Exam Practice Problems
Here are some problems that range in difficulty, focusing on post-midterm content.
Ensure you can solve all of these (+ the midterm prep problems) independently and confidently!
See also the textbook page for suggested textbook problems!
Question 1
A university’s IT help desk receives an average of 2.5 requests per hour.
- What is the probability that the desk receives exactly 4 requests during one hour?
Let \(X\) be the number of requests. \(X \sim {\mathrm{Poiss}}(\lambda_{\text{hour}} = 2.5)\).
We want to find the probability of observing exactly 4 requests:
\[
\begin{aligned}
\mathbb{P}(X = 4) &= \frac{e^{-2.5}(2.5)^4}{4!} \\
&= \frac{0.082085 \times 39.0625}{24} \\
&= \frac{3.206445}{24} \\
&\approx 0.1336
\end{aligned}
\]
Therefore, the probability that the desk receives exactly 4 requests during one hour is 0.1336.
- What is the probability that the desk receives between 45 and 47 (inclusive) requests in a given day?
Let \(Y\) represent the daily counts.
\(\lambda_{\text{day}} = 2.5 \text{ requests/hour} \times 24 \text{ hours/day} = 60\)
We want to find the probability of observing between [45, 47] calls per day:
\[
\begin{aligned}
\mathbb{P}(45 \le Y \le 47) &= \mathbb{P}(Y = 45) + \mathbb{P}(Y = 46) + \mathbb{P}(Y = 47)\\
&= \frac{e^{-60} 60^{45}}{45!} + \frac{e^{-60}60^{46}}{46!} + \frac{e^{-60}60^{47}}{47!} \\
&= 0.0076+0.0099+0.0127 \\
&= 0.0302
\end{aligned}
\]
Therefore, for the probability of observing between 45 and 47 calls (inclusive) is 0.0302.
Note: we could also solve this by solving for the CDF and using \(F_Y(47) - F_Y(44) = \mathbb{P}(Y \le 47) - \mathbb{P}(Y \le 44)\) (we use \(F_Y(44)\) to remove all probability from Y = 44 and below as we want to include \(\mathbb{P}(Y = 45)!\))
- Use the moment generating function (MGF) to show that the expect number of requests in a given day is 60. Hint: You will use the Maclaurin series expansion which says \(e^z = \sum_{n = 0}^{\infty}\frac{z^n}{n!}\)
We will show that \(\mathbb{E}[Y] = \lambda_{day}\) for any general \(Y \sim {\mathrm{Poiss}}(\lambda)\). You can also show for the specific case where \(\lambda = 60\).
\[
\begin{aligned}
m_{Y}(t) &= \mathbb{E}[e^{ty}]\\
m_{Y}(t) &= \sum_{y = 0}^{\infty}e^{ty}\frac{e^{-\lambda}\lambda^y}{y!}\\
&= e^{-\lambda}\sum_{y = 0}^{\infty}\frac{(\lambda e^t)^y}{y!}\\
&= e^{-\lambda}e^{\lambda e^t}\\
&= e^{\lambda(e^t - 1)}
\end{aligned}
\]
Then,
\[
\begin{aligned}
\mathbb{E}[Y] &= \frac{d}{dt}e^{\lambda(e^t - 1)}|_{t = 0}\\
&= \lambda(e^t)e^{\lambda(e^t - 1)}|_{t = 0}\\
&= \lambda(e^0)e^{\lambda(e^0 - 1)}\\
&= \lambda
\end{aligned}
\] As \(Y \sim {\mathrm{Poiss}}(\lambda_{day} = 60)\), we have shown that \(\mathbb{E}[Y] = 60\).
Question 2
Show that for any two random variables \(X,Y\), \(|\operatorname{Corr}(X,Y)| \leq 1\).
Recall \(\operatorname{Corr}(X,Y) = \frac{\operatorname{Cov}(X,Y)}{\sqrt{\operatorname{Var}X\cdot \operatorname{Var}Y}}\)
By Cauchy-Schwarz,
\[
\begin{aligned}
|\operatorname{Cov}(X,Y)| &= |\mathbb{E}(X-\mathbb{E}X)(Y-\mathbb{E}Y)| \\
&\leq \sqrt{\mathbb{E}(X-\mathbb{E}X)^2 \cdot \mathbb{E}(Y-\mathbb{E}Y)^2} \\
&= \sqrt{\operatorname{Var}X \operatorname{Var}Y}.
\end{aligned}
\]
Therefore
\[
\begin{aligned}
|\operatorname{Corr}(X,Y)| &= \frac{|\operatorname{Cov}(X,Y)|}{\sqrt{\operatorname{Var}(X)\cdot \operatorname{Var}(Y)}} \\
&\leq \frac{\sqrt{\operatorname{Var}(X)\cdot \operatorname{Var}(Y)}}{\sqrt{\operatorname{Var}(X)\cdot \operatorname{Var}(Y)}}\\
&= 1.
\end{aligned}
\]
Question 3
Let \(f_X(x) = \frac{16}{6}x^3e^{-2x}I_{(0,\infty)}(x)\). Find \(\mathbb{E}[X]\) without using the MGF nor using the general expected values or variances found on distributions sheet provided.
\[
\begin{aligned}
\mathbb{E}[X] &=\int_{0}^{\infty}x\times \frac{16}{6}x^3e^{-2x}dx\\
&= \frac{16}{6}\int_{0}^{\infty}x^4e^{-2x}dx\\
&= \frac{16}{6}\int_{0}^{\infty}x^{5-1}e^{-2x}dx & \text{looks like Gamma(5, 2)}\\
&= \frac{16}{6}\frac{\Gamma(5)}{(2)^5}(1) &\text{since PDF of Gamma(5,2)}\\
&= \frac{16}{6}\frac{4\times3\times2\times1}{32}\\
&= 2
\end{aligned}
\]
Question 4
A box contains 3 coins. One is a fair coin, one has heads on both sides, and one has tails on both sides.
