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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

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!

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  • $\begingroup$ Why do you have the second $0.7$ factor in the numerator in the last formula? You have $P(E \mid D) = 0.6, P(D) = 0.7$, so it should be $0.7 \cdot 0.6$ and not $0.7 \cdot 0.6 \cdot 0.7$. Without it the answer is indeed $0.93$. $\endgroup$ Commented May 4, 2016 at 22:11
  • $\begingroup$ I think I got confused with what I was doing in the denominator because for finding the probability that it does have an emergency locator, we multiply 0.7 to 0.6. I guess I don't really get the difference between P(E|D) and the probability that it does have an emergency locator? Because then it that case shouldn't my denominator also just be 0.6 + (0.3)(0.1)? $\endgroup$
    – NookLines
    Commented May 4, 2016 at 22:17

3 Answers 3

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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.

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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 $$P(D \mid E) = \frac {P( D \cap E)}{P(E)}= \frac {(.7)(.6)}{(.7)(.6)+(.3)(.1)}$$

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$P(D \mid E)$

$= P(D \cap E)/P(E)$

$= P(D \cap E)/P((E \cap D) \cup (E \cap \overline{D}))$

$= P(D \cap E)/\{P(E \cap D) + P(E \cap \overline{D})\}$

$= \{P(D) \cdot P(E \mid D)\}/\{(P(D) \cdot P(E \mid D)) + P(\overline{D}) \cdot P(E \mid \overline{D})\}$

$= (0.70 \cdot 0.60)/((0.70 \cdot 0.60) + (0.30 \cdot 0.10))$

$= 0.93$

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