Solutions for Probability

1. Sample Space and Events: Rolling two fair six-sided dice. A = sum is 7, B = first die is 4.

(a) The sample space $\Omega$ consists of all possible pairs of outcomes $(d_1,d_2)$ where $d_1$ is the result of the first die and $d_2$ is the result of the second die. $\Omega =\{(1,1),(1,2),...,(1,6),$ $(2,1),(2,2),...,(2,6),$ $...$ $(6,1),(6,2),...,(6,6)\}$ The total number of outcomes is $|\Omega |=6\times 6=36$.

(b) Event A (sum is 7): The outcomes are pairs $(d_1,d_2)$ such that $d_1+d_2=7$. $A=\{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\}$. The number of outcomes in A is $|A|=6$. Since the dice are fair, all outcomes in $\Omega$ are equally likely. $P(A)=\frac{|A|}{|\Omega |}=\frac{6}{36}=\frac{1}{6}$.

(c) Event B (first die shows a 4): The outcomes are pairs $(d_1,d_2)$ such that $d_1=4$. $B=\{(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\}$. The number of outcomes in B is $|B|=6$. $P(B)=\frac{|B|}{|\Omega |}=\frac{6}{36}=\frac{1}{6}$.

(d) $A\cap B$ is the event where the sum is 7 AND the first die is 4. The only outcome satisfying both conditions is $(4,3)$. $A\cap B=\{(4,3)\}$. $P(A\cap B)=\frac{|A\cap B|}{|\Omega |}=\frac{1}{36}$. $P(A\cup B)$ is the probability that the sum is 7 OR the first die is 4. We use the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. $P(A\cup B)=\frac{1}{6}+\frac{1}{6}-\frac{1}{36}=\frac{6}{36}+\frac{6}{36}-\frac{1}{36}=\frac{11}{36}$.

---

2. Axioms and Conditional Probability: Box: 5 Red (R), 3 Blue (B). Total = 8. Two balls drawn without replacement.

(a) Let $R_1$ be the event the first ball is red, and $B_2$ be the event the second ball is blue. We want $P(B_2|R_1)$. If the first ball drawn was red, there are 7 balls remaining: 4 Red and 3 Blue. The probability that the second ball drawn is blue is $P(B_2|R_1)=\frac{\text{Number of Blue balls remaining}}{\text{Total number of balls remaining}}=\frac{3}{7}$.

(b) Let $R_1$ be the event the first ball is red, and $R_2$ be the event the second ball is red. We want $P(R_1\cap R_2)$. Using the multiplication rule for conditional probability: $P(R_1\cap R_2)=P(R_2|R_1)P(R_1)$. $P(R_1)=\frac{5}{8}$. Given the first was red, there are 4 red balls left out of 7 total. So, $P(R_2|R_1)=\frac{4}{7}$. $P(R_1\cap R_2)=\frac{4}{7}\times \frac{5}{8}=\frac{20}{56}=\frac{5}{14}$.

(c) Let $E_R$ be the event “first ball red” and $E_B$ be the event “first ball blue”. This refers to the sample space of the first draw. Let this sample space be ${\Omega }_1=\{R,B\}$. $P(E_R)=5/8$ and $P(E_B)=3/8$.

• Axiom 1 (Non-negativity): $P(E_R)=5/8\ge 0$ and $P(E_B)=3/8\ge 0$. Satisfied.
• Axiom 2 (Normalization): The sample space for the first draw is ${\Omega }_1=E_R\cup E_B$. $P({\Omega }_1)=P(E_R)+P(E_B)$ (since $E_R$ and $E_B$ are mutually exclusive). $P({\Omega }_1)=5/8+3/8=8/8=1$. Satisfied.
• Axiom 3 (Additivity): For mutually exclusive events (like $E_R$ and $E_B$ in ${\Omega }_1$), the probability of their union is the sum of their probabilities. This was used in verifying Axiom 2. Satisfied.

