In a probability experiment, A, B and C are independent events. The probability that A will occur is r, the probability that B will occur is s, and the probability that C will occur is t whereas the r, s , and t are grater than zero.
Now I want to find a probability that, either A, B or C can occur but they can not occur together (like AAB or AAC is not allowed but AAA or BBB is allowed.)

In this case I used exclusion-inclusion formula to calculate the probability in this way:

$$P = P(A \cup B \cup C) - P(A\cap C) - P(B\cap C) -P(A\cap B) -P(A\cap B\cap C) $$

Is it a correct way?

  • $\begingroup$ It almost sounds as if you are asking for the probability that exactly one of A, B, C occurs, but then the AAB not allowed but AAA allowed makes me wonder what the problem is about, $\endgroup$ – André Nicolas Dec 12 '15 at 6:44
  • $\begingroup$ Yes only one of A, B or C will occur. $\endgroup$ – Rayan Ahmed Dec 12 '15 at 6:51
  • 1
    $\begingroup$ I don't think the formula in the OP is right, but it is late and my blood caffeine level is low. The probability of exactly one is $r(1-s)(1-t)+s(1-r)(1-t)+t(1-r)(1-s)$. $\endgroup$ – André Nicolas Dec 12 '15 at 6:58
  • $\begingroup$ @ André Nicolas, you are right and I got you that how you have done it but how to do it using set formula. $\endgroup$ – Rayan Ahmed Dec 12 '15 at 7:10

There is an error in your set formula. Draw a Venn diagram to understand that it should be

$P = P(A \cup B \cup C) - P(A\cap C) - P(B\cap C) -P(A\cap B) +2*P(A\cap B\cap C)$

  • $\begingroup$ There will 2 in front of $P(A\cap B\cap C)$ $\endgroup$ – Rayan Ahmed Dec 12 '15 at 7:54
  • $\begingroup$ Quite - right !! $\endgroup$ – true blue anil Dec 12 '15 at 9:17

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