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This may be a silly question but I am confused on how to proceed. I have a family A of items: A1, A2, A3, A4 and A5 which were being randomly stress tested for failure. I computed the probability of failure of family A (consisting of A1 through A5) as the number of items that failed during the stress testing period divided by the total population of the family.

Now, how can I compute the probability that two items will fail together assuming both dependent and independent failures?

I am thinking that if they are independent then the estimate would be:

$P(A1) * P(A2) * (1-P(A3)) * (1-P(A4)) * (1-P(A5))$ $=P(A)^2 * (1-P(A))^3$

I am not sure if this is close to what I want though. However, I am also curious to know how to solve this if the events are dependent. Any suggestions?

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As you assuming there is no difference between items in a family? E.g. $A1$ is as likely to fail as $A4$, each with probability $P(A)$? – Henry Jan 20 '12 at 7:51
@Henry: Yes. At the current moment, I am assuming that the items are identical. – Legend Jan 20 '12 at 8:17
up vote 2 down vote accepted

Assuming the failures are independent and identically distributed, your expression is correct for the probability that an identified pair fail.

The probability that any two fail is therefore ${5 \choose 2} = 10$ times this.

If failures are not independent then your expression becomes much more complicated, as the multiplication of a sequence of conditional probabilities.

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Thank you. Can you please give me some hints on the expression when they are not independent? – Legend Jan 20 '12 at 17:06
For your original example it is $P(A_1)P(A_2|A_1)P(A_3^C|A_1,A_2)P(A_4^C|A_1,A_2,A_3^C)P(A_5^C|A_1,A_2,A_3^C,A_4‌​^C)$ where $A_i^C$ means $A_i$ does not happen: $P(A_i^C)=1-P(A_i)$. Then you need to deal with the other nine possibilities, or if there is symmetry then multiply by 10 again. – Henry Jan 20 '12 at 18:20

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