# Thread: equally likely events

1. ## equally likely events

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?

2. $P(A \cap B \cap C)>0.95
\iff P(A^c \cup B^c \cup C^c) \leq 0.05$
(use DeMorgan's Laws)

To find the smallest value of $P(A)$ is equivalent to finding the largest value of $P(A^c)$.

Now the union of these three equal sets is largest when they are all disjoint (This can be proven formally using $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.)

Hence, the largest value of $P(A^c)$ is $\frac{0.05}{3}$, and it follows that the smallest value of $P(A)$ is $(1 - \frac{0.05}{3})$.

3. Originally Posted by stressedout
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?
Since the events are independent, $\Pr(A \cap B \cap C) = \Pr(A) \cdot \Pr(B) \cdot \Pr(C)$.
Since the events are equally likely, $\Pr(A) = \Pr(B) = \Pr(C) = p$, say.

Therefore $\Pr(A \cap B \cap C) = p^3$. You require the value of p such that $p^3 = 0.95$.

4. Originally Posted by mr fantastic
Since the events are independent, $\Pr(A \cap B \cap C) = \Pr(A) \cdot \Pr(B) \cdot \Pr(C)$.
Since the events are equally likely, $\Pr(A) = \Pr(B) = \Pr(C) = p$, say.

Therefore $\Pr(A \cap B \cap C) = p^3$. You require the value of p such that $p^3 = 0.95$.
Hmm I think I'm missing something - why are the events independent?

5. Originally Posted by h2osprey
Hmm I think I'm missing something - why are the events independent?
My mistake. You haven't missed anything.