Answer:
a. 0.067
b. 0.509
Step-by-step explanation:
We define the events as
[tex]L=Locator\\D=Discovered[/tex]
[tex]L'=No[/tex] [tex]Locator[/tex]
[tex]D'=Not[/tex] [tex]Discovered[/tex]
The known probabilities are,
[tex]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[/tex]
Also,
[tex]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[/tex]
And,
[tex]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[/tex]
a. The probability that it will not be discovered given that it has an emergency locator is,
[tex]P(D'|L)= \frac{P(D'\cap L)}{P(L)} =\frac{0.03}{0.45} =0.067[/tex]
b. The probability that it will be discovered given that it does not have an emergency locator is,
[tex]P(D|L')=\frac{P(D\cap L')}{P(L')} =\frac{0.03}{0.55}=0.509[/tex]
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
