Question

A coin is flipped eight times where each flip comes up either heads or tails.

How many possible outcomes $A)$ are there in total? $B)$ contain exactly three heads? $C)$ contains at least three heads? $D)$ contain the same number of heads and tails?

Answer 1

Multiplication rule of probability:
Consider two independent events associated with the same experiment.

The probability of the two events occurring in the same random experiment is the product of the probability of the first event occurring individually and the probability of the second event occurring individually.

A) The total number of possible outcomes:
Tossing each coin gives two possible outcomes: Heads, or tails
A coin is flipped eight times.
The possible outcomes for tossing each coin is $2$.

Total number of possible outcomes $=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2$

$=2^8$

$=256$

B) Contains exactly three heads:

$8$ coins are tossed.

Need to calculate the possible ways to get exactly $3$ heads.

Exactly $3$ coins mean $5$ tails exactly.

Use the combination formula to calculate the possible outcomes that contain exactly three heads.

${}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}=\frac{\mathrm{n}!}{\mathrm{r}!(\mathrm{n}-\mathrm{r})!}$

Where $\mathrm{n}$ is the total number of coins and $\mathrm{r}$ is the possible ways to get exactly $3$ heads.

$\begin{array}{rcl}{{}^{}}^8{\mathrm{C}}_3& =& \frac{8!}{3!(8-3)!}\\ & =& \frac{8!}{3!5!}\\ & =& 56\end{array}$

C) Contains at least three heads:

Total possible outcomes$=256$

Find the possible outcomes for the number of heads $0$, $1$ and $2$.

Number of outcomes with zero heads $={}^8{\mathrm{C}}_0$

$=1$

Number of outcomes with one head $={}^8{\mathrm{C}}_1$

$=8$

Number of outcomes with one head $={}^8{\mathrm{C}}_2$

$=\frac{8!}{2!6!}$

$=28$

Possible outcomes contain at least three heads $=256-(1+8+28)$

$$\begin{gathered} =256-37 \\ =219 \end{gathered}$$

D) Contain the same number of heads and tails

The same number of heads and tails is the possible outcome with $4$ heads and $4$ tails.

Possible outcomes for the same number of heads and tails $={}^8{\mathrm{C}}_4$

$$\begin{gathered} =\frac{8!}{4!(8-4)!} \\ =\frac{8!}{4!4!} \\ =70 \end{gathered}$$

Hence, the possible outcomes are there in total is $256$ .

The possible outcomes that contain exactly three heads is $56$.

The possible outcomes that contain at least three heads is $219$.

The possible outcomes that contain the same number of heads and tails is $70$.