What is the conditional probability (up to2 decimals) that the plane is in region 1?

Question

A plane is missing, and it is presumed that it was equally likely to have gone down in any of 3 possible regions. If the plane is in a given region, the conditional probability that the plane will be found upon a search is 0.6 for region 1, 0.7 for region 2 and 0.8 for region 3. Given that a search in region 1 is unsuccessful, what is the conditional probability (up to 2 decimals) that the plane is in region 1?

I think that I should use the Bayer's Law, the result should be 0.17. Can someone explain to me how can I compute the two probabilities that I need, if my method is correct? Thanks in advance! P(1)=P(2)=P(3)=1/3
P(F∣1)=0.6
P(F∣2)=0.7
P(F∣3)=0.8
P(1|$F_1^c$)=?
P($F_1^c$|1) = P($F^c$|1)
P($F^c$∣1)=0.4 (1-0.6)
P($F^c$∣2)=0.3 (1-0.7)
P($F^c$∣3)=0.2 (1-0.8)
P(1∣$F_1^c$)=$\frac{P(F_1^c∣1)P(1)}{P(F_1^c∣1)P(1)+P(F_1^c∣2)P(2)+P(F_1^c∣3)P(3)}$=$\frac{0.4/3}{0.4/3+P(F_1^c∣2)/3+P(F_1^c∣3)/3}$=??

I don't know how to compute $P(F_1^c∣2)$ and $P(F_1^c∣3)$

Answer 1

A search in region $1$ is unsuccessful. No information about searches in the other regions is provided. Your notation doesn't distinguish between a search failure in region $1$ and an overall search failure. That's OK in the case $\mathsf P(\overline F\mid 1)=\mathsf P(\overline F_1\mid 1)$, since conditional on the plane being in region $1$, the search fails exactly if it fails in region $1$; but the same isn't true in the case $\mathsf P(1\mid\overline F)\neq\mathsf P(1\mid\overline F_1)$.

Update in response to the edit in the question:

$P(\overline{F_1}\mid2)$ is the probability that a search in region $1$ fails, given that the plane is in region $2$. That probability is $1$.