COMP 2711 HW 3

1

a

Denote $W_i$ as the event that the prize is behind door $i$
Denote $M_i$ as the event that the organizer opens door $i$

Suppose we open door 1

$$P(W_1)=P(W_2)=P(W_3)=\frac{1}{3}$$ $$\begin{array}{rl}& P(M_2\mid W_1)=P(M_3\mid W_1)=\frac{1}{2}\\ & P(M_3\mid W_2)=1\\ & P(M_2\mid W_3)=1\end{array}$$

The probability for winning if we stick to the original choice is

$$\begin{array}{rl}P(W_1\mid M_2)& =\frac{P(M_2\mid W_1)P(W_1)}{P(M_2\mid W_1)P(W_1)+P(M_2\mid W_2)P(W_2)+P(M_2\mid W_3)P(W_3)}\\ & =\frac{\frac{1}{2}\times \frac{1}{3}}{\frac{1}{2}\times \frac{1}{3}+0\times \frac{1}{3}+1\times \frac{1}{3}}\\ & =\frac{1}{3}\end{array}$$

b

$$\begin{array}{rl}& E(\text{Stick})=\frac{1}{3}\times 3000+\frac{2}{3}\times 0=1000\\ & E(\text{Change})=\frac{2}{3}\times (3000-400)+\frac{1}{3}\times -400=1600\end{array}$$

It would be logical to change the door since $E(\text{Change})>E(\text{Stick})$

c

$$\begin{array}{rl}& E(\text{Stick})=\frac{1}{3}\cdot 3000+\frac{2}{3}\times 0=1000\\ & E(\text{Change})=\frac{2}{3}\times (3000-2100)+\frac{1}{3}\times -2100=-100\end{array}$$

It would be logical to stick to the original door since $E(\text{Stick})>E(\text{Change})$

d

Denote $W_i$ as the event that the prize is behind door $i$
Denote $D_i$ as the event that the wind opens door $i$

Suppose we chose door 1

$$\begin{array}{r}P(W_1)=P(W_2)=P(W_3)=\frac{1}{3}\end{array}$$ $$\begin{array}{rl}& P(D_2\mid W_1)=P(D_3\mid W_1)=\frac{1}{2}\\ & P(D_3\mid W_2)=\frac{1}{2}\\ & P(D_2\mid W_3)=\frac{1}{2}\end{array}$$

The probability for winning if we stick to the original choice is

$$\begin{array}{rl}P(W_1\mid D_2)& =\frac{P(D_2\mid W_1)P(W_1)}{P(D_2\mid W_1)P(W_1)+P(D_2\mid W_2)P(W_2)+P(D_2\mid W_3)P(W_3)}\\ & =\frac{\frac{1}{2}\times \frac{1}{3}}{\frac{1}{2}\times \frac{1}{3}+0\times \frac{1}{3}+\frac{1}{2}\times \frac{1}{3}}\\ & =\frac{1}{2}\end{array}$$ $$\begin{array}{rl}& E(\text{Stick})=\frac{1}{2}\times 3000+\frac{1}{2}\times 0=1500\\ & E(\text{Change})=\frac{1}{2}\times (3000-100)+\frac{1}{2}\times -100=1400\end{array}$$

It would be logical to stick to the original door since $E(\text{Stick})>E(\text{Change})$

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2

a

TT
HTT
HHTT
THTT
HHHTT
HTHTT
THHTT
HHHHTT
HHTHTT
HTHHTT
THHHTT
THTHTT

The expected number of coin tosses is

$$\begin{array}{rl}E(X)& =\frac{1}{2^2}\cdot 2+\frac{1}{2^3}\cdot 3+2\left(\frac{1}{2^4}\cdot 4\right)+3\left(\frac{1}{2^5}\cdot 5\right)+5\left(\frac{1}{2^6}\cdot 6\right)\\ & =2.3125\end{array}$$

b

TH
TTTH
TTTTTH
TTTTTTTH
TTTTTTTTTH
...

