# In a statistics class there are 11 juniors and 6 seniors; 4 of the seniors are females, 6 of the juniors are males. If a student is selected at random, what is the probability that the student is either a senior or a male?

Aug 10, 2016

65.74%

#### Explanation:

let $A$ be the event of being a senior and $B$ the event of male
We know
$p \left(A\right) = \frac{6}{17}$
$p \left(\neg B | A\right) = \frac{4}{6}$
$p \left(B | \neg A\right) = \frac{6}{11}$

What we can figure out is

$p \left(\neg A\right) = \frac{11}{17}$
$p \left(B | A\right) = \frac{2}{6}$
$p \left(\neg B | \neg A\right) = \frac{5}{11}$

We are then asked to sovle $p \left(A \mathmr{and} B\right) = p \left(A\right) + p \left(B\right) - p \left(A\right) \cdot p \left(B\right)$.
we just need to figure out $p \left(B\right)$ and solve. Based on the law of total probability $p \left(B\right) = {\sum}_{i}^{n} p \left(B , {A}_{i}\right)$ in our case

$p \left(B\right) = p \left(B | A\right) p \left(A\right) + p \left(B | \neg A\right) p \left(\neg A\right)$
$p \left(B\right) = \frac{2}{6} \cdot \frac{6}{17} + \frac{6}{11} \cdot \frac{11}{17} = \frac{8}{17}$

now we solve
$p \left(A \mathmr{and} B\right) = \frac{6}{17} + \frac{8}{17} - \frac{6}{17} \cdot \frac{8}{17} = .6574$