3.1.6 Solved Problems:
Discrete Random Variables


Let $X$ be a discrete random variable with the following PMF \begin{equation} \nonumber P_X(x) = \left\{ \begin{array}{l l} 0.1 & \quad \text{for } x=0.2\\ 0.2 & \quad \text{for } x=0.4\\ 0.2 & \quad \text{for } x=0.5\\ 0.3 & \quad \text{for } x=0.8\\ 0.2 & \quad \text{for } x=1\\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}

  1. Find $R_X$, the range of the random variable $X$.
  2. Find $P(X\leq 0.5)$.
  3. Find $P( 0.25 < X < 0.75)$.
  4. Find $P(X=0.2 | X < 0.6)$.

  • Solution
      1. The range of $X$ can be found from the PMF. The range of $X$ consists of possible values for $X$. Here we have $$R_X=\{0.2, 0.4, 0.5, 0.8, 1\}.$$
      2. The event $X\leq 0.5$ can happen only if $X$ is $0.2, 0.4,$ or $0.5$. Thus,
        $P(X\leq 0.5)$ $= P(X \in \{0.2, 0.4, 0.5\})$
        $= P(X=0.2)+P(X=0.4)+P(X=0.5)$

      3. Similarly, we have
        $P(0.25 < X < 0.75)$ $= P(X \in \{0.4, 0.5\})$
        $= P(X=0.4)+P(X=0.5)$

      4. This is a conditional probability problem, so we can use our famous formula $P(A|B)=\frac{P(A \cap B)}{P(B)}$. We have
        $P(X=0.2 | X < 0.6)$ $= \frac{P\big((X=0.2) \textrm{ and } (X<0.6)\big)}{P(X<0.6)}$
        $= \frac{P(X=0.2)}{P(X<0.6)}$


I roll two dice and observe two numbers $X$ and $Y$.

  1. Find $R_X, R_Y$ and the PMFs of $X$ and $Y$.
  2. Find $P(X=2, Y=6)$.
  3. Find $P(X>3|Y=2)$.
  4. Let $Z=X+Y$. Find the range and PMF of $Z$.
  5. Find $P(X=4|Z=8)$.

  • Solution
      1. We have $R_X=R_Y=\{1,2,3,4,5,6\}$. Assuming the dice are fair, all values are equally likely so \begin{equation} \nonumber P_X(k) = \left\{ \begin{array}{l l} \frac{1}{6}& \quad \text{for } k=1,2,3,4,5,6\\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} Similarly for $Y$, \begin{equation} \nonumber P_Y(k) = \left\{ \begin{array}{l l} \frac{1}{6}& \quad \text{for } k=1,2,3,4,5,6\\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}
      2. Since $X$ and $Y$ are independent random variables, we can write
        $P(X=2,Y=6)$ $=P(X=2)P(Y=6)$
        $=\frac{1}{6} \cdot \frac{1}{6}=\frac{1}{36}$.

      3. Since $X$ and $Y$ are independent, knowing the value of $Y$ does not impact the probabilities for $X$,
        $P(X > 3|Y=2)$ $=P(X > 3)$

      4. First, we have $R_Z=\{2,3,4,...,12\}$. Thus, we need to find $P_Z(k)$ for $k=2,3,..., 12$. We have
        $P_Z(2)$ $=P(Z=2)=P(X=1,Y=1)$
        $=P(X=1)P(Y=1) \textrm{ (since $X$ and $Y$ are independent)}$
        $=\frac{1}{6} \cdot \frac{1}{6}=\frac{1}{36}$;
        $P_Z(3)$ $=P(Z=3)=P(X=1,Y=2)+P(X=2,Y=1)$
        $=P(X=1)P(Y=2)+ P(X=2)P(Y=1)$
        $=\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}=\frac{1}{18}$;
        $P_Z(4)$ $=P(Z=4)=P(X=1,Y=3)+P(X=2,Y=2)+P(X=3,Y=1)$
        $=3 \cdot \frac{1}{36}=\frac{1}{12}$.

