# Bayes rule word problem help.

On my assignment I have been given the word problem

Seventy percent of the aircraft that disappear in flight the Bermuda Triangle are recovered (P(r) = 0.7). Sixty percent of the recovered aircraft have an emergency locator (P(e|r) = 0.60). Unfortunately, 90% of the aircraft not recovered do not have such a locator. Suppose that an aircraft with a locator has disappeared. What is the probability that it will be recovered (P(r|e))?

From this problem I have found that the following are given

$$P(R) = 0.7$$ $$P(E|R) = 0.6$$ $$P( \neg E| \neg R) = 0.9$$

Using Bayes Rules I can solve for $P(R|E)$ the following way

$$P(R|E)= \dfrac{P(E|R)P(R)}{P(E)}$$ $$P(R|E)= \dfrac{0.6*0.7}{P(E)} = \dfrac{0.42}{P(E)}$$

To find that actual probability $P(E)$, I figured that I might need to work with the last given probability from the problem and so I devised

$$P(\neg R|\neg E) = \dfrac{P(\neg E|\neg R)P(\neg R)}{P(\neg E)}$$ $$P(\neg R|\neg E) = \dfrac{0.9*0.3}{P(\neg E)} = \dfrac{0.27}{P(\neg E)}$$ $$P(\neg R|\neg E) * P(\neg E) = 0.27$$ $$P(\neg E) = \dfrac{0.27}{P(\neg R|\neg E)}$$ $$Using \space P(E) = 1 - P(\neg E)$$ $$P(E) = 1 - \dfrac{0.27}{P(\neg R|\neg E)}$$

Stealing from the initial application of Bayes Rule

$$P(R|E) * P(E)= 0.42$$ $$P(E) = \dfrac{0.42}{P(R|E)}$$

Combining the last two

$$1 - \dfrac{0.27}{P(\neg R|\neg E)} = \dfrac{0.42}{P(R|E)}$$

And, uhh, well, I'm not sure where I can go from here. I know that I need to determine $P(E)$ somehow to answer the problem, but I don't see how I can actually resolve out $P(E)$ or something that I need to resolve it.

Did I possible read the problem wrong?

Should the third sentence equate to $P(E) = 0.9$? I feel like that is incorrect though. Any suggestions..?

Begin, as always, with the Law of Total Probability, then invoke complementation to express it with the three known measures.

\begin{align}\def\P{\mathop{\mathbb P}}\P(E) &= \P(E\mid R)\P(R)+\P(E\mid \neg R)\P(\neg R) \\[1ex] &= \P(E\mid R)\P(R)+\big(1-\P(\neg E\mid \neg R)\big)\big(1-\P(R)\big) \\[1ex] &= 0.6\cdot 0.7+(1-0.9)(1-0.7)\\[1ex] &= 0.45\end{align}

That is all.

• Assuming there was another variable in the domain, would the first line be extended for it. For example is some variable $X$ exists would we then add $P(E|X)P(X)+P(E|\neg X)P(\neg X)$? Would we then need a clause to relate the "total probability" between $X$ and $R$ too? // This seems oddly similar to the Chain Rule?... Mar 3, 2017 at 1:28
• No, the important detail is that $R$ and $\neg R$ are mutually exclusive and exhaustive; that is their intersection is empty and their union is the entire sample space. They form a partition. The full LoTP is usable with any partition of the sample space. If $(B_k)_{k=1}^n$ is a sequence of $n$ mutually exclusive and exhaustive events,$^\dagger$ then for any event $A$ we have: $$\P(A)~=~\sum_{k=1}^n \P(A\mid B_k)\P(B_k)$$ Note: $\dagger$ ... it is actually sufficient that $A\subseteq \bigcup_{k=1}^n B_k$ . Mar 3, 2017 at 2:53
• So the partition formed from $X,R$ would be $\{X\cap R , X\cap\neg R , \neg X\cap R , \neg X\cap\neg R\}$ and so:\begin{align}\P(E) ~&=~ {{\P(E\mid X, R)\P(X,R)}+{\P(E\mid X,\neg R)\P(X,\neg R)}+{\P(E\mid \neg X, R)\P(\neg X, R)}+{\P(E\mid \neg X, \neg R)\P(\neg X, \neg R)}}\end{align} Mar 3, 2017 at 3:01

Let

• $a$ be the fraction of disappearing planes that are recovered and have a locator.
• $b$ be the fraction of disappearing planes that are recovered and have no locator.
• $c$ be the fraction of disappearing planes that are not recovered and have a locator.
• $d$ be the fraction of disappearing planes that are not recovered and have no locator.

Your first three statements then read $$a+b = \frac73(c+d)\\ a=\frac32 b \\ d=9c$$ and the probelm is asking you to find the probability that a disappeared aircraft with a locator ($a$ or $c$) will be recovered ($a$), that is, to find $$\frac{a}{a+c}$$ Trot out the algebra:

Using $d=9c$ we have $$a+b= 70c$$ Using $b=\frac23 a$ we have $$a+\frac23 a = 70c\\ \frac53 a = 70c \\ a = 42c \\\frac{a}{a+c} = \color{red}{\frac{42}{43}}$$

You could do this problem by careful application of Bayes' theorem; it will give the same manipulations and the same result. But if you are getting confused, that may be a symptom that you are trying to use a mis-matched tool for the problem.

BTW, since $a+b+c+d=1$, you can go on to solve for all the AND probabilities: $$42c+28c+c+9c = 1 \\ a = \frac{42}{80}\\ b = \frac{28}{80} \\ c = \frac{1}{80} \\ d=\frac{9}{80}$$

And then your $P(E)$ can be seen to be $\frac{43}{80}$.