---

3. Law of Total Probability: Urn 1 (U1): 2W, 3B (Total 5). Urn 2 (U2): 4W, 1B (Total 5). An urn is chosen at random, $P(U1)=P(U2)=0.5$. Let W be the event that the drawn ball is white. We need $P(W)$. Using the Law of Total Probability: $P(W)=P(W|U1)P(U1)+P(W|U2)P(U2)$. From Urn 1, $P(W|U1)=\frac{2}{5}$. From Urn 2, $P(W|U2)=\frac{4}{5}$. $P(W)=\left(\frac{2}{5}\right)(0.5)+\left(\frac{4}{5}\right)(0.5)=\frac{1}{5}+\frac{2}{5}=\frac{3}{5}=0.6$.

---

4. Bayes’ Theorem: Using the setup from Problem 3. A white ball (W) was drawn. What is the probability it came from Urn 1 ($U1$)? We need $P(U1|W)$. Using Bayes’ Theorem: $P(U1|W)=\frac{P(W|U1)P(U1)}{P(W)}$. From Problem 3, we have: $P(W|U1)=2/5$ $P(U1)=0.5$ $P(W)=3/5$ $P(U1|W)=\frac{(2/5)(0.5)}{3/5}=\frac{1/5}{3/5}=\frac{1}{3}$.

---

5. Independent Events: Given $P(A)=0.4$, $P(B)=0.5$, $P(A\cup B)=0.7$. Events A and B are independent if $P(A\cap B)=P(A)P(B)$. First, find $P(A\cap B)$ using the inclusion-exclusion principle: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ $0.7=0.4+0.5-P(A\cap B)$ $0.7=0.9-P(A\cap B)$ $P(A\cap B)=0.9-0.7=0.2$. Now check the condition for independence: $P(A)P(B)=(0.4)(0.5)=0.20$. Since $P(A\cap B)=0.2$ is equal to $P(A)P(B)=0.2$, the events A and B are independent.

---

6. Binomial Distribution: Coin flipped $n=5$ times, $P(H)=p=0.6$. $X$ = number of heads. $X\sim Bin(n=5,p=0.6)$.

(a) Probability of getting exactly $k=3$ heads: $P(X=k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k(1-p{)}^{n-k}$ $P(X=3)=\left(\genfrac{}{}{0ex}{}{5}{3}\right)(0.6{)}^3(1-0.6{)}^{5-3}=\left(\genfrac{}{}{0ex}{}{5}{3}\right)(0.6{)}^3(0.4{)}^2$ $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=\frac{5!}{3!2!}=\frac{5\times 4}{2\times 1}=10$. $P(X=3)=10\times (0.216)\times (0.16)=10\times 0.03456=0.3456$.

(b) Expected number of heads: $E[X]=np=5\times 0.6=3$.

(c) Variance of the number of heads: $Var(X)=np(1-p)=5\times 0.6\times (1-0.6)=5\times 0.6\times 0.4=3\times 0.4=1.2$.

---

7. Geometric Distribution: Free throws success probability $p=0.7$. $Y$ = number of attempts needed for the first success. $Y\sim Geom(p=0.7)$.

(a) Probability that the first success occurs on the $k=3$rd attempt: $P(Y=k)=(1-p{)}^{k-1}p$ $P(Y=3)=(1-0.7{)}^{3-1}(0.7)=(0.3{)}^2(0.7)=0.09\times 0.7=0.063$.

(b) Expected number of attempts needed: $E[Y]=\frac{1}{p}=\frac{1}{0.7}=\frac{10}{7}\approx 1.4286$.

---

8. Poisson Distribution: Average rate $\lambda =4$ customers per hour. Let $X$ be the number of arrivals in a given hour. $X\sim Pois(\lambda =4)$.

(a) Probability that exactly $k=2$ customers arrive: $P(X=k)=\frac{e^{-\lambda }{\lambda }^k}{k!}$ $P(X=2)=\frac{e^{-4}{4}^2}{2!}=\frac{e^{-4}\times 16}{2}=8e^{-4}$. (Approximate value: $8\times 0.0183156\approx 0.1465$)

(b) Probability that at least one customer arrives: $P(X\ge 1)=1-P(X=0)$. $P(X=0)=\frac{e^{-4}{4}^0}{0!}=\frac{e^{-4}\times 1}{1}=e^{-4}$. $P(X\ge 1)=1-e^{-4}$. (Approximate value: $1-0.0183156\approx 0.9817$)

(c) Mean and variance of the number of arrivals per hour: For a Poisson distribution, the mean and variance are both equal to the rate parameter $\lambda$. $E[X]=\lambda =4$. $Var(X)=\lambda =4$.

