Suppose S and T are mutually exclusive events. Find P(S or T) if P(S) = 1/3 and P(T) = 5/12?

3/4 7/18 5/36

Apr 20, 2017

In an OR situation you may ADD

Explanation:

$P \left(S \mathmr{and} T\right) = P \left(S\right) + P \left(T\right) = \frac{1}{3} + \frac{5}{12} = \frac{4}{12} + \frac{5}{12} = \frac{9}{12} = \frac{3}{4}$

Apr 24, 2017

$P \left(S \cup T\right) = \frac{3}{4}$

Explanation:

We use the following fundamental definition from Probability and Set Theory:

$P \left(A \cup B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cap B\right)$

And if we apply this to our problem we have:

$P \left(S \cup T\right) = P \left(S\right) + P \left(T\right) - P \left(S \cap T\right)$
$\text{ } = \frac{1}{3} + \frac{5}{12} - P \left(S \cap T\right)$
$\text{ } = \frac{3}{4} - P \left(S \cap T\right)$

Now we are told that $S$ and $T$ are mutually exclusive, and so:

$P \left(S \cap T\right) = 0$

Hence,

$P \left(S \cup T\right) = \frac{3}{4}$