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Suppose there are 9 events, that have a probability of 10%, 20%, 30%, ..., 90% of being a success.

How would I find the probability of exactly n number of these events succeeding?

For n = 1, I'm thinking it is something like

(first succeeds) -> (1-0.9) * 0.8 * 0.7 * 0.6 ... 0.1
(second succeeds) -> 0.9 * (1-0.8) * 0.7 * 0.6 ... 0.1
(...)
(last)

and then adding them up.

The probabilities might not always be an an Arithmetic Progression, hoping to find a solution that doesn't depend on that.

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3 Answers 3

The generating function for the probabilities $a_n$ for $n$ successes is $$\sum_{n=0}^9 a_n t^n=\prod_{k=1}^9(p_k t+(1-p_k))$$ where $p_k$ is the probability of the $k$-th event occuring, in this example $p_k=k/10$. Maybe in this case the product won't simplify so nicely, but the method is quite general.

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I tried searching, but couldn't understand anything. What does t here mean? –  Adam Oct 22 '10 at 22:43

Take the events in order. After the first, you have 0.1 chance of 1 success and 0.9 chance of 0 successes. Then add on the second. The chance of one success after two events is 0.2*the chance of failure on the first + 0.8*the chance of success on the first. A spreadsheet is your friend for this. After all the events, you will have a column with the chance of each possible number of successes. Adding to see if you still get 1 is a good check that your equations are right. If you use the absolute/relative reference codes correctly, you can copy down and right most of the equations.

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Let the probability that the $i$th event occurs be $p_i = i/10$ and let the probability that exactly $n$ events occur be $P(n).$

So the probability that no events occur is $P(0) = \prod_{i=1}^9 (1-p_i) = 3.6288 \times 10^{-4}.$

Let $S= \lbrace 1,2,3,\ldots,9 \rbrace $ then for $n \ge 1$ we have

$$P(n) = P(0) \times \sum_{i_1,i_2,\ldots,i_n \in S, \quad i_k \textrm { distinct} } \frac{p_{i_1} p_{i_2} \ldots p_{i_n} } {(1-p_{i_1})(1-p_{i_2}) \ldots (1-p_{i_n}) },$$

where there are ${9 \choose n}$ terms in the summation.

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