Seventy-one percent of the light aircraft that disappear while in flight in a certain country are subsequently discovered. Of the aircraft that are discovered, 69% 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.)

The question is not complete so I am writing the complete question below:

Seventy-one percent of the light aircraft that disappear while in flight in a certain country are subsequently discovered. Of the aircraft that are discovered, 69% 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?

a) The probability that the aircraft will not be discovered, given that it has an emergency locator is 0.066.

b) The probability that the aircraft will be discovered, given that it does not have an emergency locator is 0.463.

Step-by-step explanation:

71% aircraft which disappear while in flight are discovered. Let D denote the probability of discovering a lost aircraft. So, P(D) = 0.71.

Of the aircraft discoverd, 69% have an emergency locator. Let EL denote emergency locator. P(EL|D) = 0.69

88% of aircraft which are not discovered (ND) do not have such a locator (NEL). So, P(NEL|ND) = 0.88

a) We need the probability that the aircraft will not be discovered, given that it has an emergency locator. So we need P(ND|EL).

P(ND|EL) = P(ND ∩ EL)/P(EL)

                = P(EL|ND)P(ND)/P(EL)

                = [P(EL|ND)*P(ND)]/[P(EL|D)*P(D) + P(EL|ND)*P(ND)]

We need P(EL | ND) and P(ND). So,

P(EL | ND) = 1 - P(NEL | ND)

                 = 1 - 0.88

P(EL | ND) = 0.12

P(ND) = 1 - P(D)

          = 1 - 0.71

P(ND) = 0.29

P(ND|EL) = [P(EL|ND)*P(ND)]/[P(EL|D)*P(D) + P(EL|ND)*P(ND)]

                = (0.12*0.29)/[(0.69*0.71) + (0.12*0.29)]

                = 0.0348/0.5249

P(ND|EL) = 0.066

b) Here we need to compute the probability that the aircraft will be discovered, given that it does not have an emergency locator i.e. P(D | NEL).

P(D | NEL) = [P(NEL | D)*P(D)]/[P(NEL | ND)*P(ND) + P(NEL | D)*P(D)]

We need P(NEL | D) So,

P(NEL | D) = 1 - P(EL | D)

                = 1 - 0.69

P(NEL | D) = 0.31

P(D | NEL) = [P(NEL | D)*P(D)]/[P(NEL | ND)*P(ND) + P(NEL | D)*P(D)]

                 =(0.31*0.71)/[(0.88*0.29) + (0.31*0.71)]

                = 0.2201/0.4753

P(D | NEL) = 0.463

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation: