# Trying to figure out probability problem for a series

So essentially I would have 10 items/figures, each with the option of being either A or B (for simplicity).
Each of the 10 figures has a set pre-determined probability that A or B will happen, for example:
Item 1 - 80% probability of A
Item 2 - 30% probability of A
Item 3 - 60% probability of A, etc...
How could I get the overall probability for each possible scenario, such as the chance that 0/10 are A, 1/10 are A, ..... or all 10/10 are A
I have R if that helps, I don't necessarily need to learn how to do it, just calculate it

• Bounty seems too low.

• Two questions. First, if A doesn't happen, must B happen? As in, is it always the case that if A has an x% chance case of happening, B has a (100-x)% chance of happening? Second, if Item 1 happens, does it affect whether or not Item 2 happens?

• Yes either A or B happens, they are mutually exclusive. You could also just forget B, if you know A has a 70% chance then B has a 30% chance, but I want to measure each item by the probability that A happens. Additionally, each item is independent of one another.

You mentioned that every item is independent of eachother. So, either Item i happens or does not. We could take 10 "bits" and give them a 1 if Item i happens and a 0 if Item i does not happen. For instance, 1010000000 would be the probability only Item 1 and Item 3 are $A$, and the rest are $B$.

Represent each probability of Item i being A as a number between 0 and 1, call it $p_i$. For each 10-bit number (and there are $2^{10} = 1024$ such numbers) do the following two calculations:

• Multiply all the probabilities together for the given number. For instance, given the number 1010000000, we calculate $p_{1010000000} = p_1\cdot (1-p_2) \cdot p_3 \cdot (1-p_4)\cdot (1-p_5)\cdot (1-p_6)\cdot (1-p_7)\cdot (1-p_8)\cdot (1-p_9)\cdot (1-p_{10})$
• Count all the ones in the number. For 1010000000, there are $2$ ones.
Once they're calculated, do a sanity check that the probabilities add to one, or something close enough to one.

I'm assuming your doing this in an excel spreadsheet or some programming language, so you have one "column" with the probability multiplied and another with all the bit counts. Group together all the bit counts that are the same and add them together. There will be only 1 even where 0/10 A events happen, that's the 0000000000 event. 10 events will have exactly 1/10 A events happening. For k/10 A events happening, there will be $\binom{10}{k}$ events, which is to say that's the binomial number. The sum of the independent probabilities will give you the probability that so many A events happen.