Solution: 2018 Fall Final - 17

Author: Michiel Smid

Question

Let $A$ and $B$ be two independent events in some sample space $S$. You are given that $\Pr(A) = 1/4$ and $\Pr(B) = 2/3$. What is $\Pr(A \cup B)$?

(a)

5/6

(b)

1/3

(c)

2/3

(d)

3/4

The equation is $ Pr(A \cup B) = Pr(A) + Pr(B) - Pr(A \cap B) $

Since they’re independent, $ Pr(A \cap B) = Pr(A) \cdot Pr(B) $

$ Pr(A \cap B) = \frac{1}{4} \cdot \frac{2}{3} $

$ Pr(A \cap B) = \frac{2}{12} $

$ Pr(A \cap B) = \frac{1}{6} $

Now, we can find $ Pr(A \cup B) $

$ Pr(A \cup B) = Pr(A) + Pr(B) - Pr(A \cap B) $

$ Pr(A \cup B) = \frac{1}{4} + \frac{2}{3} - \frac{1}{6} $

$ Pr(A \cup B) = \frac{3}{12} + \frac{8}{12} - \frac{2}{12} $

$ Pr(A \cup B) = \frac{9}{12} $

$ Pr(A \cup B) = \frac{3}{4} $