---

9. Uniform Distribution: $X$ is uniformly distributed on $[a,b]=[5,15]$.

(a) Probability Density Function (PDF): $f_X(x)=\{\begin{array}{ll}\frac{1}{b-a}& \text{for }a\le x\le b\\ 0& \text{otherwise}\end{array}$ $f_X(x)=\{\begin{array}{ll}\frac{1}{15-5}=\frac{1}{10}& \text{for }5\le x\le 15\\ 0& \text{otherwise}\end{array}$

(b) Cumulative Distribution Function (CDF): $F_X(x)=P(X\le x)={\int }_{-\infty }^x{f}_X(t)dt$. $F_X(x)=\{\begin{array}{ll}0& \text{for }x<5\\ {\int }_5^x\frac{1}{10}dt=\frac{x-5}{10}& \text{for }5\le x\le 15\\ 1& \text{for }x>15\end{array}$

(c) Calculate $P(7\le X\le 12)$: Using the CDF: $P(7\le X\le 12)=F_X(12)-F_X(7)=\frac{12-5}{10}-\frac{7-5}{10}=\frac{7}{10}-\frac{2}{10}=\frac{5}{10}=0.5$. Alternatively, using the PDF: $P(7\le X\le 12)={\int }_7^{12}\frac{1}{10}dx=\frac{1}{10}[x{]}_7^{12}=\frac{1}{10}(12-7)=\frac{5}{10}=0.5$.

(d) Calculate $E[X]$ and $Var(X)$: $E[X]=\frac{a+b}{2}=\frac{5+15}{2}=\frac{20}{2}=10$. $Var(X)=\frac{(b-a{)}^2}{12}=\frac{(15-5{)}^2}{12}=\frac{10^2}{12}=\frac{100}{12}=\frac{25}{3}\approx 8.333$.

---

10. Exponential Distribution: Lifetime $X$ follows an exponential distribution with parameter $\lambda =0.2$. $X\sim Exp(0.2)$.

(a) Probability Density Function (PDF): $f(x)=\{\begin{array}{ll}\lambda e^{-\lambda x}& \text{for }x\ge 0\\ 0& \text{for }x<0\end{array}$ $f(x)=\{\begin{array}{ll}0.2e^{-0.2x}& \text{for }x\ge 0\\ 0& \text{for }x<0\end{array}$

(b) Probability that the component lasts more than 5 years, $P(X>5)$: $P(X>x)=e^{-\lambda x}$ (Survival function) $P(X>5)=e^{-0.2\times 5}=e^{-1}\approx 0.3679$.

(c) Expected lifetime $E[X]$: $E[X]=\frac{1}{\lambda }=\frac{1}{0.2}=5$ years.

(d) Using the memoryless property: $P(X>8|X>5)$. The memoryless property states $P(X>s+t|X>s)=P(X>t)$. Here $s=5$ and $t=3$. $P(X>5+3|X>5)=P(X>3)$. $P(X>3)=e^{-0.2\times 3}=e^{-0.6}\approx 0.5488$.

---

11. Normal Distribution: $X\sim N(\mu =10,{\sigma }^2=4)$. This means the standard deviation is $\sigma =\sqrt{4}=2$.

(a) Standardize the random variable $X$: $Z=\frac{X-\mu }{\sigma }=\frac{X-10}{2}$. $Z$ follows the standard normal distribution, $Z\sim N(0,1)$.

