$F=(A^c \cup B \cup C)\cap(A\cup C^c)$

Question

If the independently held events A, B e C are defined in the same probability space. Calculate P(F) with $F=(A^c \cup B \cup C)\cap(A\cup C^c)$, $P(A)=0,2$ , $P(B)=0,3$ , $P(C)=0,4$.

I am having problems with the simplification of the expression.

Answer 1

It's easier to consider the probability of $F^c$ instead of $F$, which can be calculated by de Morgan's laws: \begin{align*} F^c &= [(A^c \cup B \cup C) \cap (A \cup C^c)]^c \\ &= (A^c \cup B \cup C)^c \cup (A \cup C^c)^c \\ &= (A \cap B^c \cap C^c) \cup (A^c \cap C). \end{align*} The events $A \cap B^c \cap C^c$ and $A^c \cap C$ are mutually exclusive and have respective probabilities $0.2 \times 0.7 \times 0.6 = 0.084$ and $0.2 \times 0.4 = 0.08$, so $\operatorname{Pr}(F) = 1 - 0.084 - 0.08 = 0.836$.

Answer 2

$P(F)$ is not determined by $P(A),P(B),P(C)$. Consider a uniform probability space with $10$ elements (aka $A$ has $2$ elements, $B$ has $3$ and $C$ has $4$). $F$ can have different number of elements (so different probabilities):

• consider $A = \{1,2\} \subset B = \{1,2,3\}\subset C= \{1,2,3,4\}$. Then:

$$F =(\{3,\ldots,10\} \cup \{1,2,3\} \cup \{1,2,3,4\}) \cap (\{1,2\} \cup \{5, \ldots 10\}) = \\ = \{1,2\} \cup \{5, \ldots 10\} \Rightarrow P(F) = 0.8$$

• consider $A = \{1,2\}, B = \{3,4,5\}, C= \{6,7,8,9\}$. Then:

$$F = (\{3,\ldots,10\} \cup \{3,4,5\} \cup \{6,7,8,9\}) \cap (\{1,2\} \cup \{1, \ldots 5,10\}) = \\ = \{3,4,5,10\} \Rightarrow P(F)= 0.4$$

Another example is $A = \{1,2\}, B=\{2,3,4\}, C = \{4,5,6,7\}$.

EDIT: If you add hypothesis on $A,B,C$, then $P(F)$ might be determined.

$$F = (A^C \cup B \cup C) \cap (A \cup C^C) = [(A^C \cup B \cup C)\cap A] \cup [(A^C \cup B \cup C)\cap C^C] = \\ = (A^C \cap A) \cup (B \cap A) \cup (C \cap A) \cup (A^C \cap C^C) \cup (B \cap C^C) \cup (C \cap C^C) = \\ = (B \cap A) \cup (C \cap A) \cup (A^C \cap C^C) \cup (B \cap C^C)$$

It is easy to check that $B \cap A \subseteq (C \cap A) \cup (A^C \cap C^C) \cup (B \cap C^C)$ then

$$F = (C \cap A) \cup (A^C \cap C^C) \cup (B \cap C^C)$$

Remembering that $P(X\cup Y) = P(X) + P(Y) - P(X\cap Y)$, it is possible to split the probabilities:

$$P(F) = P(C \cap A) + P((A^C \cap C^C) \cup (B \cap C^C)) - P((C \cap A) \cap [(A^C \cap C^C) \cup (B \cap C^C)]) = \\ = P(C \cap A) + P((A^C \cap C^C) \cup (B \cap C^C)) = \\ = P(C \cap A) + P(A^C \cap C^C) + P(B \cap C^C) - P((A^C \cap C^C) \cap (B \cap C^C)) = \\ = P(A \cap C) + P(A^C \cap C^C) + P(B \cap C^C) - P(A^C \cap B \cap C^C)$$

Now $P(B \cap C^C) - P(A^C \cap B \cap C^C) = P(A \cap B \cap C^C)$, so

$$P(F) = P(A \cap C) + P(A^C \cap C^C) + P(A \cap B \cap C^C)$$

If you assume $A,B,C$ pairwise independent, you can have problems with $P(A \cap B \cap C^C)$. If $P(A \cap B \cap C) = P(A)P(B)P(C)$ (mutual independence), then $P(A \cap B \cap C^C) = P(A \cap B) - P(A \cap B \cap C)$ and you can conclude.