Properties of set using probabilities

Question

If $P(A \cap (C \cup B^c))=0.2$, $P(A^c)=0.2$, and $P(A^c \cap B \cap C^c)=0.2$, find $P(B \cap C^c)$.

My attempt so far: $P(A)= 1-P(A^c) = 1-0.2=0.8$. $P(A \cap B^c \cap C)= 1- P(A^c \cap B \cap C^c)=1-0.2=0.8$. It appears I have been going in circles in search of $P(B \cap C^c)$. What am I missing?

Answer 1

Reminders of properties:

• $P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)$
• $P(X) = 1 - P(X^c)$
• $X = (X^c)^c$
• $(X\cup Y)^c = X^c\cap Y^c$
• $(X\cap Y)^c = X^c\cup Y^c$

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We know that $P(A^c\cap B\cap C^c) = 0.2$ so we know that $0.8 = 1 - P(A^c\cap B\cap C^c) = P((A^c\cap B\cap C^c)^c) = P(A\cup B^c\cup C)$

Next, let us apply some conveniently placed parentheses and then apply inclusion-exclusion on the result:

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

We recognize now that two of these were given to us in the problem details:

$ = 0.8 + P(B^c\cup C)-0.2$

Now... simplifying everything and moving things around this tells us that $P(B^c\cup C) = 0.2$

Now, its just one more step to get the value of $P(B\cap C^c)$