If $P(A) = \frac{1}{3}, P(B) = \frac{1}{2}$, and $P(A \cup B) = \frac{3}{4}$ Find:

$P(A \cap B),\\ P(A^\complement \cup B^\complement) \\ P(A^\complement \cap B)$

Here is what I did:

$P(A \cap B) = P(A) + P(B) - P(A \cup B) = \frac{1}{12}$

$P(A^\complement \cup B^\complement) = P( [A \cap B]^\complement ) = 1 - \frac{1}{12} = \frac{11}{12}$ (Not sure if this is correct)

$P(A^\complement \cap B) = ...$ (I'm not quite sure how to approach this)

  • $\begingroup$ This is really a problem in set theory - how can you rewrite $A^\complement \cap B$ in terms of $A$, $B$, $A \cap B$ etc. As $A \cup A^\complement$ is the whole space, and the union is disjoint, you also have for every subset $B$: $B = (B \cap A) \cup (B \cap A^\complement)$ and the union is disjoint. This will lead you to Ahmad Bazzi's answer. $\endgroup$ – mathguy Sep 6 '18 at 3:55
  • $\begingroup$ Try drawing a venn diagram with 2 circles, one for A and one for B. $\endgroup$ – DanielV Sep 6 '18 at 4:01

Ponder the following diagram, where $()$ denotes $A$ and $[]$ denotes $B$:

$$( *** [ * ) ***** ] ***$$

$A$ has four stars, out of twelve, i.e. $1/3$.

$B$ has six stars, out of twelve, i.e. $1/2$.

$A\cup B$ has nine stars, out of twelve, i.e. $3/4$.

You now have the complete Venn diagram, and can answer all the questions directly. $B\cap A^c$ has five stars.

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All is correct. For the last one you'd need:

$$P(A^\complement \cap B) = P(B) - P(A \cap B) = \frac{1}{2} - \frac{1}{12} = \frac{5}{12}$$

Image to help visualize stuff

enter image description here

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  • $\begingroup$ How did you arrive to $P(A^\complement \cap B)$ = $P(B) - P(A \cap B)$. Is this a general rule? Or something that should be intuitive? $\endgroup$ – AznBoyStride Sep 6 '18 at 3:55
  • $\begingroup$ @AznBoyStride Take a look at the image that i just added, can you see why ? $\endgroup$ – Ahmad Bazzi Sep 6 '18 at 4:01
  • $\begingroup$ Yes makes perfect sense now! $\endgroup$ – AznBoyStride Sep 6 '18 at 4:10
  • 1
    $\begingroup$ Its just a rearrangement of the Law of Total Probability. $\mathsf P(B)=\mathsf P(A\cap B)+\mathsf P(A^\complement\cap B)$ $\endgroup$ – Graham Kemp Sep 6 '18 at 4:11

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