If it has an emergency locator, what is the probability that it will be discovered? Okay so here's the question 

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 that a
  light aircraft has disappeared. If it has an emergency locator, what
  is the probability that it will be discovered?

Anndd here's my answer

The answer to this question was, however, given as 93%. I don't understand how they got that answer and I was pretty confident in my solution. Can someone either tell me the answer given in the text is incorrect or what's wrong with my solution?
Thanks so much!
 A: Let's use a frequency table of a deterministic cohort of light aircraft that disappear.  Suppose there are $N = 100$ such aircraft.  As $70\%$ of these are discovered, this means $70$ aircraft belong in the group $D$, indicating that they are subsequently discovered, and $30$ aircraft belong in the group $\bar D$, indicating they are not discovered.
Among the $70$ discovered aircraft, $60\%$ have an emergency locator, so $$D \cap L = (70)(0.6) = 42$$ where $L$ represents the event that an aircraft has an emergency locator.  Thus there are $$D \cap \bar L = (70)(0.4) = 28$$ that were discovered but had no emergency locator.
Similarly, among the $30$ undiscovered aircraft, $$\bar D \cap \bar L = (30)(0.9) = 27$$ had no emergency locator; and $$\bar D \cap L = (30)(0.1) = 3$$ had an emergency locator.
We summarize the above in the following table:
$$\begin{array}{c|c|c|c} & L & \bar L & \\ \hline D & 42 & 28 & 70 \\ \hline \bar D & 3 & 27 & 30 \\ \hline & 45 & 55 & 100 \end{array}$$ 
Therefore, given that an aircraft has an emergency locator--that is to say, is one of the $45$ aircraft in column $L$--the number of discovered aircraft is $42$, thus the proportion of such aircraft is $42/45 \approx 0.933$.  

In the language of probability, where $D$ and $L$ are events, we are given $$\Pr[D] = 0.7, \quad \Pr[L \mid D] = 0.6, \quad \Pr[\bar L \mid \bar D] = 0.9,$$ and we wish to compute $$\Pr[D \mid L] = \frac{\Pr[L \mid D]\Pr[D]}{\Pr[L]}.$$  Then $$\Pr[L] = \Pr[L \mid D]\Pr[D] + \Pr[L \mid \bar D]\Pr[\bar D] = (0.6)(0.7) + (1 - 0.9)(1 - 0.7) = 0.45,$$ and $$\Pr[D \mid L] = \frac{(0.6)(0.7)}{0.45} = \frac{42}{45} \approx 0.933,$$ as we found using the deterministic cohort above.
A: Intuitively, you can think of $P(E \mid D)$ in the numerator as "given that you are already inside the $70 \%$ that has been discovered, what is the probability that inside that $70 \% $ it does have an emergency locator?" so you don't take $.7$ into account in the calculation, you only take $.6$.
You can also use the formula 
$$P(D \mid E) = \frac {P( D \cap E)}{P(E)}= \frac {(.7)(.6)}{(.7)(.6)+(.3)(.1)}$$
A: $P(D \mid E)$ 
$= P(D \cap E)/P(E)$
$= P(D \cap E)/P((E \cap D) \cup (E \cap \overline{D}))$
$= P(D \cap E)/\{P(E \cap D) + P(E \cap \overline{D})\}$
$= \{P(D) \cdot P(E \mid D)\}/\{(P(D) \cdot P(E \mid D)) + P(\overline{D}) \cdot P(E \mid \overline{D})\}$
$= (0.70 \cdot 0.60)/((0.70 \cdot 0.60) + (0.30 \cdot 0.10))$
$= 0.93$