- A coin is selected at random from the box and flipped. If the result is heads, what is the probability that the coin was the fair coin?
Let \(F\) be the event that the fair coin is selected, \(M\) (Medusa) be the event that the two-headed coin is selected, \(Z\) be the event that the two-tailed coin is selected, and let \(H\) be the event that the result is heads. We want to find \(\mathbb{P}(F \mid H)\). By Bayes’ theorem, we have
\[
\begin{aligned}
\mathbb{P}(F \mid H) &= \frac{\mathbb{P}(H \mid F)\mathbb{P}(F)}{\mathbb{P}(H \mid F)\mathbb{P}(F) + \mathbb{P}(H \mid M)\mathbb{P}(M) + \mathbb{P}(H \mid Z)\mathbb{P}(Z)} \\
&= \frac{\frac{1}{2} \cdot
\frac{1}{3}}{\frac{1}{2} \cdot \frac{1}{3} + 1 \cdot \frac{1}{3} + 0 \cdot
\frac{1}{3}} \\
&= \frac{1}{3}.
\end{aligned}
\]
- A coin is selected at random from the box and flipped twice. What is the probability that the result is heads both times?
- Let \(A\) be the event that the result is heads both times. We want to find \(\mathbb{P}(A)\). We can compute this using the law of total probability:
\[
\begin{aligned}
\mathbb{P}(A) &= \mathbb{P}(A\mid F)\mathbb{P}(F) + \mathbb{P}(A\mid M)\mathbb{P}(M) + \mathbb{P}(A\mid Z)\mathbb{P}(Z) \\
&= \frac{1}{4} \cdot \frac{1}{3} + 1 \cdot \frac{1}{3} + 0 \cdot \frac{1}{3} \\
&= \frac{5}{12}.
\end{aligned}
\]
Question 5
Let \(X,Y\) be absolutely continuous random variables with joint probability density function
\[
f_{X,Y}(x, y) = x \, e^{-x(y+1)} I_{[0,\infty)}(x)I_{[0,\infty)}(y).
\]
- Find the marginal probability density function of \(X\) and identify the distribution of \(X\).
The marginal PDF of \(X\) is given by
\[
\begin{aligned}
f_X(x) &= \int_0^\infty f_{X,Y}(x,y) dy \\
&= \int_0^\infty x e^{-x(y+1)} dy
= x e^{-x} \int_0^\infty e^{-xy} dy \\
&= x e^{-x} \cdot \frac{1}{x} \\
&= e^{-x}I_{[0,\infty)}(x).
\end{aligned}
\]
So \(X\) is an exponential random variable with rate parameter 1.
- Find the conditional probability density function of \(Y\) given \(X=x\) and identify the distribution of \(Y\) given \(X=x\).
The conditional PDF of \(Y\) given \(X=x\) is given by
\[
\begin{aligned}
f_{Y|X}(y|x) &= \frac{f_{X,Y}(x,y)}{f_X(x)} \\
&= \frac{x e^{-x(y+1)}}{e^{-x}} \\
&= xe^{-xy} I_{[0,\infty)}(y).
\end{aligned}
\]
So \(Y\) given \(X=x\) is an exponential random variable with rate parameter \(x\).
- Find \(\mathbb{E}(Y)\). Is \(\mathbb{E}(Y)\) finite?
We can compute \(\mathbb{E}[Y]\) using the law of total expectation:
\[
\begin{aligned}
\mathbb{E}[Y] &= \mathbb{E}[\mathbb{E}[Y|X]] \\
&= \mathbb{E}\left[\frac{1}{X}\right] \\
&= \int_0^\infty \frac{1}{x}e^{-x}dx \\
&=\infty
\end{aligned}
\]
So \(\mathbb{E}[Y]\) is undefined.
Question 6
Let \(X_1, X_2, \ldots, X_n, \ldots\) be independent and identically distributed random variables, each with CDF
\(F_X(x) = \left(1-\frac{1}{x}\right)I_{[1,\infty)}(x)\)
- Let \(Y_n = \max \lbrace X_1,\ldots, X_n\rbrace.\) Find the CDF of \(Y_n\).
We have
\[
\begin{aligned}
F_{Y_n}(y) &= \mathbb{P}(Y_n \leq y) \\
&= \mathbb{P}(X_1 \leq y, \ldots, X_n \leq y) \\
&=\prod_{i=1}^n \mathbb{P}(X_i \le y) \\
&= (F_X(y))^n \\
&= \left(1-\frac{1}{y}\right)^nI_{[1,\infty)}(y)
\end{aligned}
\]
- Let \(T_n = Y_n/n.\) Find the CDF of \(T_n\).
\[
\begin{aligned}
F_{T_n}(t) &= \mathbb{P}(T_n \le t) \\
&= \mathbb{P}(Y_n \le nt) \\
&= F_{Y_n}(nt) \\
&= \left(1-\frac{1}{nt}\right)^n I_{[1/n,\infty)}(t)
\end{aligned}
\]
- Show that \(T_n\) converges in distribution to a random variable \(T\), and find its PDF.
\[
\begin{aligned}
\lim_{n \to \infty} F_{T_n}(t) &= \lim_{n \to \infty} \left(1-\frac{1}{nt}\right)^n I_{[1/n,\infty)}(t) \\
&= e^{-1/t} I_{(0,\infty)}(t).
\end{aligned}
\]
Thus, \(T_n\) converges in distribution to a random variable \(T\) with CDF
\(F_T(t) = e^{-1/t} I_{(0,\infty)}(t).\) To find the PDF of \(T\), we differentiate the CDF and get \(f_T(t) = F_T'(t) = \frac{1}{t^2}e^{-1/t} I_{(0,\infty)}(t)\).
Question 7
Suppose you are given a sequence of random variables \(X_n\) with moment generating functions that satisfies \[
m_{X_n}(t) = \frac{n+1}{n} + n\log\left(1-p/n\right) + pe^{(t^2n/(n+1)^2 + t)}.
\] Show that \(X_n\) converges in distribution to a \({\mathrm{Bern}}(p)\) random variable.
Hint: the MGF of \(Z \sim {\mathrm{Bern}}(p)\) is \(m_Z(t) = 1 - p + pe^t\).
Recall that \(X_n\overset{d}{\to}X\) if there is an \(s>0\) such that for all \(t\in(-s,s)\), \(m_{X_n}(t) \to m_X(t)\).
So we check the limit of \(m_{X_n}(t)\).
\[
\begin{aligned}
\lim_{n\to\infty} m_{X_n}(t) &= \lim_{n\to\infty} \frac{n+1}{n} + n\log\left(1-p/n\right) + pe^{(t^2n/(n+1)^2 + t)}\\
&= 1 + \log \lim_{n\to\infty}(1-p/n)^n + pe^{(\lim_{n\to\infty} t^2n/(n+1)^2 + t)}\\
&= 1 + \log e^{-p} + pe^{t}\\
&= 1 - p + pe^{t}
\end{aligned}
\]
Note: since \(e^x\) is continuous everywhere, the limit can be moved inside \(e^x\). Similarly, \(\log\) is continuous on \((0, \infty)\) and \((1−p/n)n(1 - p/n)^n\) stays positive for large enough \(n\), so \(\log \lim_{n\to\infty}(1-p/n)^n = \lim_{n\to\infty} \log(1-p/n)^n\)
Therefore \(X_n\overset{d}{\to}X\sim{\mathrm{Bern}}(p)\).
Question 8
Let \(X\) represent the number of customer complaints received in a day (0, 1, or 2) and \(Y\) indicate whether a supervisor was called (0 = no, 1 = yes).The joint PMF of discrete random variables \(X\) and \(Y\) is given below:
| \(X=0\) |
0.24 |
0.06 |
|
| \(X=1\) |
0.28 |
0.12 |
|
| \(X=2\) |
0.18 |
0.12 |
|
- Compute the conditional distribution \(P_{Y \mid X}(y \mid X = 2)\) for all \(y \in \mathbb{R}\).
By definition, \(P_{Y \mid X}(y \mid X = x) = \dfrac{P(X = x, Y = y)}{P(X = x)}\). For \(X = 2\), we have \(P(X = 2) = 0.30\):
\[
\begin{aligned}
P_{Y \mid X}(0 \mid X = 2) &= \frac{P(X = 2, Y = 0)}{P(X = 2)} = \frac{0.18}{0.30} = \frac{3}{5} \\[6pt]
P_{Y \mid X}(1 \mid X = 2) &= \frac{P(X = 2, Y = 1)}{P(X = 2)} = \frac{0.12}{0.30} = \frac{2}{5}
\end{aligned}
\]
(Check: \(\dfrac{3}{5} + \dfrac{2}{5} = 1\).)
- Calculate \(P(Y = 1 \mid X \leq 1)\). Show your work.
\[
P(Y = 1 \mid X \leq 1) = \frac{P(Y = 1 \text{ and } X \leq 1)}{P(X \leq 1)}
\]
\(P(X \le 1) = 0.24+ 0.06 + 0.28+0.12 = 0.70\). Then,
\[
P(Y = 1 \text{ and } X \leq 1) = P(X=0, Y=1) + P(X=1, Y=1) = 0.06 + 0.12 = 0.18
\]
Thus,
\[
P(Y = 1 \mid X \leq 1) = \frac{0.18}{0.70} = \frac{9}{35} \approx 0.2571
\]
- Compute \(\text{Cov}(X, Y)\).
Recall that \(\text{Cov}(X, Y) = E(XY) - E(X)\,E(Y)\).
\[
E(XY) = \sum_{x}\sum_{y} xy \, P(X = x, Y = y)
\]
Only cells with \(y = 1\) contribute (all others have \(xy = 0\)):
\[
\begin{aligned}
E(XY) &= 1 \cdot 1 \cdot P(X=1,Y=1) + 2 \cdot 1 \cdot P(X=2,Y=1) \\
&= (1)(0.12) + (2)(0.12) \\
&= 0.12 + 0.24 \\
&=0.36
\end{aligned}
\]
\[
E(X) = 0(0.30) + 1(0.40) + 2(0.30) = 0 + 0.40 + 0.60 = 1.00
\]
\[
E(Y) = 0(0.70) + 1(0.30) = 0.30
\]
Thus, \[
\text{Cov}(X, Y) = E(XY) - E(X)\,E(Y) = 0.36 - (1.00)(0.30) = 0.06
\]
Question 9
Suppose \(X \sim{\mathrm{Unif}}[0,1]\) and \(Y|X \sim{\mathrm{Exp}}(X)\).
- What is the joint PDF of \(X,Y\)?
