Question

a fair coin is flipped three times.
a) state the sample space. list outcomes separated by commas.
use notation like htt to mean the first toss was heads, the second toss was tails and the third toss was tails.
{ }
b) find the probability of getting all heads.

c) find the probability of getting at least 2 tails.

d) find the probability of getting an even number of tails.

e) find the probability of getting all heads or all tails.

f) find the probability of getting exactly two heads or exactly two tails.

g) find the probability of not getting at least 2 tails.

question help: video message instructor

Explanation:

Step1: Determine sample - space

Each coin - flip has 2 possibilities. For 3 flips, the sample space has $2\times2\times2 = 8$ elements. The sample space $S=\{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.

Step2: Probability formula

The probability of an event $E$ is $P(E)=\frac{n(E)}{n(S)}$, where $n(E)$ is the number of elements in event $E$ and $n(S)$ is the number of elements in the sample space $n(S) = 8$.

Step3: Probability of all heads

The event of getting all heads is $E_1=\{HHH\}$, $n(E_1)=1$. So $P(E_1)=\frac{1}{8}$.

Step4: Probability of at least 2 tails

The event of getting at least 2 tails is $E_2=\{HTT, THT, TTH, TTT\}$, $n(E_2)=4$. So $P(E_2)=\frac{4}{8}=\frac{1}{2}$.

Step5: Probability of an even number of tails

The event of getting an even number of tails is $E_3=\{HHH, HTH, THH, TTT\}$, $n(E_3)=4$. So $P(E_3)=\frac{4}{8}=\frac{1}{2}$.

Step6: Probability of all heads or all tails

The event of getting all heads or all tails is $E_4=\{HHH, TTT\}$, $n(E_4)=2$. So $P(E_4)=\frac{2}{8}=\frac{1}{4}$.

Step7: Probability of exactly 2 heads or exactly 2 tails

The event of getting exactly 2 heads is $\{HHT, HTH, THH\}$ and the event of getting exactly 2 tails is $\{HTT, THT, TTH\}$. The combined event $E_5=\{HHT, HTH, THH, HTT, THT, TTH\}$, $n(E_5)=6$. So $P(E_5)=\frac{6}{8}=\frac{3}{4}$.

Step8: Probability of not getting at least 2 tails

The event of not getting at least 2 tails is the complement of the event of getting at least 2 tails. If $E_2$ is the event of getting at least 2 tails, then the complement $\overline{E_2}$ has $n(\overline{E_2})=8 - 4=4$. So $P(\overline{E_2})=\frac{4}{8}=\frac{1}{2}$.

Answer:

a) HHH, HHT, HTH, THH, HTT, THT, TTH, TTT
b) $\frac{1}{8}$
c) $\frac{1}{2}$
d) $\frac{1}{2}$
e) $\frac{1}{4}$
f) $\frac{3}{4}$
g) $\frac{1}{2}$