(b) Express $P(8\le X\le 11)$ in terms of the standard normal CDF $\Phi (z)$: First, standardize the bounds: Lower bound: $z_1=\frac{8-10}{2}=\frac{-2}{2}=-1$. Upper bound: $z_2=\frac{11-10}{2}=\frac{1}{2}=0.5$. $P(8\le X\le 11)=P(-1\le Z\le 0.5)$. Using the CDF $\Phi (z)=P(Z\le z)$: $P(-1\le Z\le 0.5)=P(Z\le 0.5)-P(Z\le -1)=\Phi (0.5)-\Phi (-1)$.

(c) Calculate $E[2X+5]$ and $Var(2X+5)$: Using properties of expectation and variance: $E[aX+b]=aE[X]+b$ $E[2X+5]=2E[X]+5=2(10)+5=20+5=25$. $Var(aX+b)=a^{2}Var(X)$ $Var(2X+5)=2^{2}Var(X)=4\times Var(X)=4\times 4=16$.

---

12. Expected Value and Variance: Discrete random variable $X$ with PMF: $P(X=0)=0.2,P(X=1)=0.5,P(X=2)=0.3$.

(a) Calculate $E[X]$: $E[X]=\sum xP(X=x)=(0)(0.2)+(1)(0.5)+(2)(0.3)=0+0.5+0.6=1.1$.

(b) Calculate $E[X^2]$: $E[X^2]=\sum x^{2}P(X=x)=(0^2)(0.2)+(1^2)(0.5)+(2^2)(0.3)=0\times 0.2+1\times 0.5+4\times 0.3=0+0.5+1.2=1.7$.

(c) Calculate $Var(X)$ using $Var(X)=E[X^2]-(E[X]{)}^2$: $Var(X)=1.7-(1.1{)}^2=1.7-1.21=0.49$.

(d) Calculate $E[5X-3]$ and $Var(5X-3)$: $E[5X-3]=5E[X]-3=5(1.1)-3=5.5-3=2.5$. $Var(5X-3)=5^{2}Var(X)=25\times Var(X)=25\times 0.49=12.25$.

---

13. Bivariate Discrete Distribution: Joint PMF $p_{X,Y}(x,y)$ given in the table.

(a) Find the marginal PMFs $p_X(x)$ and $p_Y(y)$: The marginal PMF $p_X(x)$ is found by summing the joint probabilities over $y$ for each $x$. $p_X(0)=p_{X,Y}(0,0)+p_{X,Y}(0,1)=0.1+0.2=0.3$. $p_X(1)=p_{X,Y}(1,0)+p_{X,Y}(1,1)=0.3+0.4=0.7$. Check: $0.3+0.7=1$. The marginal PMF $p_Y(y)$ is found by summing the joint probabilities over $x$ for each $y$. $p_Y(0)=p_{X,Y}(0,0)+p_{X,Y}(1,0)=0.1+0.3=0.4$. $p_Y(1)=p_{X,Y}(0,1)+p_{X,Y}(1,1)=0.2+0.4=0.6$. Check: $0.4+0.6=1$. Completed table:

	y=0	y=1	$p_X(x)$
x = 0	0.1	0.2	0.3
x = 1	0.3	0.4	0.7
$p_Y(y)$	0.4	0.6	1

(b) Find the conditional PMF $p_{Y|X}(y|X=1)$: $p_{Y|X}(y|x)=\frac{p_{X,Y}(x,y)}{p_X(x)}$. We need $x=1$, and $p_X(1)=0.7$. $p_{Y|X}(0|X=1)=\frac{p_{X,Y}(1,0)}{p_X(1)}=\frac{0.3}{0.7}=\frac{3}{7}$. $p_{Y|X}(1|X=1)=\frac{p_{X,Y}(1,1)}{p_X(1)}=\frac{0.4}{0.7}=\frac{4}{7}$. The conditional PMF for $Y$ given $X=1$ is $P(Y=0|X=1)=3/7$ and $P(Y=1|X=1)=4/7$.

(c) Calculate $E[XY]$: $E[XY]={\sum }_x{\sum }_{y}xy\cdot p_{X,Y}(x,y)$ $E[XY]=(0)(0)p_{X,Y}(0,0)+(0)(1)p_{X,Y}(0,1)+(1)(0)p_{X,Y}(1,0)+(1)(1)p_{X,Y}(1,1)$ $E[XY]=0+0+0+(1)(1)(0.4)=0.4$.