The joint PDF of \(X,Y\) is the marginal times the conditional:
\(f_{X,Y}(x,y) = f_{X}(x) f_{Y|X}(y|x) = I_{[0,1]}(x) xe^{-xy}I_{[0,\infty)}(y)\)
- What is the marginal density of \(X\)? Do you recognize the distribution?
The marginal PDF of \(X\) is the integral of the joint:
\[
\begin{aligned}
f_X(x) &= \int f_{X,Y}(x,y)d y\\
&= \int f_{X}(x) f_{Y|X}(y|x)d y\\
&= \int I_{[0,1]}(x) xe^{-xy}I_{[0,\infty)}(y) d y\\
&= I_{[0,1]}(x).
\end{aligned}
\] So \(X \sim {\mathrm{Unif}}[0,1]\) (we could have also seen this just by inspection).
- What is the marginal density of \(Y\)? Do you recognize the distribution?
The marginal PDF of \(Y\) is the integral of the joint:
\[
\begin{aligned}
f_Y(y) &= \int f_{X,Y}(x,y)d x\\
&= \int f_{X}(x) f_{Y|X}(y|x)d x\\
&= \int I_{[0,1]}(x) xe^{-xy}I_{[0,\infty)}(y) d x\\
&= \int_0^1 xe^{-xy} d x I_{[0,\infty)}(y)
\end{aligned}
\]
We could recognize the Gamma density integral, but the limits don’t quite work for that (they aren’t 0 to infinity).
So we need to use integration by parts:
\[
\begin{aligned}
&= \left(\left.-\frac{x}{y} e^{-x y}\right|_0^1 - \int_0^1 \frac{-1}{y}e^{-xy} d x\right) I_{[0,\infty)}(y)\\
&= \left(-\frac{1}{y} e^{-y} +\frac{1}{y} \left(-\frac{1}{y}e^{-x y}\right|_0^1\right) I_{[0,\infty)}(y)\\
&= \left(-\frac{1}{y} e^{-y} -\frac{1}{y^2} \left(e^{-y}-1\right)\right) I_{[0,\infty)}(y)\\
&= \frac{1}{y}\left(\frac{1-e^{-y}}{y}-e^{-y} \right) I_{[0,\infty)}(y).
\end{aligned}
\]
We do not recognize this as a named distribution from our list.
Question 10
Let \(X \sim {\mathrm{Exp}}(\lambda)\), where \(\lambda > 0\).
- Let \(s,t > 0\). Find \(\mathbb{P}(X > s + t \mid X > t)\).
Since \(X \sim {\mathrm{Exp}}(\lambda)\), we have \(\mathbb{P}(X > x) = e^{-\lambda x}\) for \(x > 0\). Then:
\[
\begin{aligned}
\mathbb{P}(X > s + t \mid X > t) &= \frac{\mathbb{P}(X > s + t, X > t)}{\mathbb{P}(X > t)} \\
&= \frac{\mathbb{P}(X > s + t)}{\mathbb{P}(X > t)} \\
&= \frac{e^{-\lambda (s + t)}}{e^{-\lambda t}} \\
&= e^{-\lambda s} \\
&= \mathbb{P}(X > s)
\end{aligned}
\] where we used \(\{X > s + t\} \cap \{X > t\} = \{X > s + t\}\) since \(s +t > t\).
This is referred to as the memoryless property.
- Find the conditional pdf of \(X\) given \(X > t\) for \(t > 0\).
Start from the conditional CDF:
\[
\begin{aligned}
F_{X|X>t}(x) &= \mathbb{P}(X \leq x \mid X > t) \\
&= \frac{\mathbb{P}(X \leq x, X > t)}{\mathbb{P}(X > t)}\\
&= \frac{\mathbb{P}(t < X \leq x)}{\mathbb{P}(X > t)},
\end{aligned}
\] which is 0 for \(x \leq t\).
For \(x > t\), we have
\[
\begin{aligned}
F_{X|X>t}(x) &= \frac{\mathbb{P}(X \leq x) - \mathbb{P}(X \leq t)}{\mathbb{P}(X > t)} \\
&= \frac{(1 - e^{-\lambda x}) - (1 - e^{-\lambda t})}{e^{-\lambda t}} \\
&= 1 - e^{-\lambda (x - t)}.
\end{aligned}
\]
Differentiating gives the conditional PDF:
\[
f_{X|X>t}(x) = \frac{\mathsf{d}}{\mathsf{d}x} F_{X|X>t}(x) = \lambda e^{-\lambda (x - t)}
I_{(t,\infty)}(x).
\]
- Find \(\mathbb{E}[X \mid X > t]\) for \(t > 0\).
\[
\begin{aligned}
\mathbb{E}[X \mid X > t] &= \int_t^\infty x \cdot \lambda e^{-\lambda(x-t)} \, dx \\
& = \int_0^\infty (u + t) \cdot \lambda e^{-\lambda u} \, du &\text{substituting } u = x - t\\
&= \int_0^\infty u \cdot \lambda e^{-\lambda u} \, du + t \int_0^\infty \lambda e^{-\lambda u} \, du.
\end{aligned}
\]
The first integral is \(\mathbb{E}[X] = 1/\lambda\) and the second integrates to 1, so:
\[\mathbb{E}[X \mid X > t] = \frac{1}{\lambda} + t\]
Question 11
Let \(X\) be a random variable with PDF \[f_X(x) = \left(\frac{1}{2}e^{-x} + e^{-2x}\right)I_{(0,\infty)}(x).\]
- Find the moment generating function of \(X\).
