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A number of independent events, say $A$, $B$, $\ldots\,$, $E$, can happen with associated probabilities $P(A)$, $P(B)$, $\ldots$

For each event that happens I have to pay £10. The likelihood I have to pay at least £10 is straightforward to work out: $$1 - \bigl(1-P(A)\bigr)\bigl(1-P(B)\bigr)\bigl(1-P(C)\bigr)\bigl(1-P(D)\bigr)\bigl(1-P(E)\bigr)$$

But what is the likelihood that I have to pay out at least £20? ie what is the likelihood of at least two events occuring?

For 5 events I could list out all 32 possible outcomes (from none of the events happen through to all 5 events happen) and work out the probability of each outcome. However this approach is not feasible if I have 10s of events.

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The complement of "at least two" is "none or exactly one". –  David Mitra Jan 31 '13 at 14:54
    
The answers here may help math.stackexchange.com/questions/151635/… –  Byron Schmuland Jan 31 '13 at 15:08
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1 Answer

We first give an answer to the question as put. Then we deal with $n$ events. In a sense, the general case turns out to be simpler than the $A,\dots,E$ case, since the structure is more evident.

For simplicity let $\Pr(A)=a$, $\Pr(b)=b$, and so on. The probability that at least $2$ of $A,\dots, E$ happen is $1$ minus the probability that at most $1$ of our events happens.

The probability of at most $1$ is the sum of two probabilities: (i) the probability of none and (ii) the probability of exactly $1$,

(a) You calculated this already: it is $(1-a)(1-b)(1-c)(1-d)(1-e)$.

(b) We get exactly $1$ if $A$ happens but the others don't, or if $B$ happens but the others don't, and so on. These events are pairwise disjoint, so the probability of exactly $1$ is $$a(1-b)(1-c)(1-d)(1-e)+(1-a)(b)(1-c)(1-d)(1-e)+\cdots $$ (the sum contains $5$ terms, of which we have shown the first two).

The general case: Let the independent events be $A_1,A_2,\dots,A_n$ Let $\Pr(A_i)=p_i$. Then the probability $A_i$ doesn't happen is $1-p_i$. It is convenient to let $1-p_i=q_i$. (This is a standard abbreviation.)

The probability that $0$ of the events happen is then given by the pleasant expression $q_1q_2q_3\cdots q_n$. Call this $P_0$.

The probability exactly $1$ of the $A_i$ happens is the sum of $n$ terms, where the $i$-th term is $q_1q_2\cdots q_{i-1}p_iq_{i+1}\cdots q_n$. Call this sum $P_1$.

Then the probability that $2$ or more of our events happen is $1-P_0-P_1$.

Computational note: The calculation is not nearly as ugly as it seems. Assume that none of the $q_i$ is $0$. Then the probability of exactly $1$ can be rewritten as $$\left(q_1q_2\cdots q_n\right)\left(\frac{p_1}{q_1}+\frac{p_2}{q_2}+\cdots+\frac{p_n}{q_n} \right).$$

So the probability of at least $2$ is

$$1-\left(q_1q_2\cdots q_n\right)\left(1+\frac{p_1}{q_1}+\frac{p_2}{q_2}+\cdots+\frac{p_n}{q_n} \right).$$

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