(d) Calculate $Cov(X,Y)$: $Cov(X,Y)=E[XY]-E[X]E[Y]$. First find $E[X]$ and $E[Y]$ using the marginals: $E[X]=\sum x{p}_X(x)=(0)(0.3)+(1)(0.7)=0.7$. $E[Y]=\sum y{p}_Y(y)=(0)(0.4)+(1)(0.6)=0.6$. $Cov(X,Y)=0.4-(0.7)(0.6)=0.4-0.42=-0.02$.

(e) Are X and Y independent? Check if $p_{X,Y}(x,y)=p_X(x)p_Y(y)$ for all $(x,y)$. Let’s check $(0,0)$: $p_{X,Y}(0,0)=0.1$. $p_X(0)p_Y(0)=(0.3)(0.4)=0.12$. Since $p_{X,Y}(0,0)\ne p_X(0)p_Y(0)$, $X$ and $Y$ are not independent.

---

14. Bivariate Continuous Distribution: $f_{X,Y}(x,y)=c(x+y)$ if $0\le x\le 1,0\le y\le 1$, and 0 otherwise.

(a) Find the value of $c$: The total integral of the PDF must be 1. ${\int }_0^1{\int }_0^{1}c(x+y)dxdy=1$ ${\int }_0^{1}c{\left[\frac{x^2}{2}+xy\right]}_{x=0}^{x=1}dy={\int }_0^{1}c\left(\frac{1}{2}+y\right)dy=1$ $c{\left[\frac{1}{2}y+\frac{y^2}{2}\right]}_{y=0}^{y=1}=c\left((\frac{1}{2}+\frac{1}{2})-(0)\right)=c(1)=1$. So, $c=1$. The joint PDF is $f_{X,Y}(x,y)=x+y$ for $0\le x\le 1,0\le y\le 1$.

(b) Find the marginal PDF $f_X(x)$: Integrate the joint PDF over $y$. For $0\le x\le 1$: $f_X(x)={\int }_0^1{f}_{X,Y}(x,y)dy={\int }_0^1(x+y)dy={\left[xy+\frac{y^2}{2}\right]}_{y=0}^{y=1}$ $f_X(x)=(x+\frac{1}{2})-(0)=x+\frac{1}{2}$. So, $f_X(x)=x+\frac{1}{2}$ for $0\le x\le 1$, and 0 otherwise.

(c) Find the conditional PDF $f_{Y|X}(y|x)$: $f_{Y|X}(y|x)=\frac{f_{X,Y}(x,y)}{f_X(x)}$ for $f_X(x)>0$. For $0\le x\le 1$ and $0\le y\le 1$: $f_{Y|X}(y|x)=\frac{x+y}{x+1/2}$. The conditional PDF is $f_{Y|X}(y|x)=\frac{x+y}{x+1/2}$ for $0\le y\le 1$, and 0 otherwise.

(d) Are X and Y independent? Check if $f_{X,Y}(x,y)=f_X(x)f_Y(y)$. By symmetry, the marginal PDF for Y is $f_Y(y)=y+1/2$ for $0\le y\le 1$. $f_X(x)f_Y(y)=(x+1/2)(y+1/2)=xy+\frac{x}{2}+\frac{y}{2}+\frac{1}{4}$. The joint PDF is $f_{X,Y}(x,y)=x+y$. Since $x+y\ne xy+\frac{x}{2}+\frac{y}{2}+\frac{1}{4}$ generally, $X$ and $Y$ are not independent.