The moment generating function of \(X\) is given by
\[
\begin{aligned}
M_X(t)
&= \mathbb{E}[e^{tX}] = \int_0^\infty e^{tx} \left(\frac{1}{2}e^{-x} + e^{-2x}\right) \, dx \\
&= \int_0^\infty \frac{1}{2}e^{(t-1)x} \, dx + \int_0^\infty e^{(t-2)x} \, dx \\
&= \frac{1}{2} \cdot \frac{1}{1-t} + \frac{1}{2-t} \\
&= \frac{1}{2(1-t)} + \frac{1}{2-t}.
\end{aligned}
\]
- Find \(\operatorname{Var}(X)\) using the moment generating function.
To find \(\operatorname{Var}(X)\), we first compute \(M_X'(t)\) and \(M_X''(t)\):
\[
\begin{aligned}
M_X'(t) &= \frac{1}{2(1-t)^2} + \frac{1}{(2-t)^2}, \text{ and }\\\quad
M_X''(t) &= \frac{1}{(1-t)^3} + \frac{2}{(2-t)^3}.
\end{aligned}
\] Then we have \(\mathbb{E}[X] = M_X'(0) = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}\) and \(\mathbb{E}[X^2] = M_X''(0) = 1 + \frac{1}{4} = \frac{5}{4}\). Finally, we compute \(\operatorname{Var}X = \mathbb{E}[X^2] - (\mathbb{E}[X])^2 = \frac{5}{4} - \left(\frac{3}{4}\right)^2 = \frac{5}{4} - \frac{9}{16} = \frac{11}{16}\).
Question 12
Let \(Y_i\) for \(i=1,2,\ldots\) be a sequence of independent random variables such that \(\mathbb{E}[Y_i] = \beta x_i\), where \(\beta\) and \(x_1, x_2, \ldots\) are constants satisfying \(\lim_{n \to \infty} \frac{1}{n}\sum_{i=1}^n \frac{1}{x_i^2} = c\) for some constant \(c > 0\).
Define \(Z_n = n^{-1/2}\sum_{i=1}^n (Y_i / x_i - \beta)\).
- Find \(\mathbb{E}[Z_n]\).
We have
\[
\begin{aligned}
\mathbb{E}[Z_n] &= \mathbb{E}\left[n^{-1/2}\sum_{i=1}^n (Y_i / x_i - \beta)\right] \\
&=n^{-1/2}\sum_{i=1}^n \mathbb{E}\left[\frac{Y_i}{x_i} - \beta\right] \\
&= n^{-1/2}\sum_{i=1}^n \left(\frac{\mathbb{E}[Y_i]}{x_i} - \beta\right) \\
&= n^{-1/2}\sum_{i=1}^n \left(\frac{\beta x_i}{x_i} - \beta\right) \\
&= 0.
\end{aligned}
\]
- Suppose that \(\operatorname{Var}(Y_i) = \sigma^2 < \infty\) for all \(i\). Find \(\operatorname{Var}(Z_n)\).
We have:
\[
\begin{aligned}
\operatorname{Var}(Z_n) &= \operatorname{Var}\left(n^{-1/2}\sum_{i=1}^n (Y_i / x_i - \beta)\right) \\
&= n^{-1}\sum_{i=1}^n \operatorname{Var}\left(\frac{Y_i}{x_i} - \beta\right) \\
&= n^{-1}\sum_{i=1}^n \frac{\operatorname{Var}(Y_i)}{x_i^2} \\
&=
n^{-1}\sigma^2 \sum_{i=1}^n \frac{1}{x_i^2}.
\end{aligned}
\]
Question 13
Let \(X_1, X_2, \ldots, X_n\) be independent \({\mathrm{Unif}}(-\theta,0)\) random variables.
- Let \(Y_n = \min\{X_1, X_2, \ldots, X_n\}\). Find the CDF of \(Y_n\).
We have
\[
\begin{aligned}
F_{Y_n}(y) &= \mathbb{P}(Y_n \leq y) \\
&= 1 - \mathbb{P}(Y_n > y) \\
&= 1 - \mathbb{P}(X_1 > y, \ldots, X_n > y) \\
&= 1 - \prod_{i=1}^n \mathbb{P}(X_i > y) \\
&= 1 - \prod_{i=1}^n \int_y^0 \frac{1}{\theta} \, d x \\
&= 1 - (-y/\theta)^n.
\end{aligned}
\]
Now, to deal with the support. For \(Y_n\) has support \((-\theta,0)\), so the CDF can be written as
\[
F_{Y_n}(y) = \begin{cases} 0 & y\leq -\theta \\ 1 - (-y/\theta)^n & y \in
(-\theta, 0) \\ 1 & y \geq 0,\end{cases}
\]
or equivalently
\[F_{Y_n}(y) = I_{(-\theta,\infty)}(y) - (-y/\theta)^n I_{(-\theta,0)}(y).\]
- Show that \(Y_n \overset{p}{\to}-\theta\).
We want to show that for any \(\epsilon > 0\), \(\mathbb{P}(|Y_n + \theta| > \epsilon) \to 0\) as \(n \to \infty\). We have
\[
\begin{aligned}
\mathbb{P}(|Y_n + \theta| > \epsilon) &= \mathbb{P}(Y_n > -\theta + \epsilon) + \mathbb{P}(Y_n < -\theta - \epsilon)\\
& = 1 - F_{Y_n}(-\theta + \epsilon) + 0 \\
&= 1 - \left(1-\left(\frac{-(-\theta + \epsilon)}{\theta}\right)^n \right) \\
&= \left(\frac{\theta-\epsilon}{\theta}\right)^n \to 0.
