# Independent vs. Depedent

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• Aug 14th 2013, 07:02 AM
TexasTony
Independent vs. Depedent
Mathematicians:

The following question is in my text book (forgive the Latex formatting):

If \$A, B,\$ and \$C\$ are three equally likely events, what is the smallest value for \$P(A)\$ such that \$P(A \cap B \cap C)\$ always exceeds 0.95?

My scratch work to solve this question has largely used the Multiplicative Law of Probability and the Bonferroni inequality, and is attached below, but is also incorrect. The text book provides the answer of P(A) ≥ .9833. This is equal to .95^(1/3), which would imply that P(A \cap B \cap C)=P(A)^(3)≥.95. This solution defeats me... the question in the text book, featured above, in no way implies that the events are independent, a necessity to invoke the idea that P(A \cap B \cap C)=P(A)^3. How can this be? What subtle (or obvious) rule of Probability is eluding me?

Much obliged for your time,
TexasTony
• Aug 14th 2013, 11:58 AM
Lord Voldemort
Re: Independent vs. Depedent
I can't figure it out myself but I wanted to let you know that you probably did something wrong in the solution you posted since we must have P(A) > .95 in all cases, because we require P(A and B and C) > .95 which implies P(A)*P(B and C | A) > .95 ; maybe it's an easy mistake that can be fixed?
• Aug 14th 2013, 02:05 PM
TexasTony
Re: Independent vs. Depedent
@Lord Voldemort: Are you referring to the first line? I was trying to use the Bonferroni inequality, assuming that it could be employed in such a manner. The only example I have of it in my book only uses to variables, so I was uncertain of its use with three, but I took my best guess. However, my answer is clearly wrong. I had tried to use the conditional probability theorem, but it did not get me any where. Its clear by the answer that you have to assume that: P(A and B and C)>P(A)^3, which would imply that the events are all independent. This fits your case as well, as P(A)*P(B and C|A)->P(A)*P(B and C) if all the events are independent. What I do not understand is why we can make that assumption?

And I just got it. Since the book asked specifically for the smallest value of P(A) so that the statement is always true (regardless of Independence or Dependence), you have to figure out the probabilities for both cases, and then take the larger one, which in this case, and potentially all, is the Independent one.

Tom Riddle, I am much obliged. Happy muggle hunting,
TexasTony
• Aug 14th 2013, 02:27 PM
Lord Voldemort
Re: Independent vs. Depedent
I didn't mean to make any conclusions - I just thought there may have been a slight mistake in the work you submitted since it gives the obviously nonsensical answer of P(A) = .65 ; I'm unfamiliar with Bonferroni inequality so I couldn't follow your work completely.

"And I just got it. Since the book asked specifically for the smallest value of P(A) so that the statement is always true (regardless of Independence or Dependence), you have to figure out the probabilities for both cases, and then take the larger one, which in this case, and potentially all, is the Independent one."

This is true, but can you show that the smallest value of P(A) for the independent case is less than or equal to the smallest value of P(A) for the dependent case? I was trying to do this and wasn't able to come up with anything.
• Aug 16th 2013, 12:00 PM
TexasTony
Re: Independent vs. Depedent
I believe (and I could be wrong) that the dependent case would have to be a range, dependent on how A and B and C intersect. This would is represented by the relationship with both have given P(A and B) = P(A)P(B|A). Since the relationship could be anything in this case, the best we might be able to do is find a lower bound for the dependent case of P(A and B and C).

P((A and B) and (A and C)= P(A and B) x P(A and C| A and B)
P(A)P(B|A) x P(A and C| A and B)... this is as far as I can go, but I think it's far enough to prove my point. Consider just the terms: P(A)P(B|A). In order for them to produce >.95 (which they have to, as the next term is certainly less than or equal to 1), the lower bound for P(A) is .95. In this case P(B|A)=1. In order to this to carry through, we can assume that P(C|A)=1 as well, then the lower bound is still 1. (Note: A venn diagram for this case would have the sections A, B, and C all completely overlapping, so that if an event is in A, it is necessarily in B and C as well.) The upper bound for P(A) is therefore 1, and then the P(B|A)<1, but not so low as to bring the probability below .95 (because again, P(A and C| A and B) will be less than or equal to 1). From this explanation, we can conclude that the lower bound for P(A) in the dependent case is .95, and the upper bound 1.

Given that nonsense is correct, than we can conclude that the dependent case is less than the independent case, and that P(A)>.983 in order to be true for all cases.

Allright Voldy, have I convinced you?

Also, I now realize that this should have been in the advanced statistics section, as it is a college text book I am reading... my bad.