Which is the probability of JohnCalls given Burglary? Why?

[alima]

Homework Statement

Which is the probability of JohnCalls given Burglary? Why?

Relevant Equations

Bayes' Theorem = P (A|B) = ( P(B|A) * P(A) ) / P(B)

Questions:

1. P (JohnCalls|Burglary) ?
2. Why?

Source of the image: Artificial Intelligence: A Modern Approach - Third Edition, by Stuart Russell and Peter Norvig.

My attempt at solving: using Bayes' Theorem = P (A|B) = ( P(B|A) * P(A) ) / P(B)

P(JohnCalls|Burglary) = P(J|B) = ( P(B|J) * P(J) / P(B) )

• P(B): 0.001
• P(J): it depends on the Alarm ringing or not!
• P(B|J): how I am supposed to deduce this???

 

[Mark44]

Homework Statement: Which is the probability of JohnCalls given Burglary? Why?
Relevant Equations: Bayes' Theorem = P (A|B) = ( P(B|A) * P(A) ) / P(B)

Questions:

1. P (JohnCalls|Burglary) ?
2. Why?

View attachment 344732
Source of the image: Artificial Intelligence: A Modern Approach - Third Edition, by Stuart Russell and Peter Norvig.

My attempt at solving: using Bayes' Theorem = P (A|B) = ( P(B|A) * P(A) ) / P(B)

P(JohnCalls|Burglary) = P(J|B) = ( P(B|J) * P(J) / P(B) )

• P(B): 0.001
• P(J): it depends on the Alarm ringing or not!
• P(B|J): how I am supposed to deduce this???

What does the illustration represent, particularly the JohnCalls and MaryCalls parts? Is the idea that if there is a burglary or earthquake, then the alarm goes off, after which either John or Mary calls someone? This isn't clear to me from the drawing.

There is a formula for conditional probability that you don't show.
$$P(A | B) = \frac{P(A \cap B)}{P(B)}, \text{ if } P(B) \ne 0$$

 

[alima]

What does the illustration represent, particularly the JohnCalls and MaryCalls parts? Is the idea that if there is a burglary or earthquake, then the alarm goes off, after which either John or Mary calls someone? This isn't clear to me from the drawing.

Sorry for not being clear enough. Here is the description of the book in my own words:

1. There is a house, and this house has an Alarm;
2. The Alarm detects Burglary, but also sometimes respond to minor Earthquakes;
3. John and Mary are neighbours of said house and will call you, the resident, if they hear the Alarm ringing;
4. The neighbours are not 100% accurate. John sometimes thinks the telephone ring is the Alarm ring (false positive), and Mary does not always hear the Alarm because she likes to hear to loud music (false negative).

There is a formula for conditional probability that you don't show.
$$P(A | B) = \frac{P(A \cap B)}{P(B)}, \text{ if } P(B) \ne 0$$

I didn't mention it because I don't know it.
Is it relevant to the problem?
My professor did not present this equation to my class.

 

[Mark44]

The formula I showed seems to me to be very relevant to your problem, as it is for conditional probability. The formula appears on this wikipedia page -- https://en.wikipedia.org/wiki/Bayes'_theorem

 

[FactChecker]

There is an important point that you need to notice. According to that diagram, the event of who calls is independent of whether the alarm is due to a burglary or an earthquake. Reversing that, whether the alarm was due to a burglary or an earthquake is independent of who calls. (Independence is reflexive or you can prove this with another application of Bayes Rule.) So P(B|J) = P(B).
WAIT! I might have to think about this because I am assuming from the diagram the exact thing the problem is asking about.

 

[alima]

There is an important point that you need to notice. According to that diagram, the event of who calls is independent of whether the alarm is due to a burglary or an earthquake.

That is totally correct.
The diagram can be simplified like that, right, since Earthquake isn't relevant to the question?

 

[alima]

There is a formula for conditional probability that you don't show.
$$P(A | B) = \frac{P(A \cap B)}{P(B)}, \text{ if } P(B) \ne 0$$

Fine, applying it to the problem:
$$P(J | B) = \frac{P(J \cap B)}{P(B)}$$

1. P(R) is 0.001, great!;
2. P(J ∩ R) I don't know how to deduce. Could you give me a hint?

I mean, there is no direct relationship between J and R, because there is A (Alarm) in the middle of the path.

 

[Mark44]

The diagram can be simplified like that, right, since Earthquake isn't relevant to the question?

I don't know whether you can disregard earthquake events. P(A) = .95 if both burglary and earthquake are true, but P(A) = .94 if only a burglary event occurred. If we assume that event B (burglary) means that only a burglary has occurred, then the probability .94 seems appropriate. But we also have to consider that the alarm goes off with probability .001 even if no burglary has occurred.

Also, you have to consider that John calls 90% of the time when he hears the alarm, and also 5% of the time otherwise (like if the telephone rings but he thinks it's the alarm).

 

[FactChecker]

I don't know whether you can disregard earthquake events. P(A) = .95 if both burglary and earthquake are true, but P(A) = .94 if only a burglary event occurred.

Good catch! I didn't realize that the probabilities of the alarm might create a dependence between (burglary,earthquake) and (JohnCalls, MaryCalls). I think it does in this case.

 

[FactChecker]

Good catch! I didn't realize that the probabilities of the alarm might create a dependence between (burglary,earthquake) and (JohnCalls, MaryCalls). I think it does in this case.

I should say that the alarm probabilities creates a dependence between (burglary,earthquake) and (JohnCalls, MaryCalls, nobody calls).

 

[pbuk]

There is a formula for conditional probability that you don't show.
$$P(A | B) = \frac{P(A \cap B)}{P(B)}, \text{ if } P(B) \ne 0$$

I don't think that helps here because

$$P(J | B) = \frac{P(J \cap B)}{P(B)}$$

I mean, there is no direct relationship between J and B, because there is A (Alarm) in the middle of the path.

... in other words we don't know $P(J \cap B)$.

But we do know $P(J | A)$ and $P(J | \neg A)$, and we should be able to work out $P(A | B)$ (note we are given $P(A | (B \cap E)$ and $P(A | (B \cap \neg E) )$).

Can you see how to combine these?

If you are a visual thinker you might find it helps to draw a probability tree.

 

[pbuk]

But we also have to consider that the alarm goes off with probability .001 even if no burglary has occurred.

No we don't have to consider that, because we are given that a burglary has occurred.

What we do have to consider though is the possibility that an earthquake has also occurred, making the probability that the alarm goes off 0.95 instead of 0.94. How can we account for this? Is the effect smaller than the precision appropriate to the problem?

 

[alima]

I don't think that helps here because [...] we don't know $P(J \cap B)$.

So, I must use just Bayes Theorem instead of the formula that Mark44 shown?

We do know $P(J | A)$ and $P(J | \neg A)$, and we should be able to work out $P(A | B)$ (note we are given $P(A | (B \cap E)$ and $P(A | (B \cap \neg E) )$).

Can you see how to combine these?

If you are a visual thinker you might find it helps to draw a probability tree.

On my first post, I pointed out that my doubt is how to discover $P(B | J)$ and $P(J)$ , in order to complete the Bayes Theorem . Am I focusing wrongly?

 

[alima]

I don't know whether you can disregard earthquake events.

You are right. I reincluded the Earthquake on my simplified diagram.

What else should I change?

 

[pbuk]

So, I must use just Bayes Theorem instead of the formula that Mark44 shown?

No, Bayes' Theorem is no help to you here.

On my first post, I pointed out that my doubt is how to discover $P(B | J)$ and $P(J)$ , in order to complete the Bayes Theorem . Am I focusing wrongly?

Yes you are focusing wrongly because as you point out:

we don't know $P(J \cap B)$.

Instead, focus on the probabilities in this:

we do know $P(J | A)$ and $P(J | \neg A)$, and we should be able to work out $P(A | B)$ (note we are given $P(A | (B \cap E)$ and $P(A | (B \cap \neg E) )$).

and

If you are a visual thinker you might find it helps to draw a probability tree.

 

[pbuk]

What else should I change?

Nothing, the diagram in the first post in this thread was already fine.

