Let $A$,$B$,$C$ be three events in a probability space. Suppose that \begin{align} \mathbb P(A) &= 0.5\\ \mathbb P(B) &= 0.3\\ \mathbb P(C) &= 0.2\\ \mathbb P(A\cap B) &= 0.15\\ \mathbb P(A\cap C) &= 0.1\\ \mathbb P(B\cap C) &= 0.06. \end{align} The smallest possible value of $P(A^\complement\cap B^\complement\cap C^\complement)$ is:
$(A)\ 0.31$
$(B)\ 0.25$
$(C)\ 0$
$(D)\ 0.26$
\begin{align} P(A^\complement\cap B^\complement\cap C^\complement)=1&-P(A\cup B\cup C)\\ =1&-P(A)-P(B)-P(C)\\ +&P(A\cap B)+P(B\cap C)+P(C\cap A)-P(A\cap B\cap C)\\ =1&-0.5-0.3-0.2+0.15+0.1+0.06-P(A\cap B\cap C)\\ =0&.31-P(A\cap B\cap C) \end{align}
Now I am stuck. Can someone guide me to the right answer $(B)$ given in my book?
Thanks
Your answer is great. In the last part, Note the fact that the probability of the intersection of A, B and C is less or equal than that of A and B. Similarly B,C or C,A are same. In your problem to minimize the probability, maximize P(ABC) with the above fact. Therefore the maximum of P(ABC) is P(BC)=0.06