The probability that the first occurrence of “head” appears on an even-numbered toss is

$$\begin{array}{rl}P(E)& =\sum _{i=1}^{\infty }{\left(\frac{1}{2}\right)}^{2i}\\ & =\sum _{i=1}^{\infty }{\left(\frac{1}{4}\right)}^i\\ & =\frac{\frac{1}{4}}{1-\frac{1}{4}}\\ & =\frac{1}{3}\end{array}$$

c

Probability of changing from H to T is $\frac{1}{4}$
Probability of changing from T to H is $\frac{1}{4}$

Let $X_i$ be the random variable that indicates the coin changing sides from position $i$ to $i+1$

e.g. $X_1$ is the coin changing side from position 1 to 2 and $X_{n-1}$ is the coin changing side from position $n-1$ to $n$

Let $X$ be the random variable that indicates the number of side changes, i.e. $X=X_1+X_2+\cdots +X_{n-1}$

$$\begin{array}{rl}E(X)& =E(X_1+X_2+\cdots +X_{n-1})\\ & =E(X_1)+E(X_2)+\cdots +E(X_{n-1})\\ & =\sum _{i=1}^{n-1}\left(\frac{1}{4}+\frac{1}{4}\right)\\ & =(n-1)\left(\frac{1}{4}+\frac{1}{4}\right)\\ & =\frac{n-1}{2}\end{array}$$

d

Probability of changing from H to T is $p(1-p)$
Probability of changing from T to H is $(1-p)p$

$$\begin{array}{rl}E(X)& =\sum _{i=1}^{n-1}(p(1-p)+(1-p)p)\\ & =(n-1)(p(1-p)+(1-p)p)\\ & =2p(1-p)(n-1)\end{array}$$

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3

a

$$P(C=P\mid T=R)=0.6$$

b

$$\begin{array}{rl}P(C=S)& =P(C=S\cap T=R)+P(C=S\cap T=P)+P(C=S\cap T=S)\\ & =P(C=S\mid T=R)P(T=R)+P(C=S\mid T=P)P(T=P)\\ & +P(C=S\mid T=S)P(T=S)\\ & =(1-0.6)\times \frac{1}{3}+0.8\times \frac{1}{3}+0.8\times \frac{1}{3}\\ & =\frac{2}{3}\end{array}$$

c

Let $C$ be the event that Chisato wins in one turn of the game

$$\begin{array}{rl}P(C)& =P(C=P\cap T=R)+P(C=S\cap T=P)\\ & =P(C=P\mid T=R)P(T=R)+P(C=S\mid T=P)P(T=P)\\ & =0.6\times \frac{1}{3}+0.8\times \frac{1}{3}\\ & =\frac{7}{15}\end{array}$$

d

Let $T$ be the event that Chisato wins in one turn of the game

$$\begin{array}{rl}P(T)& =P(C=P\cap T=S)+P(C=S\cap T=R)\\ & =P(C=P\mid T=S)P(T=S)+P(C=S\mid T=R)P(T=R)\\ & =(1-0.8)\times \frac{1}{3}+(1-0.6)\times \frac{1}{3}\\ & =\frac{1}{5}\end{array}$$

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e

Let $I$ be the event that in one turn of the game the outcome is tie

$$P(I)=1-\frac{7}{15}-\frac{1}{5}=\frac{1}{3}$$

Let $\mathbb{C}$ be the event that Chisato wins at the end

$$\begin{array}{rl}P(\mathbb{C})& =P(C)+P(I)P(C)+P(I{)}^{2}P(C)+\dots \\ & =P(C)(1+P(I)+P(I{)}^2+\dots )\\ & =P(C)\left(\frac{1}{1-P(I)}\right)\\ & =\frac{7}{15}\left(\frac{1}{1-\frac{1}{3}}\right)\\ & =\frac{7}{10}\end{array}$$

f

$$\begin{array}{rl}E(X)& =np\\ & =365\times \frac{7}{10}\\ & =255.5\end{array}$$

g

$$\begin{array}{rl}V(X)& =np(1-p)\\ & =365\times \frac{7}{10}\times \left(1-\frac{7}{10}\right)\\ & =76.65\end{array}$$

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4

Denote every 2 letters as the ball chosen from box A and box B respectively at a certain step, e.g.

BR stands for the step

1. Choose Blue from A and Red from B

RBBR stands for the steps

1. Choose Red from A and Blue from B
2. Choose Blue from A and Red from B

a

After the first step, there is 1 case

1. BR

	Blue	Red
A	9	1
B	1	9

After the second step, there are 4 cases

1. RR

	Blue	Red
A	9	1
B	1	9

2. RB

	Blue	Red
A	10	0
B	0	10

3. BR

	Blue	Red
A	8	2
B	2	8

4. BB

	Blue	Red
A	9	1
B	1	9

The probability for each of these cases are

$$\begin{array}{r}P(RR)=\frac{1}{10}\frac{9}{10}=\frac{9}{100}\\ P(RB)=\frac{1}{10}\frac{1}{10}=\frac{1}{100}\\ P(BR)=\frac{9}{10}\frac{9}{10}=\frac{81}{100}\\ P(BB)=\frac{9}{10}\frac{1}{10}=\frac{9}{100}\end{array}$$

