# How do you calculate cumulative probability? If one tosses a coin 3 times, what are the odds that 1 toss will be "heads"?

Dec 2, 2017

$\frac{3}{8}$

#### Explanation:

There are two ways I can think of:

Method 1

In order to have exactly one head in three rows, it means the other two throws are tails. Consider how this could happen:
The 'head could be on the first, second, or third throw.

$P \left(H , T , T\right) \mathmr{and} P \left(T , H , T\right) \mathmr{and} P \left(T , T , H\right)$

$P \left(H\right) = P \left(T\right) = \frac{1}{2}$

$P \left(H , T , T\right) \mathmr{and} P \left(T , H , T\right) \mathmr{and} P \left(T , T , H\right)$

$= \left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right) + \left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right) + \left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right)$

$= \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8}$

Method 2

Write down all the $8$ possible outcomes for $3$ throws:

$H H H$

$\textcolor{b l u e}{H T T , H T H , T T H}$

$T H H , T H T , H H T$

$T T T$

Of the $8$ outcomes there are $\textcolor{b l u e}{3}$ which have one head.

$P \left(\text{one head}\right) = \frac{3}{8}$