovered) = 1 - .9 = .1P(no locator|not discovered) = .9P(discovered and locator) = .7 * .6 = .42P(discovered and no locator) = .7 * .4 = .28P(not discovered and locator) = .3 * .1 = .03P(not discovered and no locator) = .3 * .9 = .27a) The total probability that an aircraft has a locator is .42 + .03 = .45. So the probability it will not be discovered, given it has a locator, is .03/.45 = .067b) The total probability that an aircraft does not have a locator is .28 + .27 = .55. So the probability it will be discovered, given it does not have a locator, is .28/.55 = .509c) Probability that 7 are discovered = C(10,7) * P(discovered|locator)^7 * P(not discovered|locator)^3We already figured out P(not discovered|locator) = .067, so P(discovered|locator) = 1-.067 = .933. C(10,7) = 10*9*8, so we can compute total probability: 10*9*8 * .933^7 * .067^3 = .133Step-by-step explanation: