Conditional Probability Question

[lmprobability]

Hello,

I believe I have over-complicated what should have been a simple conditional probability question.

I am provided with the probabilities of three events:
p(A) = 0.52
p(B) = 0.199
p(C) = 0.18

I am also given the intersect values:
p (A intersect B) = 0.10
p(B intersect C) = 0.04
p(A intersect C) = 0.09

The question asks: In one trial, what is the probability that neither A, nor B, nor C occurs?

Because I was not told that the events are independent, I will assume that they are not. Instead, I assumed they were dependent events and started to calculate using the formula P(a)P(B given A)P(C given A and B), but I started to doubt if this was the correct way to approach the problem. I believe there may be an easier way to solve this and I am just not interpreting the question correctly.

I understand that whatever answer I get for the probability of these three events occurring needs to be subtracted from 1 to determine the probability of neither occurring. At this point, I have spent almost 2 hours trying to solve this problem. I would be so appreciative if someone could please assist me in working through this problem step-by-step so that I may also improve my understanding of how you arrived at the answer. Thank you for your consideration!

 

[Dr.Peterson]

Hello,

I believe I have over-complicated what should have been a simple conditional probability question.

I am provided with the probabilities of three events:
p(A) = 0.52
p(B) = 0.199
p(C) = 0.18

I am also given the intersect values:
p (A intersect B) = 0.10
p(B intersect C) = 0.04
p(A intersect C) = 0.09

The question asks: In one trial, what is the probability that neither A, nor B, nor C occurs?

Because I was not told that the events are independent, I will assume that they are not. Instead, I assumed they were dependent events and started to calculate using the formula P(a)P(B given A)P(C given A and B), but I started to doubt if this was the correct way to approach the problem. I believe there may be an easier way to solve this and I am just not interpreting the question correctly.

I understand that whatever answer I get for the probability of these three events occurring needs to be subtracted from 1 to determine the probability of neither occurring. At this point, I have spent almost 2 hours trying to solve this problem. I would be so appreciative if someone could please assist me in working through this problem step-by-step so that I may also improve my understanding of how you arrived at the answer. Thank you for your consideration!

I may be thinking wrongly, but I don't think there's enough information.

I tried making a Venn diagram, which has 8 regions to fill in; you have 6 facts, plus the fact that the sum of all 8 regions is 1, so in effect your could write 7 equations in 8 unknowns.

Would it be possible to show us an image of the original problem?

 

[pka]

I am provided with the probabilities of three events:
p(A) = 0.52, p(B) = 0.199. p(C) = 0.18
I am also given the intersect values: p (A intersect B) = 0.10. p(B intersect C) = 0.04, p(A intersect C) = 0.09

The question asks: In one trial, what is the probability that neither A, nor B, nor C occurs?

I agree with Dr. Peterson that at least one piece of crucial information is missing or omitted.
Are you very sure that the problem is copied completely & correctly?
You are asked to find [imath]\mathcal{P}(A^c\cap B^c\cap C^c)[/imath].
It is the case that [imath](A^c\cap B^c\cap C^c)^c=A\cup B\cup C[/imath].

[imath][/imath][imath][/imath][imath][/imath]

 

[lmprobability]

Ah, my apologies! I forgot to attach the diagram provided. My intuition was to rely on the diagram and argue that if neither A nor B occur, then C cannot occur because it is comprised of the overlaps of A and B. I will include the question again in the attachment.

I do appreciate your time and consideration.

Kind regards.

 

[Dr.Peterson]

Ah, my apologies! I forgot to attach the diagram provided. My intuition was to rely on the diagram and argue that if neither A nor B occur, then C cannot occur because it is comprised of the overlaps of A and B. I will include the question again in the attachment.

I do appreciate your time and consideration.

Kind regards.

​

It looks like that gives us one more fact, which should be enough: [imath]C\subset A\cup B[/imath], which is equivalent to what you said.

So one of the 8 regions of a traditional Venn diagram can be filled with 0.

Now please make an attempt so we can see where you need help.

(One trick that may help is to let x be the probability of the intersection of all three sets, and work out the other parts from that, with the goal of writing one equation in x.)

 

[lmprobability]

View attachment 32949​

It looks like that gives us one more fact, which should be enough: [imath]C\subset A\cup B[/imath], which is equivalent to what you said.

So one of the 8 regions of a traditional Venn diagram can be filled with 0.

Now please make an attempt so we can see where you need help.

(One trick that may help is to let x be the probability of the intersection of all three sets, and work out the other parts from that, with the goal of writing one equation in x.)

Ok, here are my attempts. I appreciate your patience as I try to explain these approaches!

I have taken two approaches. I believe the second approach to be incorrect, however, because I cannot assume the events are independent. I believe the first approach is the best option given the information that I know (I do not know the value of A intersect B intersect C and am uncertain how to solve for that value unless I attempt in the way I did for approach 2).

Approach 1:

p(A union B union C) = p(A) + p(B) + p(C) – p(A intersect B) – p(A intersect C) – p(B intersect C)
= 0.52 + 0.199 + 0.18 - (0.10 + 0.04 + 0.09)
= 0.899 - 0.23 = 0.669
= 1- 0.669
= 0.331

Approach 2: (trying to calculate for the intersection of A, B, and C assuming they are independent events):

p(A union B union C) = p(A) + p(B) + p(C) – p(A) x p(B) – p(A) x p(C) – p(B) x p(C) + p(A) x p(B) x p(C)

= 0.52 + 0.199 + 0.18 – (0.52 x 0.199) – (0.52 x 0.18) – (0.199 x 0.18) + (0.52 x 0.199 x 0.18)

= 0.52 + 0.199 + 0.18 – 0.10348 – 0.0936 – 0.03582 + 0.0186264

= 0.6847264 = approx. 0.685

= 1-0.685

=0.315

 

[Cubist]

(One trick that may help is to let x be the probability of the intersection of all three sets, and work out the other parts from that, with the goal of writing one equation in x.)

The above is great advice. Here's my attempt...



Either I've made a mistake OR the question is flawed. OP please check this.

EDIT: For their diagram to be correct, then surely p(C) ought to be less than or equal to p(B intersect C) + p(A intersect C)

 

[Dr.Peterson]

The above is great advice. Here's my attempt...

View attachment 32950

Either I've made a mistake OR the question is flawed. OP please check this.

EDIT: For their diagram to be correct, then surely p(C) ought to be less than or equal to p(B intersect C) + p(A intersect C)

That's what I did initially (almost exactly). I added the probabilities shown and found that the requested probability (for the outside region) is 0.331 - x, but we don't know x. That's what led to my post #2.

With the new information that [imath]C\subset A\cup B[/imath], we can set 0.05+x = 0, and conclude that x = -0.05. Which is a problem!

So something is wrong there, and your explanation makes it obvious. But it's possible that they solved this by a method that didn't reveal the flaw, and think the answer is 0.331-(-0.05) = 0.381.

Approach 1:

p(A union B union C) = p(A) + p(B) + p(C) – p(A intersect B) – p(A intersect C) – p(B intersect C)
= 0.52 + 0.199 + 0.18 - (0.10 + 0.04 + 0.09)
= 0.899 - 0.23 = 0.669
1-p(A union B union C) = 1- 0.669
= 0.331

I've added something you omitted, to make it sensible.

But to make this work, you would also need to add p(A intersect B intersect C); do you see why?

And this leads to exactly what I said, giving 0.331 - x.

 

[lmprobability]

I do see why! This is starting to make much more sense now.

 

[Steven G]

Sorry but p(A union B union C) = p(A) + p(B) + p(C) – p(A intersect B) – p(A intersect C) – p(B intersect C) is not true. Draw a diagram and see why not.

 

[Dr.Peterson]

Let's change something to make it solvable. Suppose that

p(A) = 0.52​

p(B) = 0.199​

p(C) = 0.10​

​

p(A ∩ B) = 0.10​

p(B ∩ C) = 0.04​

p(A ∩ C) = 0.09​

​

p(C ∩ not A ∩ not B) = 0 from the picture​

Then you can use what you know about C and its intersections to find p(A ∩ B ∩ C); this is particularly easy to see on a standard Venn diagram. No variable is needed.

From there, you can fill in everything else, either using Venn, or using the inclusion-exclusion principle that you almost had right.

 

[lmprobability]

Let's change something to make it solvable. Suppose that

p(A) = 0.52​

p(B) = 0.199​

p(C) = 0.10​

​

p(A ∩ B) = 0.10​

p(B ∩ C) = 0.04​

p(A ∩ C) = 0.09​

​

p(C ∩ not A ∩ not B) = 0 from the picture​

Then you can use what you know about C and its intersections to find p(A ∩ B ∩ C); this is particularly easy to see on a standard Venn diagram. No variable is needed.

From there, you can fill in everything else, either using Venn, or using the inclusion-exclusion principle that you almost had right.

I understand what you're trying to demonstrate. I appreciate your time and consideration. I will apply this approach.