i don't have a background in Probability or Mathematics so i may be looking at a simple problem without knowing it. I have the following independent Events and their probabilities:

  • Event $A \rightarrow 83\%$
  • Event $B \rightarrow 25\%$
  • Event $C \rightarrow 41\%$
  • Event $D \rightarrow 68\%$
  • Event $E \rightarrow 11\%$
  • Event $F \rightarrow 47\%$

I know that if the events where all equal probability (for example $50\%$), the probability of getting them all right would be:

  • $0.5^6 = 1.56\%$

I also understand that if we wanted to know the probability of getting $5$ out of $6$ right the probability would be:

  • $0.5^5 \cdot 0.5^1 \cdot \dfrac{6!}{5! \cdot 1!} = 9.38\%.$

And the same reasoning goes for getting $4$ out of $6$ right:

  • $0.5^4\cdot0.5 ^ 2\cdot \dfrac{6!}{4! \cdot 2!} = 23.44\%.$

But how do we compute it when the events have a different set of known probabilities? For example, how do we compute the probability of getting $4$ out of $6$ events right with the above set of probabilities. Thanks

  • $\begingroup$ Are these meant to be independent events? If so, then you have to consider the probability of each specific $5$ (or $4$) and add. $\endgroup$ – lulu Mar 1 '18 at 15:49
  • $\begingroup$ Sorry, forgot to mention, they are independent. I do understand the computation, and i could program a computer to add all the different probability paths, but although i mentioned 6 events i'm handling more than 100 events in my real case scenario, and the possibilities for a simple question like "whats the probability of answering right 98 out of 100)" are a lot. I thought their could be a generalized way to answer this. $\endgroup$ – user2198816 Mar 1 '18 at 15:49
  • $\begingroup$ That should be added to your post, it's critical information. $\endgroup$ – lulu Mar 1 '18 at 15:51
  • $\begingroup$ I have added to the question, thanks. $\endgroup$ – user2198816 Mar 1 '18 at 15:56
  • $\begingroup$ If there is no real structure in the probabilities, you will have to consider all combinations where $4$ answers are right an $2$ answers are wrong, multiply the respective probabilities for such a combination, and add them all up. But that's what computers are good for. $\endgroup$ – Luke Mar 1 '18 at 15:56

If I were doing this by hand, here's what I'd try: First group A, B, C together and find the probabilities of $0, 1, 2$, or $3$ correct for A, B, C. This is fairly manageable.

Then I'd do the same for a grouping of D, E, F.

Let's denote the probabilities for $0, 1, 2$, or $3$ correct for A, B, C by $a_0, a_1, a_2, a_3$ respectively; and similarly $d_0, d_1, d_2, d_3$ for the D, E, F probabilities.

Then overall: Probability of $0$ correct is $a_0\cdot d_0$.

Probability of $1$ correct is $a_0\cdot d_1+a_1\cdot d_0$.

Probability of $2$ correct is $a_0\cdot d_2+a_1\cdot d_1+a_2\cdot d_0$.


I think this is easier than the total brute-force approach.

Of course having a computer do it for you is even easier!


A probability generating function (PGF) will solve the problem, but it's not a labor-saving device unless you have access to a computer algebra system. The PGF for the number of successes in your example is $$f(x) = (0.89\, +0.11 x) (0.75\, +0.25 x) (0.59\, +0.41 x) (0.53\, +0.47 x) (0.32\, +0.68 x) (0.17\, +0.83 x)$$ Upon expansion (needless to say, I used a computer algebra system), $$f(x) = 0.0113548\, +0.102715 x+0.295685 x^2+0.353916 x^3+0.19047 x^4+0.0428685 x^5+0.0029909 x^6$$ So the probability of $0$ successes is $0.0113548$, the probability of exactly $1$ success is $0.102715$, the probability of exactly $2$ successes is $0.295685$, etc.

Even if you don't have access to a computer algebra system, a PGF is a good way to organize your work, but you may have to do some tedious computations.

  • $\begingroup$ awkward, this looks great, but what does X stand for? I thought it was the number of successes but i cant get your results. Thanks. $\endgroup$ – user2198816 Mar 2 '18 at 16:41
  • $\begingroup$ @user2198816 $x$ doesn't stand for anything. The coefficient of $x^n$ is the probability of $n$ successes. See en.wikipedia.org/wiki/Probability-generating_function $\endgroup$ – awkward Mar 2 '18 at 18:01
  • $\begingroup$ Oh, understood, thanks, i just have to build the PGF, expand it to get the powers and x^n is the probability of exactly n successes. $\endgroup$ – user2198816 Mar 2 '18 at 22:21

Hint:$$\sum^6_{i=1}\frac{A\land B\land C\land D\land E\land F}{\lnot e_{i\pmod6}\land\lnot e_{i+1\pmod6}}$$.

Where $A,B,C,\dots$ are the respective probabilities, $e_1$ is $A$, $e_2$ is $B$, and so on till $e_0$ is $F$.


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