---

15. Independence Check: Joint PDF $f_{X,Y}(x,y)=e^{-(x+y)}$ for $x>0,y>0$.

Find the marginal PDFs: For $x>0$: $f_X(x)={\int }_0^{\infty }{e}^{-(x+y)}dy={\int }_0^{\infty }{e}^{-x}{e}^{-y}dy=e^{-x}{\int }_0^{\infty }{e}^{-y}dy$ $f_X(x)=e^{-x}[-e^{-y}{]}_0^{\infty }=e^{-x}(0-(-e^0))=e^{-x}(1)=e^{-x}$. So, $f_X(x)=e^{-x}$ for $x>0$. (X follows Exp(1)). For $y>0$: $f_Y(y)={\int }_0^{\infty }{e}^{-(x+y)}dx={\int }_0^{\infty }{e}^{-x}{e}^{-y}dx=e^{-y}{\int }_0^{\infty }{e}^{-x}dx$ $f_Y(y)=e^{-y}[-e^{-x}{]}_0^{\infty }=e^{-y}(0-(-e^0))=e^{-y}(1)=e^{-y}$. So, $f_Y(y)=e^{-y}$ for $y>0$. (Y follows Exp(1)).

Check independence: $f_X(x)f_Y(y)=(e^{-x})(e^{-y})=e^{-x-y}=e^{-(x+y)}$. This is equal to the joint PDF $f_{X,Y}(x,y)$. Also, the support region $x>0,y>0$ is a product space. Therefore, $X$ and $Y$ are independent.

---

16. Covariance and Correlation: Given $E[X]=2,E[Y]=3,Var(X)=4,Var(Y)=9,E[XY]=7$.

Calculate Covariance: $Cov(X,Y)=E[XY]-E[X]E[Y]=7-(2)(3)=7-6=1$.

Calculate Correlation Coefficient ${\rho }_{X,Y}$: ${\rho }_{X,Y}=\frac{Cov(X,Y)}{{\sigma }_X{\sigma }_Y}$. We need standard deviations: ${\sigma }_X=\sqrt{Var(X)}=\sqrt{4}=2$. ${\sigma }_Y=\sqrt{Var(Y)}=\sqrt{9}=3$. ${\rho }_{X,Y}=\frac{1}{2\times 3}=\frac{1}{6}$.

Are X and Y uncorrelated? Two variables are uncorrelated if their covariance is 0. Since $Cov(X,Y)=1\ne 0$, $X$ and $Y$ are not uncorrelated (they are correlated).

---

17. Conditional Expectation: Using the joint PMF from Problem 13. Calculate $E[Y|X=0]$. $E[Y|X=x]={\sum }_{y}y\cdot P(Y=y|X=x)={\sum }_{y}y\frac{p_{X,Y}(x,y)}{p_X(x)}$. From Problem 13a, $p_X(0)=0.3$. $E[Y|X=0]={\sum }_{y}y\frac{p_{X,Y}(0,y)}{p_X(0)}$ $E[Y|X=0]=0\cdot \frac{p_{X,Y}(0,0)}{p_X(0)}+1\cdot \frac{p_{X,Y}(0,1)}{p_X(0)}$ $E[Y|X=0]=0\cdot \frac{0.1}{0.3}+1\cdot \frac{0.2}{0.3}=0+\frac{0.2}{0.3}=\frac{2}{3}$.

---

18. Law of Total Expectation: Given $E[Y|X=x]=2x+1$ and $E[X]=5$. Calculate $E[Y]$. By the Law of Total Expectation (also known as the Tower Property): $E[Y]=E[E[Y|X]]$. Substitute the given conditional expectation: $E[Y]=E[2X+1]$. Using linearity of expectation: $E[Y]=2E[X]+E[1]=2E[X]+1$. Substitute $E[X]=5$: $E[Y]=2(5)+1=10+1=11$.

---

19. Multivariate Normal Distribution: $X=[X_1,X_2{]}^T\sim N(\mu ,\Sigma )$ with $\mu =\left[\begin{array}{c}1\\ 2\end{array}\right]$ and $\Sigma =\left(\begin{array}{cc}4& 1\\ 1& 9\end{array}\right)$.

(a) What are $E[X_1],E[X_2],Var(X_1),Var(X_2)$, and $Cov(X_1,X_2)$? These are read directly from the mean vector $\mu$ and covariance matrix $\Sigma$. $E[X_1]={\mu }_1=1$. $E[X_2]={\mu }_2=2$. $Var(X_1)={\Sigma }_{11}=4$. $Var(X_2)={\Sigma }_{22}=9$. $Cov(X_1,X_2)={\Sigma }_{12}={\Sigma }_{21}=1$.