        We can continue similarly:
        $P_Z(5)$ $=\frac{4}{36}=\frac{1}{9}$;
        $P_Z(6)$ $=\frac{5}{36}$;
        $P_Z(7)$ $=\frac{6}{36}=\frac{1}{6}$;
        $P_Z(8)$ $=\frac{5}{36}$;
        $P_Z(9)$ $=\frac{4}{36}=\frac{1}{9}$;
        $P_Z(10)$ $=\frac{3}{36}=\frac{1}{12}$;
        $P_Z(11)$ $=\frac{2}{36}=\frac{1}{18}$;
        $P_Z(12)$ $=\frac{1}{36}$.

        It is always a good idea to check our answers by verifying that $\sum_{z \in R_Z} P_Z(z)=1$. Here, we have
        $\sum_{z \in R_Z} P_Z(z)$ $=\frac{1}{36}+\frac{2}{36}+\frac{3}{36}+\frac{4}{36}+\frac{5}{36}+\frac{6}{36}$

      5. Note that here we cannot argue that $X$ and $Z$ are independent. Indeed, $Z$ seems to completely depend on $X$, $Z=X+Y$. To find the conditional probability $P(X=4|Z=8)$, we use the formula for conditional probability
        $P(X=4|Z=8)$ $=\frac{P(X=4,Z=8)}{P(Z=8)}$
        $=\frac{P(X=4)P(Y=4)}{P(Z=8)} \textrm{ (since $X$ and $Y$ are independent)}$
        $\frac{\frac{1}{6} \cdot \frac{1}{6}}{\frac{5}{36}}$


I roll a fair die repeatedly until a number larger than $4$ is observed. If $N$ is the total number of times that I roll the die, find $P(N=k)$, for $k=1,2,3, ...$.

  • Solution
    • In each trial, I may observe a number larger than $4$ with probability $\frac{2}{6}=\frac{1}{3}$. Thus, you can think of this experiment as repeating a Bernoulli experiment with success probability $p=\frac{1}{3}$ until you observe the first success. Thus, $N$ is a geometric random variable with parameter $p=\frac{1}{3}$, $N \sim Geometric(\frac{1}{3})$. Hence, we have \begin{equation} \nonumber P_N(k) = \left\{ \begin{array}{l l} \frac{1}{3}(\frac{2}{3})^{k-1}& \quad \text{for } k=1,2,3,...\\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}


You take an exam that contains $20$ multiple-choice questions. Each question has $4$ possible options. You know the answer to $10$ questions, but you have no idea about the other $10$ questions so you choose answers randomly. Your score $X$ on the exam is the total number of correct answers. Find the PMF of $X$. What is $P(X>15)$?

  • Solution
    • Let's define the random variable $Y$ as the number of your correct answers to the $10$ questions you answer randomly. Then your total score will be $X=Y+10$. First, let's find the PMF of $Y$. For each question your success probability is $\frac{1}{4}$. Hence, you perform $10$ independent $Bernoulli(\frac{1}{4})$ trials and $Y$ is the number of successes. Thus, we conclude $Y \sim Binomial(10,\frac{1}{4})$, so \begin{equation} \nonumber P_Y(y) = \left\{ \begin{array}{l l} {10 \choose y}(\frac{1}{4})^y(\frac{3}{4})^{10-y}& \quad \text{for } y=0,1,2,3,...,10\\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} Now we need to find the PMF of $X=Y+10$. First note that $R_X=\{10,11,12,...,20\}$. We can write
      $P_X(10)$ $=P(X=10)=P(Y+10=10)$
      $=P(Y=0)={10 \choose 0}\left(\frac{1}{4}\right)^0\left(\frac{3}{4}\right)^{10-0}=\left(\frac{3}{4}\right)^{10}$;
      $P_X(11)$ $=P(X=11)=P(Y+10=11)$
      $=P(Y=1)={10 \choose 1}\left(\frac{1}{4}\right)^1\left(\frac{3}{4}\right)^{10-1}=10\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^{9}$.

