Additive Principle:

$P(A \cup B)=P(A)+P(B)−P(A \cap B)$

I found the following numbers, and verified that they are indeed correct:

$P(A) = 0.2$, $P(B) = 0.3$, $P(A \cup B) = 0.5$, $P(A \cap B) = 0$(Mutually exclusive), $P(C) = ?$

The thing is, there is no $P(C)$ in the additive principle, I found all of these values, but I'm not sure how I can fit $P(C)$ into the principle, if $P(C)$ doesn't exist in the first place. If someone can just push me in the direction that'd be awesome!

Entire question BELOW.

A sample space contains only three simple events: A, B, and C. If P(A) = 0.2 and P(B) = 0.3 find:

P(A and B) if A and B are mutually exclusive

P(A or B) if a and B are mutually exclusive

P(C) if A and B are mutually exclusive

  • $\begingroup$ It looks correct, if indeed P(A∩B)=0 $\endgroup$ – CAGT Mar 22 '16 at 23:55
  • $\begingroup$ @CAGT I don't follow, I'm being asked to calculate the unknown value for P(C), the problem is in the additive principle, there is only P(A) and P(B), no P(C), so I'm not sure how you would calculate that, the answer says 0.5, but I'm not sure how they got that? Also I apologize, I'm not familiar with properly getting the write text format for equations and proper symbols :) $\endgroup$ – Xor Mar 22 '16 at 23:57
  • $\begingroup$ I guess P(C) is the result for mutually exclusive events. So it should read P(C)=P(A∪B)=P(A)+P(B)−P(A∩B), if you still don't feel comfortable, then if P(C) is a third event, you need different formulas. $\endgroup$ – CAGT Mar 23 '16 at 0:02
  • 1
    $\begingroup$ If the space contains only three simple events: $\mathsf P(A\cup B\cup C) = 1$ $\endgroup$ – Graham Kemp Mar 23 '16 at 0:19
  • 1
    $\begingroup$ Yeah, I'm going to go with: You are not given enough information about C to find more than a lower bound on its probability $\endgroup$ – Graham Kemp Mar 23 '16 at 0:22

Your Answer

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

Browse other questions tagged or ask your own question.