Bayes theorem and conditional probability

Question

I have a problem like this:

Seventy-eight 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 85% 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?

I know D = discovered = .78, and so D-complement = not discovered = .22

Thanks

Answer 1

If we define:

$A$ - aircraft that are discovered.
$B$ - aircraft that have an emergency locator.

We know from the question that: $P(A) = 0.78$, $P(B\mid A) = 0.6$ and $P(B_c\mid A_c) = 0.85$. so using Bayes theorem we can deduce that: $$ \frac{P(B\cap A)}{P(A)} = 0.6 \Longrightarrow P(B\cap A) = 0.468$$ $$ \frac{P(B_c\cap A_c)}{P(A_c)} = 0.85 => P(B_c\cap A_c) = 0.187$$

I hope you can take it from here.

Answer 2

Use the law of total probability, where$$ P(B) = P(B|A)P(A) + P(B|A^c)P(A^c). $$