## Probability.Addition rule,inclusive or,at least one of three

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jacksmack
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Location: Italy

### Probability.Addition rule,inclusive or,at least one of three

Hi,

I have the following problem:

A student has three probability to pass three exams. Respectively, 3/4, 2/3 and 1/2.
What is the probability that she will pass at least one of these three exams?

So it is analog to say:
"What is the probability that she will pass either the first OR the second OR the third exam?"

The answer given by the book is 23/24.
But, I have obtained a different result and I would want to know why. Please can you help me?

This is my reasoning:
If I understand, this is the case of an "inclusive or" so I have done this:

P(A or B or C) =
= P(A) + P(B) + P(C) - [P(A and B and C) + P(A and B) + P(A and C) + P(B and C)] =
= 3/4 + 2/3 + 1/2 - [ ( 3/4 · 2/3 · 1/2 ) + ( 3/4 · 2/3 ) + ( 3/4 · 1/2 ) + ( 2/3 · 1/2 ) ] =
= 11/24

that it is different from 23/24. Where is the error?

Many thanks!

Nicias
Posts: 168
Joined: Tue Aug 13, 2013 4:22 pm UTC

### Re: Probability.Addition rule,inclusive or,at least one of t

You over count in the part you subtract.

When you remove P(A and B) that includes P(A and B and C) and P(A and B and ~C). You already counted the P(A and B and C)

elliptic
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Location: UK

### Re: Probability.Addition rule,inclusive or,at least one of t

With questions like this the trick is to turn it the other way around.

What's the probability that the student passes *none* of the exams?

Then the probability of any other result (ie. at least one pass) is ......

jestingrabbit
Factoids are just Datas that haven't grown up yet
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### Re: Probability.Addition rule,inclusive or,at least one of t

jacksmack wrote:P(A or B or C) =
= P(A) + P(B) + P(C) - [P(A and B and C) + P(A and B) + P(A and C) + P(B and C)]

The right rule here is

P(A or B or C) =
= P(A) + P(B) + P(C) - [P(A and B) + P(A and C) + P(B and C)] + P(A and B and C)

More generally, if you have n things that you are "or"ing, you add the probabilities of odd numbers of them happening and subtract the probabilities of even numbers happening. If you draw a Venn diagram, you should be able to convince yourself why that is.
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PM 2Ring
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