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?

  • $\begingroup$ 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 $P(A)$? $\endgroup$ – Henry Jan 20 '12 at 7:51
  • $\begingroup$ @Henry: Yes. At the current moment, I am assuming that the items are identical. $\endgroup$ – Legend Jan 20 '12 at 8:17

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.

  • $\begingroup$ Thank you. Can you please give me some hints on the expression when they are not independent? $\endgroup$ – Legend Jan 20 '12 at 17:06
  • $\begingroup$ 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. $\endgroup$ – Henry Jan 20 '12 at 18:20

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