We are given that $P(A)=.43, P(B)=.44,$ and $P(C)=.19$. Also, $B$ and $C$ are mutually exclusive but $P(A\cap B)=.07$ and $P(A\cap C)= .13$

(a) What is $P(A' \cap B')$?

Note: I know that since $A,B$ are not mutually exclusive, that $P(A \cap B)=P(A) \times P(B)$. Does this same rule apply for the complement? If so, then I can use the rule $P(A)+P(A')=1$.

(b) What is $P(A \cup B \cup C)$?

I know the formula: $P(A\cup B\cup C) = P(A) +P(B)+P(C)-P(A\cap B)-P(A\cap C)- P(B\cap C)+P(A\cap B\cap C)$.

Note: The part I am having trouble on with this formula is finding $P(A\cap B\cap C)$.



$$P(A) \times P(B) \approx 0.1892 \ne P(A \cap B)$$

$A$ and $B$ are not independent.

  • Note that $$P(A' \cap B') = 1-P(A \cup B)$$

  • $ (A \cap B \cap C) \subset (B \cap C)$, Hence $P(A \cap B \cap C)=0$.

  • Also, check your inclusion-exclusion formula.

  • $\begingroup$ that is De Morgan's law. $\endgroup$ – Siong Thye Goh Sep 10 '18 at 16:14
  • $\begingroup$ I know that De Morgan's Law states: $(A\cup B)' = A' \cap B'$ and $(A\cap B)' = A' \cup B'$. How does this translate into that probability formula? $\endgroup$ – rover2 Sep 10 '18 at 16:18
  • $\begingroup$ note: i have fixed my inclusion-exclusion formula. i copied it wrong while typing the solution however it was correct in my notes $\endgroup$ – rover2 Sep 10 '18 at 16:19
  • $\begingroup$ Since $A' \cap B' = (A \cup B)' $ we have $P(A' \cap B')= P((A \cup B)' )$. We have $P((A \cup B)' ) = 1-P(A \cup B)$. $\endgroup$ – Siong Thye Goh Sep 10 '18 at 16:20
  • $\begingroup$ got it, thank you $\endgroup$ – rover2 Sep 10 '18 at 16:22

You know that $B$ and $C$ are mutually exclusive i.e. $A\cap B$ and $C$ are mutually exclusive,

$$P(A\cap B \cap C)=0$$

Also $$P(A\cap B)\neq P(A)\cdot P(B)$$

As $A$ and $B$ are not independent events.

For Part (II),

Using De-Morgans Law,

$$P(A'\cap B')=P((A\cup B)')=1-P(((A\cup B)')')$$

$$P(A'\cap B')=1-P(A\cup B)$$

Also you know that,

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


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