      So, you get the idea. In general for $k \in R_X=\{10,11,12,...,20\}$,
      $P_X(k)$ $=P(X=k)=P(Y+10=k)$
      $=P(Y=k-10)={10 \choose k-10}\left(\frac{1}{4}\right)^{k-10}\left(\frac{3}{4}\right)^{20-k}$.

      To summarize, \begin{equation} \nonumber P_X(k) = \left\{ \begin{array}{l l} {10 \choose k-10}(\frac{1}{4})^{k-10}(\frac{3}{4})^{20-k}& \quad \text{for } k=10,11,12,...,20\\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} In order to calculate $P(X>15)$, we know we should consider $y=6,7,8,9,10$ \begin{equation} \nonumber P_Y(y) = \left\{ \begin{array}{l l} {10 \choose y}(\frac{1}{4})^y(\frac{3}{4})^{10-y}& \quad \text{for } y=6,7,8,9,10\\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} \begin{equation} \nonumber P_X(k) = \left\{ \begin{array}{l l} {10 \choose k-10}(\frac{1}{4})^{k-10}(\frac{3}{4})^{20-k}& \quad \text{for } k=16,17,...,20\\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} \begin{align}%\label{} \nonumber P(X>15) &= P_X(16)+P_X(17)+P_X(18)+P_X(19)+P_X(20)\\ \nonumber &={10 \choose 6}(\frac{1}{4})^{6}(\frac{3}{4})^{4}+ {10 \choose 7}(\frac{1}{4})^{7}(\frac{3}{4})^{3} +{10 \choose 8}(\frac{1}{4})^{8}(\frac{3}{4})^{2}\\ \nonumber &+ {10 \choose 9}(\frac{1}{4})^{9}(\frac{3}{4})^{1}+ {10 \choose 10}(\frac{1}{4})^{10}(\frac{3}{4})^{0}. \end{align}


Let $X \sim Pascal(m,p)$ and $Y \sim Pascal(l,p)$ be two independent random variables. Define a new random variable as $Z=X+Y$. Find the PMF of $Z$.

  • Solution
    • This problem is very similar to Example 3.7, and we can solve it using the same methods. We will show that $Z \sim Pascal(m+l,p)$. To see this, consider a sequence of $H$s and $T$s that is the result of independent coin tosses with $P(H)=p$, (Figure 3.2). If we define the random variable $X$ as the number of coin tosses until the $m$th heads is observed, then $X \sim Pascal(m,p)$. Now, if we look at the rest of the sequence and count the number of heads until we observe $l$ more heads, then the number of coin tosses in this part of the sequence is $Y \sim Pascal(l,p)$. Looking from the beginning, we have repeatedly tossed the coin until we have observed $m+l$ heads. Thus, we conclude the random variable $Z$ defined as $Z=X+Y$ has a $Pascal(m+l,p)$ distribution.
      Fig.3.2 - Sum of two Pascal random variables.
      In particular, remember that $Pascal(1,p)=Geometric(p)$. Thus, we have shown that if $X$ and $Y$ are two independent $Geometric(p)$ random variables, then $X+Y$ is a $Pascal(2,p)$ random variable. More generally, we can say that if $X_1, X_2, X_3,...,X_m$ are $m$ independent $Geometric(p)$ random variables, then the random variable $X$ defined by $X=X_1+X_2+...+X_m$ has a $Pascal(m,p)$ distribution.


The number of customers arriving at a grocery store is a Poisson random variable. On average $10$ customers arrive per hour. Let $X$ be the number of customers arriving from $10am$ to $11:30am$. What is $P(10 < X \leq 15)$?

  • Solution
    • We are looking at an interval of length $1.5$ hours, so the number of customers in this interval is $X \sim Poisson(\lambda=1.5 \times 10=15)$. Thus,
      $P(10 < X \leq 15)$ $= \sum_{k=11}^{15} P_X(k)$
      $= \sum_{k=11}^{15} \frac{e^{-15} 15^k}{k!}$
      $=e^{-15} \left[ \frac{15^{11}}{11!}+\frac{15^{12}}{12!}+\frac{15^{13}}{13!}+\frac{15^{14}}{14!}+\frac{15^{15}}{15!}\right]$


Let $X \sim Poisson(\alpha)$ and $Y \sim Poisson(\beta)$ be two independent random variables. Define a new random variable as $Z=X+Y$. Find the PMF of $Z$.

