0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.