Chapter 7.3, Problem 7.78E

(a)

To determine

To Calculate: the probability of each possible outcome.

Explanation of Solution

Given:

T=Teacher wins round and

S=student wins round

Formula used:

Compliment Rule

  $\text{P}\left({\text{A}}^c\right)=1-\text{P}\left(\text{A}\right)$

Multiplication Rule

  $\text{P}\left(\text{A}\cap \text{B}\right)\text{=P}\left(\text{A}\text{and}\text{B}\right)\text{=P}\left(\text{A}\right)\text{×P}\left(\text{B}\right)$

Calculation:

  $P(T)=\frac{2}{3}$

Two people play a game and each person has different dice and the person who rolls the highest wins the round. First player to win 3 rounds will win the game.

Each round has two possible options: Teacher wins or Student win

The Sample Space contains all outcomes of game:

{TTT, TTST, TSTT, STTT, TTSST, TSTST, TSSTT, STTST, STSTT, SSTTT, SSS, SSTS, STSS, TSSS, SSTTS, STSTS, STSS, TSSS, SSTTS, STSTS, STTSS, TSSTS, TSTSS, TTSSS}

Complement rule:

  $P\left(S\right)=1-P\left(T\right)=1-\frac{2}{3}=\frac{1}{3}$

Multiplication rule for independent events

  $\begin{array}{c}P\left(TTT\right)=P\left(T\right)\times P\left(T\right)\times P\left(T\right)\\ =\frac{2}{3}\times \frac{2}{3}\times \frac{2}{3}\\ =\frac{8}{27}\\ =0.2963\end{array}$

  $\begin{array}{c}P\left(TTST\right)=P\left(T\right)\times P\left(T\right)\times P(S)\times P\left(T\right)\\ =\frac{2}{3}\times \frac{2}{3}\times \frac{1}{3}\times \frac{2}{3}\\ =\frac{8}{81}\\ =0.0988\end{array}$

  $\begin{array}{c}P\left(TTST\right)=P\left(TSTT\right)=P\left(STTT\right)\\ =\frac{8}{81}\\ =0.0988\end{array}$

  $\begin{array}{c}P\left(TTST\right)=P\left(T\right)\times P\left(T\right)\times P(S)\times P\left(S\right)\times P\left(T\right)\\ =\frac{2}{3}\times \frac{2}{3}\times \frac{1}{3}\times \frac{1}{3}\times \frac{2}{3}\\ =\frac{8}{243}\\ =0.0329\end{array}$

  $\begin{array}{c}P\left(TTSST\right)=P\left(TSTST\right)=P\left(TSSTT\right)=P\left(STTST\right)=P\left(STSTT\right)=P\left(SSTTT\right)\\ =\frac{8}{243}\\ =0.0329\end{array}$

  $\begin{array}{c}P\left(SSS\right)=P\left(S\right)\times P\left(S\right)\times P\left(S\right)\\ =\frac{1}{3}\times \frac{1}{3}\times \frac{1}{3}\\ =\frac{1}{27}\\ =0.0370\end{array}$

  $\begin{array}{c}P\left(SSTS\right)=P\left(S\right)\times P\left(S\right)\times P\left(T\right)\times P\left(S\right)\\ =\frac{1}{3}\times \frac{1}{3}\times \frac{2}{3}\times \frac{1}{3}\\ =\frac{2}{81}\\ =0.0247\end{array}$

  $\begin{array}{c}P\left(SSTS\right)=P\left(STSS\right)=P\left(TSSS\right)\\ =\frac{2}{81}\\ =0.0247\end{array}$

  $\begin{array}{c}P\left(SSTTS\right)=P\left(S\right)\times P\left(S\right)\times P\left(T\right)\times P\left(T\right)\times P\left(S\right)\\ =\frac{1}{3}\times \frac{1}{3}\times \frac{2}{3}\times \frac{2}{3}\times \frac{1}{3}\\ =\frac{4}{243}\\ =0.0165\end{array}$

(b)

To determine

To Calculate: the probability that the teacher wins the next game.

Explanation of Solution

Formula used:

Compliment Rule

  $\text{P}\left({\text{A}}^c\right)=1-\text{P}\left(\text{A}\right)$

Multiplication Rule

  $\text{P}\left(\text{A}\cap \text{B}\right)\text{=P}\left(\text{A}\text{and}\text{B}\right)\text{=P}\left(\text{A}\right)\text{×P}\left(\text{B}\right)$

Calculation:

The teacher will win the game, if the last in the event is a T

  $\begin{array}{c}\text{P}\left(\text{teacher}\text{wins}\text{the}\text{game}\right)\text{=P}\left(\text{TTT}\right)\text{+P}\left(\text{TTST}\right)\text{+P}\left(\text{TSTT}\right)\text{+P}\left(\text{STTT}\right)\\ \text{+P}\left(\text{TTSST}\right)\text{+P}\left(\text{TSTST}\right)\text{+P}\left(\text{TSSTT}\right)\\ \text{+P}\left(\text{STTST}\right)\text{+P}\left(\text{STSTT}\right)\text{+P}\left(\text{SSTTT}\right)\\ \text{=P}\left(\text{TTT}\right)\text{+3×P}\left(\text{TTST}\right)\text{+6×P}\left(\text{TTSST}\right)\\ \text{=}\frac{\text{8}}{\text{27}}\text{+3×}\frac{\text{8}}{\text{81}}\text{+6×}\frac{\text{8}}{\text{243}}\\ \text{=}\frac{\text{64}}{\text{81}}\\ \text{=0}\text{.7901}\end{array}$