\end{aligned}
\]
Thus, \(Y_n \overset{p}{\to}-\theta\).
Question 14
Suppose \(X\sim{\mathrm{Gam}}(\alpha_1,\beta)\), \(Y\sim{\mathrm{Gam}}(\alpha_2,\beta)\) are independent random variables.
Find the PDF of \(W=X+Y\) using the MGF.
Because \(X,Y\) are independent,
\[
\begin{aligned}
m_W(t) &= m_X(t)m_Y(t)\\
&= (1-t/\beta)^{-\alpha_1}(1-t/\beta)^{-\alpha_2}\\
&= (1-t/\beta)^{-(\alpha_1+\alpha_2)}.
\end{aligned}
\]
We recognize this as the MGF for \({\mathrm{Gam}}(\alpha_1+\alpha_2, \beta)\). Therefore the PDF of \(W\) is
\[
f_W(w) = \frac{\beta^{\alpha_1+\alpha_2}}{\Gamma(\alpha_1+\alpha_2)}w^{\alpha_1+\alpha_2-1}e^{-\beta w} I_{[0,\infty)}(w).
\]
Question 15
Let \(X\) and \(Y\) be continuous random variables with joint probability density function
\[
f(x, y) = c \cdot x \cdot y^2, \qquad 0 < x < 2,\ 0 < y < x
\]
and \(f(x, y) = 0\) otherwise.
- Find the constant \(c\) that makes \(f\) a valid joint PDF.
Integrate over the triangular support \(\{0 < y < x < 2\}\):
\[
\int_0^2 \int_0^x c\, x\, y^2\; dy\; dx
= c \int_0^2 x \cdot \frac{x^3}{3}\; dx
= \frac{c}{3} \int_0^2 x^4\; dx
= \frac{c}{3} \cdot \frac{32}{5}
= \frac{32c}{15}
\]
Setting equal to 1 gives
\[
c = \frac{15}{32}
\]
- Derive the marginal PDFs \(f_X(x)\) and \(f_Y(y)\). State clearly the support of each marginal. Verify that each integrates to 1.
\[
f_X(x) = \int_0^x \frac{15}{32}\, x\, y^2\; dy
= \frac{15}{32}\, x \cdot \frac{x^3}{3}
= \frac{5}{32}\, x^4, \qquad 0 < x < 2
\]
Verification: \(\displaystyle\int_0^2 \frac{5}{32} x^4\, dx = \frac{5}{32} \cdot \frac{32}{5} = 1\)
For fixed \(y \in (0, 2)\), \(x\) must satisfy \(x > y\) and \(x < 2\), so \(x \in (y, 2)\):
\[
f_Y(y) = \int_y^2 \frac{15}{32}\, x\, y^2\; dx
= \frac{15}{32}\, y^2 \cdot \frac{4 - y^2}{2}
= \frac{15}{64}\, y^2(4 - y^2), \qquad 0 < y < 2
\]
Verification: \(\displaystyle\int_0^2 \frac{15}{64} y^2(4-y^2)\, dy = \frac{15}{64}\left[\frac{4y^3}{3} - \frac{y^5}{5}\right]_0^2 = \frac{15}{64} \cdot \frac{64}{15} = 1\)
- Find the conditional PDF \(f_{Y \mid X}(y \mid x)\) for a fixed value \(x \in (0, 2)\).
\[
f_{Y \mid X}(y \mid x)
= \frac{f(x,y)}{f_X(x)}
= \frac{\dfrac{15}{32}\, x\, y^2}{\dfrac{5}{32}\, x^4}
= \frac{3y^2}{x^3}, \qquad 0 < y < x
\]
- Compute \(E[Y \mid X = x]\). Then use the law of total expectation to find \(E[Y]\).
Conditional expectation:
\[
E[Y \mid X = x]
= \int_0^x y \cdot \frac{3y^2}{x^3}\; dy
= \frac{3}{x^3} \cdot \frac{x^4}{4}
= \frac{3x}{4}
\]
Then by the law of total expectation:
\[
E[Y] = E\!\left[\frac{3X}{4}\right] = \frac{3}{4}\, E[X]
= \frac{3}{4} \int_0^2 x \cdot \frac{5}{32} x^4\; dx
= \frac{3}{4} \cdot \frac{5}{32} \cdot \frac{64}{6}
= \frac{3}{4} \cdot \frac{5}{3}
= \boxed{\frac{5}{4}}
\]
- Verify your answer to (d) by computing \(\mathbb{E}[Y]\) directly from \(f_Y(y)\).
\[
\mathbb{E}[Y] = \int_0^2 y \cdot \frac{15}{64} y^2(4 - y^2)\; dy
= \frac{15}{64} \int_0^2 (4y^3 - y^5)\; dy
= \frac{15}{64} \left[y^4 - \frac{y^6}{6}\right]_0^2
= \frac{15}{64} \cdot \frac{16}{3}
= \frac{5}{4} \checkmark
\]
- Compute \(\operatorname{Var}(Y \mid X = x)\). Then use the law of total variance to find \(\operatorname{Var}(Y)\).