The expected value of the number of blue balls in box A afer the second step is

$$\begin{array}{rl}E(X)& =\frac{9}{100}\times 9+\frac{1}{100}\times 10+\frac{81}{100}\times 8+\frac{9}{100}\times 9\\ & =\frac{41}{5}\end{array}$$

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b

After the third step, there are 16 cases

1. RRRR, 9 blue balls in box A
2. RRRB, 10 blue balls in box A
3. RRBR, 8 blue balls in box A
4. RRBB, 9 blue balls in box A
5. RBRR, 10 blue balls in box A (impossible)
6. RBRB, 10 blue balls in box A (impossible)
7. RBBR, 9 blue balls in box A
8. RBBB, 10 blue balls in box A (impossible)
9. BRRR, 8 blue balls in box A
10. BRRB, 9 blue balls in box A
11. BRBR, 7 blue balls in box A
12. BRBB, 8 blue balls in box A
13. BBRR, 9 blue balls in box A
14. BBRB, 10 blue balls in box A
15. BBBR, 8 blue balls in box A
16. BBBB, 9 blue balls in box A

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The probability for each of these cases are

$$\begin{array}{rl}& P(\text{RRRR})=\frac{9}{100}\frac{1}{10}\frac{9}{10}=\frac{81}{10000}\\ & P(\text{RRRB})=\frac{9}{100}\frac{1}{10}\frac{1}{10}=\frac{9}{10000}\\ & P(\text{RRBR})=\frac{9}{100}\frac{9}{10}\frac{9}{10}=\frac{729}{10000}\\ & P(\text{RRBB})=\frac{9}{100}\frac{9}{10}\frac{1}{10}=\frac{81}{10000}\\ & \\ & P(\text{RBRR})=\frac{1}{100}\frac{0}{10}\frac{10}{10}=\frac{0}{10000}\\ & P(\text{RBRB})=\frac{1}{100}\frac{0}{10}\frac{0}{10}=\frac{0}{10000}\\ & P(\text{RBBR})=\frac{1}{100}\frac{10}{10}\frac{10}{10}=\frac{100}{10000}\\ & P(\text{RBBB})=\frac{1}{100}\frac{10}{10}\frac{0}{10}=\frac{0}{10000}\\ & \\ & P(\text{BRRR})=\frac{81}{100}\frac{2}{10}\frac{8}{10}=\frac{1296}{10000}\\ & P(\text{BRRB})=\frac{81}{100}\frac{2}{10}\frac{2}{10}=\frac{324}{10000}\\ & P(\text{BRBR})=\frac{81}{100}\frac{8}{10}\frac{8}{10}=\frac{5184}{10000}\\ & P(\text{BRBB})=\frac{81}{100}\frac{8}{10}\frac{2}{10}=\frac{1296}{10000}\\ & \\ & P(\text{BBRR})=\frac{9}{100}\frac{1}{10}\frac{9}{10}=\frac{81}{10000}\\ & P(\text{BBRB})=\frac{9}{100}\frac{1}{10}\frac{1}{10}=\frac{9}{10000}\\ & P(\text{BBBR})=\frac{9}{100}\frac{9}{10}\frac{9}{10}=\frac{729}{10000}\\ & P(\text{BBBB})=\frac{9}{100}\frac{9}{10}\frac{1}{10}=\frac{81}{10000}\end{array}$$

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The expected value of the number of blue balls in box A after the third step is

$$\begin{array}{rl}E(X)& =\frac{81}{10000}\times 9+\frac{9}{10000}\times 10+\frac{729}{10000}\times 8+\frac{81}{10000}\times 9\\ & \\ & +\frac{0}{10000}\times 10+\frac{0}{10000}\times 10+\frac{100}{10000}\times 9+\frac{0}{10000}\times 10\\ & \\ & +\frac{1296}{10000}\times 8+\frac{324}{10000}\times 9+\frac{5184}{10000}\times 7+\frac{1296}{10000}\times 8\\ & \\ & +\frac{81}{10000}\times 9+\frac{9}{10000}\times 10+\frac{729}{10000}\times 8+\frac{81}{10000}\times 9\\ & \\ & =\frac{189}{25}\end{array}$$