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

Question

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

The question says it all. I know $$P(A \cap (B \cup C))=P(A \cap B)+P(A \cap C).$$ Would this mean, $$P(A \cup(B \cap C))=P(A \cup B)+P(A \cup C)?$$ Just want to make sure.

Answer 1

The equation $$ P(A\cup (B\cap C))=P(A\cup B)+P(A\cup C) $$ will almost certainly not hold. This is because $$ P(A\cup (B \cap C))=P((A\cup B)\cap (A\cup C))\leq P(A\cup B) $$ and similarly $$ P(A\cup (B \cap C))\leq P(A\cup C). $$ For the stated equation to hold, both $P(A\cup B)$ and $P(A\cup C)$ would have to be zero, showing that $$ P(A)=P(B)=P(C)=0. $$ However, if this condition is satisfied, then the stated equation will hold.

Answer 2

Things to remember:

• Intersection "distributes" over unions.

$$X\cap (Y\cup Z) = (X\cap Y)\cup (X\cap Z)$$

Note: This should remind you how $a\cdot (b+c)=(a\cdot b)+(a\cdot c)$ with multiplication and addition.

• Unions also distribute over intersections.

$$X\cup (Y\cap Z) = (X\cup Y)\cap (X\cup Z)$$

• Probabilities of a union of events can expand via inclusion-exclusion:

$$Pr(X\cup Y)=Pr(X)+Pr(Y)-Pr(X\cap Y)$$

Note: when $X\cap Y$ occurs with probability zero, for example when $X$ and $Y$ are mutually exclusive/disjoint events then this simplifies as $Pr(X\cup Y)=Pr(X)+Pr(Y)$. In general however this is not the case and their intersection is non-trivial so you must not forget the final term being subtracted.

• Probabilities of an intersection can expand via product rule:

$$Pr(X\cap Y)=Pr(X)\times Pr(Y\mid X)$$

Note: The term on the right, $Pr(Y\mid X)$ is a conditional probability and is in general not equal to $Pr(Y)$. Only when $X$ and $Y$ are independent events will you have $Pr(Y\mid X)$ can simplify to $Pr(Y)$ in which case the above can be rewritten to be $Pr(X\cap Y)=Pr(X)\times Pr(Y)$. Without knowledge of whether or not $X$ and $Y$ are dependent on one another or independent of one another, we cannot make that simplification.

So, in other words, union inside of probability acts "like" addition outside of probability while intersection inside of probability acts "like" multiplication outside of probability... though again with those extra considerations and warnings mentioned above.

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Now, we have that

$$\begin{array}{rll}Pr(A\cap (B\cup C)) &= Pr((A\cap B)\cup (A\cap C))&\text{by distributivity}\\&=Pr(A\cap B)+Pr(A\cap C)-Pr((A\cap B)\cap (A\cap C))&\text{by inclusion-exlcusion}\\&=Pr(A\cap B)+Pr(A\cap C)-Pr(A\cap B\cap C)&\text{by associativity and absorption}\end{array}$$

So, when you say "I know that...", if that was your attempt to cite some identity that you learned, you learned it incorrectly and had forgotten to subtract the final piece.

If you say "I know that..." as though it were a given hypothesis, then by the above we would learn that $Pr(A\cap B\cap C)$ is equal to zero by comparing what it should have been equal to against what you are told it was equal to.

As for the other identity,

$$\begin{array}{rll}Pr(A\cup (B\cap C))&=Pr(A)+Pr(B\cap C)-Pr(A\cap (B\cap C))&\text{by inclusion-exclusion}\\&=Pr(A)+Pr(B\cap C)-Pr(A\cap B\cap C)&\text{by associativity}\end{array}$$

You could have if you insisted broken apart a different way:

$$\begin{array}{rll}Pr(A\cup (B\cap C))&=Pr((A\cup B)\cap (A\cup C))\\&=Pr(A\cup B)\times Pr((A\cup C)\mid (A\cup B))\end{array}$$

but this seems very unhelpful.