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I want to generalize a formula and I need your help with this. This is not my homework or assignment but I need to come up with a concise formula that fits my documentation.

Background for my problem:

Considering all events to be independent of each other,

Let the probability of $Event$ $0$ be $P_{0}$ and $Event$ $1$ be $P_{1}$ and so on ...

Then the probability that two events occur simultaneously is : $P_0P_1$ . This is nothing but the area of intersection of two circles $P_0$ and $P_1$.

Continuing the same way,

The total probability of simultaneous occurrence in case of three events is:

$P_0 P_1 + P_0P_2 + P_1P_2 - 2 P_{0}P_{1}P_{2}$.

Also can be visualized by drawing three intersecting circles.

One clarification here: This gives me the total probability of any two events plus all the three occurring at the same time right?

I cannot visualize the formula by drawing circles anymore. How can the above formula be generalized to get the probability of simultaneous occurrences when there are 4, 5, ... independent events.

I have seen that the inclusion-exclusion principle is the answer. But I am not able to get an intuition for it. The inclusion-exclusion principle gives the probability of 2,3,4 sets intersecting but isn't my question different?

I get this doubt because, probability of four independent events occurring simultaneous is: $P_0P_1P_2P_3$. But what I need is a general formula for the total probability of two or more independent events at the same time.

Can anyone of you please throw some light?


Yes indeed I meant that the probability of "two or more events". The answer you have given is very precise and the one I was looking for. Yes it is boring to try to visualize circles when the number exceeds three. Instead, in general if use 1 - ($w_0 + w_1$) then we should land up correctly given that the events are independent. Thank you so much.

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It seems that by "The total probability of simultaneous occurrence " you actually mean "the ocurrence of more than two events" –  leonbloy May 30 '12 at 16:12
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5 Answers

$$1-\left(1+\sum_{i=1}^n\frac{P_i}{1-P_i}\right)\cdot\prod_{k=1}^n(1-P_k)$$

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It seems that by "The total probability of simultaneous occurrence " you actually mean "the ocurrence of two or more events" The inclusion-exclusion principle, in its most basic form, is usually apply to get the probability of "one or more" events (i.e.: an OR). It can certainly be applied here, also, but it gets a little more complicated. Your "elementary" events are now rather the pairs $A_{ij}=E_i \cap E_j$ (the intersections of two circles in your Venn diagram). Applying the inclusion-exclusion principle to the three events you'd get

$$P(A_{01} \cup A_{12} \cup A_{02}) =\\ P(A_{01})+P(A_{12})+P(A_{02})- P(A_{01} \cap A_{12})-P(A_{01} \cap A_{12})-P(A_{12} \cap A_{02}) + P(A_{12} \cap A_{02} \cap A_{02})=\\ $$

which, because $P(A_{12} \cap A_{02}) = P(E_0\cap E_1 \cap E_2)= P(A_{12} \cap A_{02} \cap A_{12})$ equals

$$P(E_0 \cap E_1) + P(E_1 \cap E_2) +P(E_0 \cap E_2) - 2 P(E_0\cap E_1 \cap E_2) $$

To extend this to 4 events is straighwforward, but borish and hard to visualize.

Of course, in your case, if the events are independent, it's more easy to compute the complement: $1-(w_0 + w_1)$ where $w_0=$ probability of no event and $w_1=$ probability of exactly one event.

$w_0 = (1-P_0)(1-P_1)(1-P_2)(1-P_3)$ $w_1 = P_0(1-P_1)(1-P_2)(1-P_3) + (1-P_0)P_1(1-P_2)(1-P_3) + (1-P_0)(1-P_1)P_2(1-P_3)+ (1-P_0)(1-P_1)(1-P_2)P_3 $

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$P[\bigcup_{k=1}^N P_k]=\sum_{k=1}^NP_k-\sum_{j<k}^NP_jP_k+\dots+(-1)^{N+1}P_1P_2\dots P_N$

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Inclusion-Exclusion formulas

Let $A_1, A_2,\dots, A_n$ be events. The probability that exactly $k$ of them occur $(k\geq 0)$ is

$$p_n(k)=\sum_{i=k}^n (-1)^{i-k}{i\choose k}S_i,$$

while the probability that $k$ or more of them occur $(k\geq 1)$ is

$$P_n(k)=\sum_{i=k}^n (-1)^{i-k}{i-1\choose k-1}S_i.$$

Here $S_0=1$, $S_1=\sum_i P(A_i)$, $S_2=\sum_{i< j} P(A_i A_j)$, $S_3=\sum_{i< j<k} P(A_i A_j A_k)$, etc.

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Just use the opposite event : one or less.

For example for 3 events (none+only $P_1$+only $P_2$+only $P_3$): $$E_3=(1-P_1)(1-P_2)(1-P_3)+P_1(1-P_2)(1-P_3)+(1-P_1)P_2(1-P_3)+(1-P_1)(1-P_2)P_3$$

Then what you want is only $1-E_n$, that is easy to compute now.

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