Bayes theorem and conditional probability 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

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.