if A, B, C are 3 equally likely events, what is the smallest value for P(A) such that P ( A ∩ B ∩ C ) always exceeds 0.95?
$\displaystyle P(A \cap B \cap C)>0.95
\iff P(A^c \cup B^c \cup C^c) \leq 0.05 $ (use DeMorgan's Laws)
To find the smallest value of $\displaystyle P(A)$ is equivalent to finding the largest value of $\displaystyle P(A^c)$.
Now the union of these three equal sets is largest when they are all disjoint (This can be proven formally using $\displaystyle P(A \cup B) = P(A) + P(B) - P(A \cap B)$.)
Hence, the largest value of $\displaystyle P(A^c)$ is $\displaystyle \frac{0.05}{3}$, and it follows that the smallest value of $\displaystyle P(A)$ is $\displaystyle (1 - \frac{0.05}{3})$.
Since the events are independent, $\displaystyle \Pr(A \cap B \cap C) = \Pr(A) \cdot \Pr(B) \cdot \Pr(C)$.
Since the events are equally likely, $\displaystyle \Pr(A) = \Pr(B) = \Pr(C) = p$, say.
Therefore $\displaystyle \Pr(A \cap B \cap C) = p^3$. You require the value of p such that $\displaystyle p^3 = 0.95$.