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!

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

**Moderators:** gmalivuk, Moderators General, Prelates

### 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)

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)

### 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 ......

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
**Posts:**5967**Joined:**Tue Nov 28, 2006 9:50 pm UTC**Location:**Sydney

### 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.

ameretrifle wrote:Magic space feudalism is therefore a viable idea.

### Who is online

Users browsing this forum: No registered users and 11 guests