\[
\mathbb{E}[Y^2 \mid X = x]
= \int_0^x y^2 \cdot \frac{3y^2}{x^3}\; dy
= \frac{3}{x^3} \cdot \frac{x^5}{5}
= \frac{3x^2}{5}
\]
Then,
\[
\operatorname{Var}(Y \mid X = x)
= \frac{3x^2}{5} - \left(\frac{3x}{4}\right)^2
= \frac{3x^2}{5} - \frac{9x^2}{16}
= x^2\left(\frac{48 - 45}{80}\right)
= \frac{3x^2}{80}
\]
\[
E\!\left[\operatorname{Var}(Y \mid X)\right]
= \frac{3}{80}\, \mathbb{E}[X^2]
= \frac{3}{80} \int_0^2 x^2 \cdot \frac{5}{32} x^4\; dx
= \frac{3}{80} \cdot \frac{5}{32} \cdot \frac{128}{7}
= \frac{3}{80} \cdot \frac{20}{7}
= \frac{3}{28}
\]
First, \(\mathbb{E}[X] = \int_0^2 x \cdot \frac{5}{32}x^4\,dx = \frac{5}{3}\).
\[
\mathbb{E}[X^2] = \frac{5}{32} \cdot \frac{2^7}{7} = \frac{20}{7}.
\]
Thus,
\[
\operatorname{Var}(X) = \frac{20}{7} - \left(\frac{5}{3}\right)^2 = \frac{20}{7} - \frac{25}{9} = \frac{180 - 175}{63} = \frac{5}{63}
\]
\[
\begin{aligned}
\operatorname{Var}\left(\mathbb{E}[Y \mid X]\right) &= \operatorname{Var}\!\left(\frac{3X}{4}\right)\\
&= \left(\frac{3}{4}\right)^2 \operatorname{Var}[X]\\
&= \left(\frac{9}{16}\right)\cdot \frac{5}{63}\\
&= \frac{5}{112}
\end{aligned}
\]
Then,
\[
\operatorname{Var}(Y)
= \frac{3}{28} + \frac{5}{112}
= \frac{12}{112} + \frac{5}{112}
= \frac{17}{112}
\]
- Verify your answer directly using \(\mathbb{E}[Y^2] - (\mathbb{E}[Y])^2\).
\[
\begin{aligned}
\mathbb{E}[Y^2] &= \int_0^2 y^2 \cdot \frac{15}{64}y^2(4-y^2)\,dy \\
&= \frac{15}{64}\left[\frac{4y^5}{5}-\frac{y^7}{7}\right]_0^2 \\
&= \frac{15}{64}\cdot\frac{256}{35} \\
&= \frac{12}{7}
\end{aligned}
\] Then,
\[
\operatorname{Var}(Y) = \frac{12}{7} - \left(\frac{5}{4}\right)^2 = \frac{12}{7} - \frac{25}{16} = \frac{192 - 175}{112} = \frac{17}{112}
\]
Question 16
Let \(X\) be a continuous random variable with PDF
\[
f_X(x) = \frac{3}{8}x^2, \qquad 0 < x < 2
\]
and \(f_X(x) = 0\) otherwise.
Define the random variable \(Y = \frac{X^3}{2}\).
- Find the range of \(Y\) and verify that \(g(x) = x^3/2\) is a strictly monotone function on the support of \(X\).
Since \(X \in (0, 2)\) and \(g(x) = x^3/2\) is strictly increasing (its derivative \(g'(x) = 3x^2/2 > 0\) for all \(x > 0\)), we can apply the change-of-variables formula directly.
The range of \(Y\) is obtained by evaluating \(g\) at the endpoints:
\[
g(0) = 0, \qquad g(2) = \frac{2^3}{2} = 4.
\]
So \(Y \in (0, 4)\).
- Find the PDF \(f_Y(y)\) of \(Y\). State clearly the support of \(Y\). What distribution does \(Y\) follow?
Since \(g\) is strictly increasing and we’re dealing with a continuous RV, the Jacobian method gives
\[
f_Y(y) = f_X\!\left(h^{-1}(y)\right) \cdot \left|\frac{d}{dy} h^{-1}(y)\right|.
\]
From \(y = x^3/2\) we get \(x = (2y)^{1/3}\), so
\[
h^{-1}(y) = (2y)^{1/3}.
\]
Then,
\[
\frac{d}{dy}(2y)^{1/3} = \frac{1}{3}(2y)^{-2/3} \cdot 2 = \frac{2}{3}(2y)^{-2/3}.
\]
Substituting this into the formula, we get:
\[
\begin{aligned}
f_Y(y) &= f_X\!\left((2y)^{1/3}\right) \cdot \frac{2}{3}(2y)^{-2/3}I_{(0, 4)}(y)\\
&= \frac{3}{8}\left[(2y)^{1/3}\right]^2 \cdot \frac{2}{3}(2y)^{-2/3}I_{(0, 4)}(y)\\
&= \frac{3}{8}(2y)^{2/3} \cdot \frac{2}{3}(2y)^{-2/3}I_{(0, 4)}(y)\\
&= \frac{1}{4}I_{(0, 4)}(y)
\end{aligned}
\]
\(Y\) follows a \({\mathrm{Unif}}(0, 4)\) distribution.
- Verify that \(f_Y(y)\) is a valid PDF by confirming it integrates to 1 and \(f_Y(y) \ge 0\) for all \(y \in {\mathbb{R}}\).
\[
\int_0^4 \frac{1}{4}\; dy = \frac{1}{4} \cdot 4 = 1
\] Further, \(f_Y(y) = \frac{1}{4}I_{(0, 4)}(y)\) is zero for \(y \le 0\) or \(y \ge 4\). For \(0 < y <4\), \(f_/Y(y) = \frac{1}{4}I_{(0, 4)}(y) = 1/4\) which is greater than zero. Therefore, \(f_Y(y) \ge 0\) for all \(y \in {\mathbb{R}}\).
