Conditional probability that aircraft will not be discovered

Question

Problem

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. If it has an emergency locator, what is the probability that it will not be discovered?

My Attempt

Let two events D, E denote

D : discovered, E : the aircraft has an emergency locator

Then we know $P(D)=0.7, P(D')=0.3$ and $P(E|D)=0.6, P(E'|D')=0.9$

What I want to know : $P(D'|E) = \frac{P(D')P(E|D')}{P(E)}$

To find $P(E)$, I get $P(E)=P(D)P(E|D)+P(D')P(E|D')$ from total probability theorem.

But how can I know $P(E|D')$ to find $P(E)$? I think I used all conditions but I couldn't find $P(E)$.

I know there is a similar question in MSE but that post does not solve my question so I upload this one.

Thanks for your help.

Answer 1

The flight data gives us this table:

P(row|column)	D	D'
E	0.6	0.1
E'	0.4	0.9

We need to adjust the columns to allow for the distribution of $D$:

P(row|column)P(column)	D	D'
E	0.42	0.03
E'	0.28	0.27

Now the table sums to $1$, and we can read the probability as:

$$\frac{0.03}{0.03+0.42} = \frac{1}{15}$$

Answer 2

But how can I know $P(E|D′)$ to find $P(E)$?

Use the rule for the Probability of Complements.

$$\begin{align}\mathsf P(E\mid D')&=1-\mathsf P(E'\mid D')\\&=100\%-90\%\\&=10\%\end{align}$$

Ie: Since $90\%$ of those not discovered do not have a locator, therefore $10\%$ of those not discovered do have a locator.