Tutorial Exercises

1

$$\frac{4}{52}=\frac{2}{26}=\frac{1}{13}$$

title: Correct answer.

2

1. $\frac{18}{38}$
2. $\frac{18}{38}\times \frac{18}{38}=\frac{324}{1444}=\frac{81}{361}$
3. $\frac{2}{38}=\frac{1}{19}$
4. ${\left(\begin{array}{c}1-\frac{2}{38}\end{array}\right)}^5=76.3\%$
5. $\frac{6}{38}\times \frac{38-6}{38}=\frac{192}{144}=\frac{48}{361}=13.3\%$

title: Correct answers.

3

Let $u$ be any of the four as likely numbers. Let $t$ be one of the two three times as likely numbers.

$$\begin{array}{rl}P(u)& =3P(t)\\ 2P(u)+4P(t)& =1\\ & \\ 2\times 3P(t)+4P(t)& =1\\ 10P(t)& =1\\ P(t)& =\frac{1}{10}\\ & \\ P(u)& =\frac{3}{10}\end{array}$$

1. Hence the probabilities are:
   • $P(1)=\frac{1}{10}$
   • $P(2)=\frac{3}{10}$
   • $P(3)=\frac{1}{10}$
   • $P(4)=\frac{3}{10}$
   • $P(5)=\frac{1}{10}$
   • $P(6)=\frac{1}{10}$
2. Probability $P(\text{even})=\frac{3}{10}+\frac{3}{10}+\frac{1}{10}=\frac{7}{10}$

title: Correct answers.

4

Everything is equally likely, hence the probability of any combination is $\frac{1}{4}$. The probability they have at least one boy is $\frac{3}{4}$, the probability they have two is $\frac{1}{4}$.

$$\begin{array}{rl}P(E|F)& =\frac{p(E\cap F)}{p(F)}\\ P(\text{two boys}|\text{have at least one boy})& =\frac{\frac{1}{4}}{\frac{3}{4}}\\ & =\frac{1}{3}\end{array}$$

See Conditional probability.

title: Correct answer.

5

$$\begin{array}{rl}p(\bar{E}|F)& =1-p(E|F)\\ p(\bar{E}|F)+p(E|F)& =1\\ \frac{p(\bar{E}\cap F)}{p(F)}+\frac{p(E\cap F)}{p(F)}& =1\\ \frac{p(\bar{E}\cap F)+p(E\cap F)}{p(F)}& =1\end{array}$$

6

Independence

1. $E_1:\{THH,THT,TTH,TTT\}$, $E_2:\{THT,HHT,THH,HHH\}$

   $E_1\cap E_2:\{THT,THH\}$

   $p(E_1\cap E_2)=\frac{2}{8}=\frac{1}{4}$

   $p(E_1)\cdot p(E_2)=\frac{4}{8}\cdot \frac{4}{8}=\frac{16}{64}=\frac{1}{4}$

   As $p(E_1\cap E_2)=p(E_1)\cdot p(E_2)$, $E_1$ and $E_2$ are independent events.

2. $E_1:\{THH,THT,TTH,TTT\}$, $E_2:\{HHT,THH\}$

   $E_1\cap E_2:\{THH\}$

   $p(E_1\cap E_2)=\frac{1}{8}$

   $p(E_1)\cdot p(E_2)=\frac{4}{8}\cdot \frac{2}{8}=\frac{1}{8}$

   As $p(E_1\cap E_2)=p(E_1)\cdot p(E_2)$, $E_1$ and $E_2$ are independent events.

3. $E_1:\{HTH,TTH,HTT,TTT\}$, $E_2:\{HHT,THH\}$

   $E_1\cap E_2:\varnothing$

   $p(E_1\cap E_2)=0$

   $p(E_1)\cdot p(E_2)=\frac{4}{8}\cdot \frac{2}{8}=\frac{1}{8}$

   As $p(E_1\cap E_2)\ne p(E_1)\cdot p(E_2)$, $E_1$ and $E_2$ are not independent events.

title: Correct answers.

7

1. $12\times {\frac{1}{12}}^2=\frac{1}{12}$

2. Let $E$ be the event that everyone has a birthday during a different month. Hence, there are $12\cdot 11\cdot 10\cdot 9=11880$ combinations.

   There is a total of $12^4$ possible combinations.

   Hence the event $E^{\prime }$ is that at least two people have birthdays in the same month.

   $$\begin{array}{rl}p(E^{\prime })& =1-p(E)\\ & =1-\frac{12\cdot 11\cdot 10\cdot 9}{12^4}\\ & =1-\frac{11\cdot 10\cdot 9}{12^3}\\ & =1-\frac{55}{96}\\ & =\frac{41}{96}\end{array}$$

title: Correct answers.

8

• Infected: $8\%$ $p(I)=0.08$ $p(\bar{I})=0.92$
• $98\%$ of those who have it test positive $p(T|I)=0.98$ $p(\bar{T}|I)=0.02$
• $3\%$ of those who don’t test positive $p(T|\bar{I})=0.03$ $p(\bar{T}|\bar{I})=0.97$

1. $p(I|T)$

   $$\begin{array}{rl}p(I|T)& =\frac{p(T|I)p(I)}{p(T|I)p(I)+p(T|\bar{I})p(\bar{I})}\\ & =\frac{0.98\cdot 0.08}{0.98\cdot 0.08+0.03\cdot 0.92}\\ & =0.74\end{array}$$

title: Correct answer.

2. $p(\bar{I}|T)$

   $$\begin{array}{rl}p(\bar{I}|T)& =1-p(I|T)\\ & =1-0.74\\ & =0.26\end{array}$$

title: Correct answer.

3. $p(I|\bar{T})$

   $$\begin{array}{rl}p(I|\bar{T})& =\frac{p(\bar{T}|I)p(I)}{p(\bar{T}|I)p(I)+p(\bar{T}|\bar{I})p(\bar{I})}\\ & =\frac{0.02\cdot 0.08}{0.02\cdot 0.08+0.97\cdot 0.92}\\ & =0.0018\end{array}$$

title: Correct answer.

4. $p(\bar{I}|\bar{T})$

   $$\begin{array}{rl}p(\bar{I}|\bar{T})& =1-p(I|\bar{T})\\ & =1-0.0018\\ & =0.998\end{array}$$

title: Correct answer.