# Another probability question

1. Mar 21, 2016

### erisedk

1. The problem statement, all variables and given/known data
Of the three independent events E1 , E2 and E3, the probability that only E1 occurs is α, only E2 occurs is β and only E3 occurs is γ. Let the probability p that none of the events E1 , E2 and E3 occurs satisfy the equations
$(α - 2β) p = αβ$ and $(β - 3γ) p = 2βγ$.
All the given probabilities are assumed to lie in the interval (0,1).
Then, (Probability of occurrence of E1) / (Probability of occurrence of E3) =

2. Relevant equations

3. The attempt at a solution
http://s3.amazonaws.com/minglebox-p...ata-0000-fdbffe7622c53ecd0122c5c50d0b0334.gif

I don't know how to use those equations. All I know is what regions α, β, γ and p represent. α ≡ region 1, β ≡ region 3, γ ≡ region 7 and
1 - regions(1+2+3+4+5+6+7) = p. How do I proceed?

Last edited: Mar 21, 2016
2. Mar 21, 2016

### andrewkirk

The key is that the events are independent, so the probabilities of the joint occurrence or non-occurrence of different events is simply the product of the individual probabilities.

To use that fact, define A, B, C to be the probabilities of occurrence of each of the three events. Then you can write $\alpha,\beta,\gamma$ and $p$ each in terms of A,B,C.

Then your two equations above will be equations in terms of unknown probabilities A, B and C and you are asked to find A/C. See if you can rearrange the two equations so that they are in terms of the two unknowns B and A/C. If you can do that, you will have two equations and two unknowns, which you can solve.

3. Mar 21, 2016

### erisedk

I did that actually, it didn't lead anywhere. Let me post it.

4. Mar 21, 2016

### andrewkirk

I tried it and it worked for me: I got 6. There's an awful lot of cancelling that happens when you write out the equations, and it simplifies nicely.
Perhaps you made a misstep somewhere that stopped things from cancelling as they should.

5. Mar 21, 2016

### erisedk

Let P(E1) = a , P(E2) = b and P(E3) = c.

α = a - ab - ac + abc
β = b - ab - bc + abc
γ = c - ac - bc + abc
p = (1-a)(1-b)(1-c)

I can't use these to solve the equations. I tried various ways, it's not working.

6. Mar 21, 2016

### erisedk

Oh perhaps. Let me check again.

7. Mar 21, 2016

### andrewkirk

Try the following. It'll be less messy, and hence less likely to generate mistakes. Write $a'$ for $1-a, b'$ for $1-b, c'$ for $1-c$.
Then you get $\alpha=ab'c',\beta=a'bc',\gamma=ab'c',p=a'b'c'$ which is much neater and easier to use.
Write your two equations out using $a,b,c,a',b',c'$ and you'll find you can do a lot of cancelling and greatly simplify them before you need to convert $a', b' c'$ back to expressions in terms of $a,b,c$.

If it doesn't work, post what you did.

8. Mar 21, 2016

### erisedk

Thank you so much!! I got it :)