  • Solution
    • First note that since $R_X=\{0,1,2,..\}$ and $R_Y=\{0,1,2,..\}$, we can write $R_Z=\{0,1,2,..\}$. We have
      $P_Z(k)$ $=P(X+Y=k)$
      $=\sum_{i=0}^{k} P(X+Y=k|X=i)P(X=i)$ $\textrm{ (law of total probability)}$
      $=\sum_{i=0}^{k} P(Y=k-i|X=i)P(X=i)$
      $=\sum_{i=0}^{k} P(Y=k-i)P(X=i)$ $\textrm{ (since $X$ and $Y$ are independent)}$
      $=\sum_{i=0}^{k} \frac{e^{-\beta}\beta^{k-i}}{(k-i)!} \frac{e^{-\alpha}\alpha^{i}}{i!}$
      $=e^{-(\alpha+\beta)}\sum_{i=0}^{k} \frac{\alpha^{i} \beta^{k-i}}{(k-i)!i!}$
      $=\frac{e^{-(\alpha+\beta)}}{k!}\sum_{i=0}^{k} \frac{k!}{(k-i)!i!} \alpha^{i} \beta^{k-i}$
      $=\frac{e^{-(\alpha+\beta)}}{k!}\sum_{i=0}^{k} {k \choose i} \alpha^{i} \beta^{k-i}$
      $=\frac{e^{-(\alpha+\beta)}}{k!}(\alpha+\beta)^k$ $\textrm{(by the binomial theorem)}$.

      Thus, we conclude that $Z \sim Poisson(\alpha+\beta)$.

Let $X$ be a discrete random variable with the following PMF \begin{equation} \nonumber P_X(k) = \left\{ \begin{array}{l l} \frac{1}{4} & \quad \text{for } k=-2\\ \frac{1}{8} & \quad \text{for } k=-1\\ \frac{1}{8} & \quad \text{for } k=0\\ \frac{1}{4} & \quad \text{for } k=1\\ \frac{1}{4} & \quad \text{for } k=2\\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} I define a new random variable $Y$ as $Y=(X+1)^2$.
  1. Find the range of $Y$.
  2. Find the PMF of $Y$.
  • Solution
    • Here, the random variable $Y$ is a function of the random variable $X$. This means that we perform the random experiment and obtain $X=x$, and then the value of $Y$ is determined as $Y=(x+1)^2$. Since $X$ is a random variable, $Y$ is also a random variable.
      1. To find $R_Y$, we note that $R_X=\{-2,-1,0,1,2\}$, and
        $R_Y$ $=\{y=(x+1)^2 | x \in R_X \}$

      2. Now that we have found $R_Y=\{0,1,4,9\}$, to find the PMF of $Y$ we need to find $P_Y(0), P_Y(1), P_Y(4)$, and $P_Y(9)$:
        $P_Y(0)$ $=P(Y=0)=P((X+1)^2=0)$
        $=P(X=-1)= \frac{1}{8}$;
        $P_Y(1)$ $=P(Y=1)=P((X+1)^2=1)$
        $=P\big((X=-2) \textrm{ or } (X=0) \big)$;
        $P_X(-2)+ P_X(0)= \frac{1}{4}+\frac{1}{8}=\frac{3}{8}$;
        $P_Y(4)$ $=P(Y=4)=P((X+1)^2=4)$
        $=P(X=1)= \frac{1}{4}$;
        $P_Y(9)$ $=P(Y=9)=P((X+1)^2=9)$
        $=P(X=2)= \frac{1}{4}$.

        Again, it is always a good idea to check that $\sum_{y \in R_Y} P_Y(y)=1$. We have $$\sum_{y \in R_Y} P_Y(y)=\frac{1}{8}+\frac{3}{8}+\frac{1}{4}+\frac{1}{4}=1.$$

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