 

[pbuk]

Hint: looking for an equation that will give you the answer will not help, the solution is easily built up from simple combinations of probabilities, either in a diagram or a table

Spoiler

The table starts like this

$B \cap \neg E \cap A \cap J$	$1 \cdot 0.998 \cdot 0.94 \cdot 0.9$	0.844308

 

[alima]

Hint: looking for an equation that will give you the answer will not help, the solution is easily built up from simple combinations of probabilities, either in a diagram or a table

Spoiler

The table starts like this

$B \cap \neg E \cap A \cap J$

$1 \cdot 0.998 \cdot 0.94 \cdot 0.9$

0.844308

Hey, a Truth Table I know how to make!
I learned it from classes about binary calculation and also philosophy classes.

Here it is:

#​	Burglary​	JohnCalls​	Alarm​	Earthquake​
1​	T​	T​	T​	T​
2​	T​	T​	T​	F​
3​	T​	T​	F​	T​
4​	T​	T​	F​	F​

Entering the numbers on it, it becomes:

#​	Burglary​	JohnCalls​	Alarm​	Earthquake​
1​	0.001​	0.90​	0.95​	0.002​
2​	0.001​	0.90​	0.94​	0.998​
3​	0.001​	0.05​	0.05​	0.002​
4​	0.001​	0.05​	0.06​	0.998​

Mutiplying each line, I get these results for each of the 4 scenarios:

1. 0.00000171
2. 0,000844308
3. 0,000000005
4. 0,000002994

Did I understand rightly so far on this post?

 

[pbuk]

Did I understand rightly so far on this post?

Almost: you don't need to multiply by 0.001 because B is true for $P(J|B)$.

 

[alima]

Almost: you don't need to multiply by 0.001 because B is true for $P(J|B)$.

You are right. The scenario already states that B happened, so it should be 1 (100%) instead of just 0.001.

Fixing:

1. 0.00171;
2. 0.844308;
3. 0.000005;
4. 0.002994.

Or could also be represented as:

1. 0.171%;
2. 84.4308%;
3. 0.0005%;
4. 0.2994%.

But I still don't see how these numbers can help me "work out $P(A | B)$". I don't see how to combine $P(A | (B \cap E)$ and $P(A | (B \cap \neg E) )$ because I'm a total noob and failed in getting more knowledge online. This forum is being the most helpful place so far.

But we do know $P(J | A)$ and $P(J | \neg A)$, and we should be able to work out $P(A | B)$ (note we are given $P(A | (B \cap E)$ and $P(A | (B \cap \neg E) )$).

Can you see how to combine these?

Is it too much to ask for another hint?

 

[pbuk]

You have four numbers representing the probabilities of all of the possible combinations of events where J is true, given that B is true.

Just add them together!

 

[alima]

You have four numbers representing the probabilities of all of the possible combinations of events where J is true, given that B is true.

Oh, now I understand why use the Truth Table!

Just add them together!

0.00171 + 0.844308 + 0.000005 + 0.002994 = 0.849017

Thank you for all the help.

 

[alima]

If you are a visual thinker you might find it helps to draw a probability tree.

I'm curious about how a decision tree of this scenario would be.
I sketched one on my agenda book, which I reproduce below.

But I could not fit Earthquake on it, because it is simultaneous with Burglary when it happens, and in the decision tree there is no space for simultaneous events.

How could I fit Earthquake on such Decision Tree?

                                         [ JohnCalls
                          /  Alarm rings [ !JohnCalls
            - happens    <               
          /                 \ !Alarm rings [ JohnCalls
         /                                 [ !JohnCalls
Burglary<
         \                                  [ JohnCalls
          \                  /  Alarm rings [ !JohnCalls
          - don't happen <                 
                          \ !Alarm rings [ JohnCalls
                                         [ !JohnCalls

 

[pbuk]

But I could not fit Earthquake on it, because it is simultaneous with Burglary when it happens, and in the decision tree there is no space for simultaneous events.

How could I fit Earthquake on such Decision Tree?

When you have simultaneous events you can put them in an arbitrary order: in this case, because we only need to consider the "Burglary happens" branch, I would do it like this:

                                                                [ JohnCalls
                                                 /  Alarm rings [ !JohnCalls
                                   - happens    <              
                                 /               \ !Alarm rings [ JohnCalls
                                /                               [ !JohnCalls
Burglary happens -> Earthquake <
                                \                                   [ JohnCalls
                                 \                   /  Alarm rings [ !JohnCalls
                                   - doesn't happen <                
                                                     \ !Alarm rings [ JohnCalls
                                                                    [ !JohnCalls