Question

A plane is missing and is presumed to have equal probabilityof going down in any of three regions. If a plane is actually downin region i, let 1-αi denote the probability thatthe plane will be found upon search of the ith region, i=1,2,3.What is the conditional probability that the plane is in
a) region 1, given that the search of region 1 wasunsuccessful
b) region 2, given that the search of region 1 wasunsuccessful
c) region 3, given that the search of region 2 wasunsuccessful

Answer 1

oh i see it says the probability is \(1-\alpha_i\) is that right?

Answer 2

Yes that is right

Answer 3

I know I have to use the Bayes Theorem, but I'm not understanding it at all when i try to apply it.

Answer 4

If I can get through A I think I'll understand the rest

Answer 5

ok now i have to think

Answer 6

dont think you need baye's for the first one
if we put A is the event the plane went down in region 1, and B is the event the search was unsuccessful, then you are looking for
\[P(A|B)=\frac{P(A\cap B)}{P(B)}\]
the denominator is \(\alpha_1\) and the numerator is \(\frac{1}{3}\times (\alpha_1)\)

Answer 7

maybe i am messing this up, let me think for another moment

Answer 8

except isn't \[(1-\alpha _{i})\] the probability of finding the plane GIVEN that it went down in the region? (P(B|A))?

Answer 9

give me a minute, if you have it

Answer 10

yeah to your answer, but the given part is "the search is unsuccessful"

Answer 11

the language is confusing, and also the fact that the condition is not "the search is unsuccessful' but "the search in region 1 is unsuccessful'

Answer 12

I think the problem is that I don't know what P(B) is, which is where we'd apply Bayes Theorem/ Law of Total Probability.

Answer 13

ok i think i got it at least for the first one

Answer 14

lets put A as the even the plane went down in region 1, B as the event the search is unsuccessful in region 1

Answer 15

then \[P(A|B)\] is what you are looking for

Answer 16

bayes gives
\[P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^c)P(A^c)}\]

Answer 17

if the plane is down in region 1, then the probability it is found is \(1-\alpha_1\) so the probability it is not found is \(\alpha_1\) i.e. \[P(B|A)=\alpha_1\]

Answer 18

and of course \(P(A)=\frac{1}{3}\)

Answer 19

\[P(B|A^c)\] is the probability that the search is region 1 is unsuccessful given that the plane is not in region 1, and that is 1

Answer 20

and \(P(A^c)=\frac{2}{3}\)

Answer 21

okay that all makes sense so far

Answer 22

we put it together and get
\[\frac{\alpha_1\times \frac{1}{3}}{\alpha_1\times \frac{1}{3}+\frac{2}{3}}\] multiply top and bottom by 3 and get
\[\frac{\alpha_1}{\alpha_1+2}\]

Answer 23

i would not bet $20 that this is right, but i would bet $12

Answer 24

Well this makes sense with my (very limited) understanding of the material

Answer 25

what about part B then?

Answer 26

ok i think this one is a bit harder, but doable

Answer 27

my professor did say something about it being a bit of a trick question

Answer 28

this time lets put A and the plane is in region 2, and B as the search in region 1 is unsuccessful

Answer 29

we want
\[P(A|B)\] and again it is
\[\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^c)P(A^c)}\]

Answer 30

\[P(B|A)\] is the probability that the search in region 1 is unsuccessful given that the plane is in region 2, and that is 1

Answer 31

the part that is not obvious is \(P(B|A^c)\) and for that we need two pieces. this is the probability that the search in region 1 is unsuccessful given that the plane is not in region 2

Answer 32

for our denominator what we really need is the probability of B, i.e. we want the probability that the search in region 1 is unsuccessful
so we need to consider all three possibilities
the plane is in region 1, the probability it is unsuccessful is \(\alpha_1\)
the plane is in region 2, the probability is 1
the plane is in region 3, the probability is also 1
multiplying that by the probability that the plane is in each region give
\[P(B)=\frac{1}{3}\alpha_1+\frac{1}{3}+\frac{1}{3}\]

Answer 33

you will notice that this is exactly the same denominator we had before, i.e. we already computed \(P(B)\) so this was kind of a waste of time

Answer 34

so for number 2, the answer is
\[\frac{\frac{1}{3}}{\frac{1}{3}\alpha_1+\frac{2}{3}}\]
\[=\frac{1}{\alpha_a+2}\]

Answer 35

why do we need P(B)?

Answer 36

because bayes formula is really nothing more than a method of computing
\[P(A|B)=\frac{P(A\cap B)}{P(B)}\]

Answer 37

the point is that sometimes you do not know \(P(B)\) directly so you have to find it somehow

Answer 38

what we did in problem 1 was find the same thing, but we didn't know \(P(B)\) directly , not did we know \(P(A\cap B)\) but it is always the case that \(P(A\cap B)=P(B|A)P(A)\)

Answer 39

in other words bayes formula is just a trick of algebra, but it allows you do change the conditioning from A given B to B given A
if you know the second, then you can compute the first

Answer 40

by the way, now that we know if B is the probability that the search in region 1 is unsuccessful, then it is pretty clear that the probability that the search in region 2 is unsuccessful is
\[\frac{1}{3}\alpha_2+\frac{2}{3}\]

Answer 41

btw we should have computed these first, it would have made life easier
let me know if i have completely lost you

Answer 42

let me think for a moment

Answer 43

ok
we could have even used a baby tree diagram to do this

Answer 44

okay I think I got it. So this time instead of changing the bottom P(B) we left it as P(B) since we already knew it. and only changed the top, which allowed us to get the probability.

Answer 45

right

Answer 46

You sir ( or madam) are a life saver.

Answer 47

and is we put B is the probability the search is unsuccessful in region 1, how to we find \(P(B)\)? the plane had to go down somewhere
in 1, 2 or 3
so we do the following
\[P(B)=P(B\cap A_1)+P(B\cap A_2)+P(B\cap A_3)\] where \(A_i\) is the event the plane goes down in region \(i\)
and these are computed via
\[P(B\cap A_i)=P(B|A_i)P(A_i)\]

Answer 48

you good for the last one?

Answer 49

Don't need to do the last one. Only A and B were assigned. Prof said C was the same as B. Which makes sense. The math would look the same.

Answer 50

ok sorry it took me so long
i will use the excuse that the wording was confusing (which it is) but once you get the hang of bayes formula it is not so bad
just keep in mind what you are really looking for

Answer 51

oh yeah 3 looks identical to 2

Answer 52

It's alright it took me a long time to even start asking on here.

Answer 53

lol come back any time

Answer 54

Thank you!

Answer 55

yw