# Conditional probability plane problem

I was presented with this problem and am not sure where to take it.

A plane is missing and is presume to have equal probability of going down in any of 3 regions. If a plane is actually down in region $i$, let $1-\alpha$ denote the probability that the plane will be found upon a search of the $ith$ region, $i=1,2,3.$ What is the conditional probability that the plane is in:

A) region 1, given the search of region 1 is unsuccessful

B) region 2, given the search of region 1 is unsuccessful

C) region 3, given the search of region 1 is unsuccessful

My idea:

So we know $P(1)=P(2)=P(3) = \frac{1}{3}$. (Probability it is that region)

We also know that $P(F_1\mid 1)=1-\alpha_1$, and similarly for the other regions.

Then for A) want $P(1\mid F_1^c)$

$$P(1\mid F_1^c) = \frac{P(F_1^c\mid 1)P(1)}{P(F_1^c\mid 1)P(1)+P(F_2^c\mid 1)P(2)+P(F_3^c\mid 1)P(3)} = \frac{\alpha_1}{\alpha_1+\alpha_2+\alpha_3}$$

However I feel this is wrong but cannot find why.

For B) and C) I am at a loss. I know we want $P(2|F_1^c)$ and $P(2\mid F_1^c)$ but I can't see how we could use the same formula to get that.

Any help would be much appreciated.

• In your denominator, it seems that the last two terms in the sum should each be 1/3 since it is certain that the search of region 1 fails if the plane is in 2 or 3 (i.e. the product is 1 x (1/3) = 1/3 for each of those terms). For (B) and (C) the denominator is the same but the numerator is the second or third term in the denominator.
– Ned
Commented Sep 20, 2014 at 17:34

$(A)$ You seem to have the Bayes formula wrong.

\begin{eqnarray*} P(1 \mid F_1^c) &=& \dfrac{P(F_1^c \mid 1)P(1)}{P(F_1^c \mid 1)P(1)+P(F_1^c \mid 2)P(2)+P(F_1^c \mid 3)P(3)} = \frac{\alpha_1}{\alpha_1+2} \\ && \\ && \mbox{since } P(F_1^c \mid 2) = P(F_1^c \mid 3) = 1 \end{eqnarray*}

$(B)$

\begin{eqnarray*} P(2 \mid F_1^c) = \dfrac{P(F_1^c \mid 2)P(2)}{P(F_1^c \mid 1)P(1)+P(F_1^c \mid 2)P(2)+P(F_1^c \mid 3)P(3)} = \frac{1}{\alpha_1+2} \end{eqnarray*}

$(C)$

\begin{eqnarray*} P(3 \mid F_1^c) = \dfrac{P(F_1^c \mid 3)P(3)}{P(F_1^c \mid 1)P(1)+P(F_1^c \mid 2)P(2)+P(F_1^c \mid 3)P(3)} = \frac{1}{\alpha_1+2} \end{eqnarray*}