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Seventy percent of the light aircraft that disappear while in flight in a certain countryare subsequently discovered. Of the aircraft that are discovered,60%have an emer-gency locator, whereas90%of the aircraft not discovered do not have such a locator.Suppose a light aircraft has disappeared.(a)[5 pts]If it has an emergency locator, what is the probability that it will not bediscovered

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Step-by-step explanation: If we denote the percentage of light aircraft subsequently discovered, and the percentage of which is 70, by the mark P(discovered) = 0.7. Further, if the percentage of those light aircraft that are discovered and have a locator is 60, that is, P(discovered / locator) = 0.6, then the percentage of light aircraft that are not discovered will be P(discovered ' ) = 1 - 0.7 = 0.3. As is the percentage of undiscovered aircraft that do not have a locator P(discovered ' / locator ') = 0.9.

What we need to do is determine the probability that the missing aircraft will not be found and has a locator. So, P(discovered ' / locator) = ?

We will do this as follows. First of all the total percentage of aircraft discovered and having a locator is 0.7 x 0.6 = 0.42. Further, the total percentage of aircraft not discovered and lacking a locator is 0.9 x 0.3 = 0.27. Then the total percentage of aircraft discovered and lacking a locator will be 0.7 - 0.42 = 0.28. Like the total percentage of aircraft not discovered that have a locator is 0.3 - 0.27 = 0.03.

To make all this more obvious, we will show in the table

Locator Locator '

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Discovered 0.42 = 0.7 x 0.6 0.28 0.70

Discovered ' 0.03 0.27 = 0.9 x 0.3 0.30

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0.45 0.55 1.00

Then what we're looking for is the probability that the aircraft will not be discovered and has a locator will be:

P = (discovered ' / locator) = P (discovered ' / locator) / P (locator), that is

P = (discovered ' / locator) = 0.03 / 0.45 = 0.06667