(b) Are $X_1$ and $X_2$ independent? Why or why not? For jointly normally distributed variables, independence is equivalent to having zero covariance. Since $Cov(X_1,X_2)=1\ne 0$, $X_1$ and $X_2$ are not independent.

(c) What is the distribution of the linear combination $W=X_1+2X_2$? A linear combination of jointly normal random variables is also normally distributed. We need to find its mean and variance. Mean: $E[W]=E[X_1+2X_2]=E[X_1]+2E[X_2]=1+2(2)=5$. Variance: $Var(W)=Var(X_1+2X_2)$ Using the formula $Var(a{X}_1+b{X}_2)=a^{2}Var(X_1)+b^{2}Var(X_2)+2abCov(X_1,X_2)$ with $a=1,b=2$: $Var(W)=1^{2}Var(X_1)+2^{2}Var(X_2)+2(1)(2)Cov(X_1,X_2)$ $Var(W)=(1)(4)+(4)(9)+(4)(1)=4+36+4=44$. Therefore, $W$ follows a normal distribution with mean 5 and variance 44. $W\sim N(\text{mean}=5,\text{variance}=44)$.

---

20. Multinomial Distribution: Rolling a fair six-sided die $n=10$ times. $X_1$ = number of times ‘1’ appears. $X_2$ = number of times ‘2’ or ‘3’ appears. $X_3$ = number of times ‘4’, ‘5’, or ‘6’ appears.

(a) What are the parameters $n,k,$ and the probabilities $p_1,p_2,p_3$? $n=10$ (number of trials/rolls). $k=3$ (number of categories). The probabilities for each category on a single roll are: $p_1=P(\text{outcome is ’1’})=1/6$. $p_2=P(\text{outcome is ’2’ or ’3’})=P(\text{’2’})+P(\text{’3’})=1/6+1/6=2/6=1/3$. $p_3=P(\text{outcome is ’4’, ’5’, or ’6’})=P(\text{’4’})+P(\text{’5’})+P(\text{’6’})=1/6+1/6+1/6=3/6=1/2$. Check: $p_1+p_2+p_3=1/6+2/6+3/6=6/6=1$.

(b) Calculate the probability of observing the outcome $(x_1=2,x_2=3,x_3=5)$. Check that $x_1+x_2+x_3=2+3+5=10=n$. The multinomial probability formula is: $P(X_1=x_1,X_2=x_2,...,X_k=x_k)=\frac{n!}{x_1!x_2!...x_k!}{p}_1^{x_1}{p}_2^{x_2}...p_k^{x_k}$. $P(X_1=2,X_2=3,X_3=5)=\frac{10!}{2!3!5!}(p_1{)}^2(p_2{)}^3(p_3{)}^5$ $P(X_1=2,X_2=3,X_3=5)=\frac{10!}{2!3!5!}{\left(\frac{1}{6}\right)}^2{\left(\frac{1}{3}\right)}^3{\left(\frac{1}{2}\right)}^5$. Calculate the multinomial coefficient: $\frac{10!}{2!3!5!}=\frac{10\times 9\times 8\times 7\times 6\times 5!}{(2\times 1)\times (3\times 2\times 1)\times 5!}=\frac{10\times 9\times 8\times 7\times 6}{2\times 6}=10\times 9\times 8\times 7\times \frac{1}{2}=2520$. Calculate the probability part: ${\left(\frac{1}{6}\right)}^2{\left(\frac{1}{3}\right)}^3{\left(\frac{1}{2}\right)}^5=\left(\frac{1}{36}\right)\left(\frac{1}{27}\right)\left(\frac{1}{32}\right)=\frac{1}{36\times 27\times 32}=\frac{1}{31104}$. $P(X_1=2,X_2=3,X_3=5)=2520\times \frac{1}{31104}=\frac{2520}{31104}$. Simplifying the fraction (dividing numerator and denominator by 72): $P(X_1=2,X_2=3,X_3=5)=\frac{35}{432}$. (Approximate value: $35/432\approx 0.0810$)

---