Question

88% of the light aircraft that disappear while in flight in a certain country are subsequently discovered. Of the aircraft that are discovered, 72% have an emergency locator, whereas 86% 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?C) If we consider 10 light aircraft that disappeared in flight with an emergency recorder, what isthe probability that 7 of them are discovered?

Answer 1

Answer:

a) $P(D'/L)=0.03$

b) $P(D/L')=0.7$

Step-by-step explanation:

From the question we are told that

Probability of discovered $P(D)=0.88$

Probability of aircraft with locator $P(L/D)=0.72$

Probability of lost aircraft without locator $P(L'/D')=0.86$

Generally the probability that aircraft are not discovered is mathematically given as

$P(D')=1-0.88$

$P(D')=0.12$

Generally the probability that aircraft with locator are discovered is mathematically given as

$P(L'/D)=1-0.72$

$P(L'/D)=0.28$

Generally the  Probability of lost aircraft without locator mathematically given as

$P(L/D')=1-0.86$

$P(L/D')=0.14$

Generally the  Probability of having a locator mathematically given as

$P(L)=P(L/D)*P(D)+P(L/D')*P(D')$

$P(L)=0.72*0.88+0.14 *0.12$

$P(L)=0.6504$

a)Generally the  Probability of getting lost even with locator mathematically given as

$P(D'/L)=\frac{P(L/D')(P(D')}{P(L)}$

$P(D'/L)=\frac{0.14*0.12}{0.6504}$

$P(D'/L)=0.0258302583$

$P(D'/L)=0.03$

b)Generally the  Probability of getting lost even with locator mathematically given as

$P(D/L')=\frac{P(L'/D)(P(D)}{P(L')}$

$P(D/L')=\frac{0.28*0.88}{1-0.6504}$

$P(D/L')=0.704805492$

$P(D/L')=0.7$