Comp Sci What is the probability that there is a burglary given John and Mary calls?

[shivajikobardan]

Homework Statement

What is the probability that there is a burglary given John and Mary calls?

Relevant Equations

bayes theorem maybe.

this is the question



Here is a tutorial video but his steps are very confusing to me. I personally know bayes theorem and have already studied probability and got good marks in it(It may not be a metric for being quality in it given that it is nepal we are talking about.)
https://courses.engr.illinois.edu/ece448/sp2020/slides/lec15.pdf
here is the slide I'm referring to. The answer seems 0.72 or 0.28 according to video.

My attempt-:
I try finding P(B/(J,M)) but I don't get a way to find it. Burglary is independent of anything else. IDK how to find this value.

 

[andrewkirk]

Try breaking it into parts, eg:

P(B | J & M) = P(B | A) x P(A | J & M)
Then use Bayes' Theorem to work out P(B | A) and P(A | J & M).

Note that the question gives you P(A | B & E), P(A | B & ~E), P(~B & E), P(~B & ~E)
[left-side table, ~ means NOT]
plus P(B), P(E) and since you said they're independent, you have P(B & E) = P(B)P(E)
[top two values]
plus P(J | A), P(J | ~A), P(M | A), P(M | ~A)
[bottom of slide]

You are going to need the value for P(J & M) which means you need to know whether J and M are independent ( in which case we'll have P(J & M) = P(J) P(M) ). Do they tell you that?

You are also going to need the value for P(J & M | A) which means you need to know whether J|A and M|A are independent ( in which case we'll have P(J & M | A) = P(J | A) P(M | A) ). Do they tell you that?

 

[shivajikobardan]

Try breaking it into parts, eg:

P(B | J & M) = P(B | A) x P(A | J & M)
Then use Bayes' Theorem to work out P(B | A) and P(A | J & M).

Note that the question gives you P(A | B & E), P(A | B & ~E), P(~B & E), P(~B & ~E)
[left-side table, ~ means NOT]
plus P(B), P(E) and since you said they're independent, you have P(B & E) = P(B)P(E)
[top two values]
plus P(J | A), P(J | ~A), P(M | A), P(M | ~A)
[bottom of slide]

..

You are going to need the value for P(J & M) which means you need to know whether J and M are independent ( in which case we'll have P(J & M) = P(J) P(M) ). Do they tell you that?

Yes J and M are independent to each other.

You are also going to need the value for P(J & M | A) which means you need to know whether J|A and M|A are independent ( in which case we'll have P(J & M | A) = P(J | A) P(M | A) ). Do they tell you that?

J is dependent on A and so is M.

 

[andrewkirk]

J is dependent on A and so is M.

Yes we know that, but that's not enough, as it still leaves different possibilities. We need to know P(J | M&A) and P(M | J & A).
Consider the following two cases:

Case 1:
A & M =>J (If the alarm goes off and Mary calls then John always calls too)
P(J | M & A) = 1
So P(J & M | A) = P(J&M&A) / P(A) = P(J | M&A) P(M&A) /P(A) = P(M&A) / P(A) = P(M|A)P(A)/P(A) = P(M|A) = 0.7

Case 2:
A & ~M => J (If alarm goes off and Mary doesn't call then John does call)
P(J | A & ~M) = 1

So P(J & ~M & A) = P(J | A & ~M) P(A & ~M) = P(J | A & ~M) P(~M | A) P(A) = 1 x (1 - 0.7) P(A) = 0.3 P(A)
So P(J & M & A) = P(J & A) - P(J & ~M & A) = P(J | A) P(A) - 0.3P(A) = 0.95 P(A) - 0.3 P(A) = 0.65 P(A)
So P(J & M | A) = 0.65 P(A) / P(A) = 0.65.

So these different dependencies give us different results.

If they don't give you any information about those dependencies, I expect they intended - but forgot to say - that the events M | A and J | A are independent, so that P(J & M | A) = P(J | A) P(M | A) = 0.95 x 0.7 = 0.665. Note how that is between the values from the above two cases.