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I have a programming problem I'm trying to solve where, in order to be efficient, I have to find the probability of some events occuring.

Lets say I have 3 transactions. At each transaction there is a 5% chance that the transaction will be declined. I want to know what are the chances of 0, 1, 2 and 3 of the transactions being declined.

I have being trying to visualise this with a venn diagram and trying to work out the areas of each event, but unfortunately when I add up the separate events I keep on getting more than or less than 1.

The event that 0 of the transactions will be declined is simply:

p(0 transactions declined) = .95 * .95* .95 = 6859/8000

The event that all 3 of the transactions will be declined should be:

p(3 transactions declined) = .05 * .05 * .05 = 1/8000

Now I start running into trouble:

p(2 transactions only declined) = (3 x (.05 * .05)) -3/8000 (because the A∩B∩C has been counted three times and we only want to know about when there are 2 transactions, and NOT when there are 2 or 3) = 57/8000

p(1 transaction only declined) = Well p(A∩(A'∪B'∪C')) x 3 would give us the answer. I make this out to be 3 x (0.05 -(2 x 20/8000) - 1/8000) = 1077/8000

If I add all those events up, then I get 7994/8000. I'm not sure what the right way to go about this is. I'd like to work it out for myself, but I really need a point in the right direction. I don't just want to make this work for three transactions, but for as many as I want so I need a general formula.

Many thanks

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3 Answers

up vote 3 down vote accepted

You need to look up the binomial distribution. One online resource is Grinstead and Snell's book.

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Thanks for your reply –  Joe Aug 26 '10 at 16:04
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http://en.wikipedia.org/wiki/Binomial_distribution

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Thanks for your reply –  Joe Aug 26 '10 at 16:03
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Your calculation for p(1), which I don't completely follow, is somewhat off. Another, more straightforward way of calculating this would be

$p(\textrm{1 transaction declined}) = {3 \choose 1} \times .05 \times .95 \times .95 = \frac{1083}{8000}$

Note that this exactly accounts for the $\frac{6}{8000}$ you are missing.

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Thanks for your reply –  Joe Aug 26 '10 at 16:04
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