Seventy percent of the light aircraft that disappear while in flight in a certain country are subsequently discovered. Of the aircraft that

Question

Seventy percent of the 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 such a locator. Suppose a light aircraft has disappeared.
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?

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Peyton 6 days 2021-10-11T16:36:03+00:00 2 Answers 0

Answers ( )

    0
    2021-10-11T16:37:36+00:00

    Answer:

    Step-by-step explanation:

    Given info

    The information is based on the disappearance of the of the light aircraft in the flight in the in a certain country . 70% of the aircraft that disapear while in the flight were discovered . The probability that there is an emergency locator        

    in the aircraft given that it is discovered is 60% and the probability that the aircraft that has not discovered have an emergenncy locator is 90%.

    a) To determine probiblity it wount be discoverd

    Calculation

    P(the aircraft is discovered and has a locator) = P(discovered) X P(locator/discovered)

                                                 = 0.7 x 0.6

                                                 = 0.42

    The probability of the flight discovered and has locator is obatained as given below:

    P(The aircraft is discoverd and has no locator) = P(discovered)  x P(no locator/discovered)

                                                 = 0.7 x 0.4

                                                 =0.28

    The probability of the flight not discovered and has a locator is obatained below.

    P(The aircraft is not discovered and has a locator) = P(not discovered) x P (locator  | not discorverd)

                                                = 1-0

    0
    2021-10-11T16:38:01+00:00

    Answer:

    a. 0.067

    b. 0.509

    Step-by-step explanation:

    We define the events as

    L=Locator\\D=Discovered

    L'=No Locator

    D'=Not Discovered

    The known probabilities are,

    P(D)=0.7\\P(D')=1-P(D)=1-0.7=0.3\\P(L|D)=0.6\\P(L'|D)=1-P(L|D)=1-0.6=0.4\\P(L'|D')=0.9\\P(L|D')=1-P(L'|D')=1-0.9=0.1

    Also,

    P(D\cap L)=P(L|D)P(D)=(0.7)(0.6)=0.42\\P(D\cap L')=P(L'|D)P(D)=(0.4)(0.7)=0.28\\P(D'\cap L)=P(L|D')P(D')=(0.1)(0.3)=0.03\\P(D'\cup L')=P(L'|D')P(D')=(0.9)(0.3)=0.27

    And,

    P(L)=P(D\cap L)+P(D'\cap L)=0.42+0.03=0.45\\P(L')=1-P(L)=1-0.45=0.55

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

    P(D'|L)= \frac{P(D'\cap L)}{P(L)} =\frac{0.03}{0.45} =0.067

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

    P(D|L')=\frac{P(D\cap L')}{P(L')} =\frac{0.03}{0.55}=0.509

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