- Find \(\mathbb{E}[Y]\) and \(\text{Var}(Y)\) using \(f_Y(y)\).
Since \(Y \sim \text{Uniform}(0, 4)\):
\[
\begin{aligned}
\mathbb{E}[Y] &= \int_0^4 y \cdot \frac{1}{4}\; dy \\
&= \frac{1}{4} \cdot \frac{y^2}{2}\Bigg|_0^4 \\
&= \frac{1}{4} \cdot 8 \\
&= 2.
\end{aligned}
\]
\[
\begin{aligned}
\mathbb{E}[Y^2] &= \int_0^4 y^2 \cdot \frac{1}{4}\; dy\\
&= \frac{1}{4} \cdot \frac{y^3}{3}\Bigg|_0^4 \\
&= \frac{1}{4} \cdot \frac{64}{3} \\
&= \frac{16}{3}.
\end{aligned}
\]
Thus,
\[
\text{Var}(Y) = \mathbb{E}[Y^2] - (\mathbb{E}[Y])^2 = \frac{16}{3} - 4 = \frac{4}{3}.
\]
- Verify \(\mathbb{E}[Y]\) by computing \(\mathbb{E}[X^3/2]\) directly from \(f_X(x)\).
\[
\begin{aligned}
\mathbb{E}[Y] &= E\!\left[\frac{X^3}{2}\right]\\
&= \int_0^2 \frac{x^3}{2} \cdot \frac{3}{8}x^2\; dx\\
&= \frac{3}{16} \int_0^2 x^5\; dx\\
&= \frac{3}{16} \cdot \frac{x^6}{6}\Bigg|_0^2\\
&= \frac{3}{16} \cdot \frac{64}{6}\\
&= \frac{3}{16} \cdot \frac{32}{3}\\
&= 2
\end{aligned}
\]
Question 17
Let \(X\) and \(Y\) be discrete random variables with the following joint PMF:
| \(X = 0\) |
0.1 |
0.2 |
0.1 |
| \(X = 1\) |
0.1 |
0.3 |
0.2 |
- Find \(\mathbb{E}[X]\), \(\mathbb{E}[Y]\), and \(\mathbb{E}[XY]\).
Marginal of \(X\):
\[
P(X = 0) = 0.1 + 0.2 + 0.1 = 0.4, \qquad P(X = 1) = 0.1 + 0.3 + 0.2 = 0.6.
\]
\[
\mathbb{E}[X] = 0(0.4) + 1(0.6) = 0.6.
\]
Marginal of \(Y\):
\[
P(Y = 0) = 0.2, \quad P(Y = 1) = 0.5, \quad P(Y = 2) = 0.3.
\]
\[
\mathbb{E}[Y] = 0(0.2) + 1(0.5) + 2(0.3) = 1.1.
\]
Computing \(\mathbb{E}[XY]\): Only the cell \((X=1, Y=1)\) and \((X=1, Y=2)\) contribute (since \(xy = 0\) whenever \(x = 0\)):
\[
\mathbb{E}[XY] = (1)(1)(0.3) + (1)(2)(0.2) = 0.3 + 0.4 = 0.7.
\]
- Compute \(\operatorname{Cov}(X, Y)\).
Using the shortcut formula:
\[
\operatorname{Cov}(X, Y) = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y] = 0.7 - (0.6)(1.1) = 0.7 - 0.66 = 0.04.
\]
Since \(\operatorname{Cov}(X,Y) > 0\), \(X\) and \(Y\) tend to increase together.
Question 18
Let \(X\) and \(Y\) be continuous random variables with joint PDF
\[
f(x, y) = \frac{15}{32}\, x\, y^2,I_{(0, 2)}(x)I_{(0, x)}(y)
\]
Compute \(\operatorname{Cov}(X,Y)\).
Recall\(\operatorname{Cov}(X, Y) = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]\).
\(\mathbb{E}[X]\):
\[
\begin{aligned}
\mathbb{E}[X] &= \int_0^2 \int_0^x x \cdot \frac{15}{32}\,xy^2\; dy\; dx\\
&= \frac{15}{32} \int_0^2 x^2 \int_0^x y^2\; dy\; dx\\
&= \frac{15}{32} \int_0^2 x^2 \cdot \frac{x^3}{3}\; dx\\
&= \frac{15}{96} \int_0^2 x^5\; dx\\
&= \frac{15}{96} \cdot \frac{64}{6}\\
&= \frac{5}{3}.\\
\end{aligned}
\]
\(\mathbb{E}[Y]\):
\[
\begin{aligned}
\mathbb{E}[Y] &= \int_0^2 \int_0^x y \cdot \frac{15}{32}\,xy^2\; dy\; dx\\
&= \frac{15}{32} \int_0^2 x \int_0^x y^3\; dy\; dx\\
&= \frac{15}{32} \int_0^2 x \cdot \frac{x^4}{4}\; dx\\
&= \frac{15}{128} \int_0^2 x^5\; dx\\
&= \frac{15}{128} \cdot \frac{64}{6}\\
&= \frac{5}{4}.
\end{aligned}
\]
\(\mathbb{E}[XY]\):
\[
\begin{aligned}
\mathbb{E}[XY] &= \int_0^2 \int_0^x xy \cdot \frac{15}{32}\,xy^2\; dy\; dx\\
&= \frac{15}{32} \int_0^2 x^2 \int_0^x y^3\; dy\; dx\\
&= \frac{15}{32} \int_0^2 x^2 \cdot \frac{x^4}{4}\; dx\\
&= \frac{15}{128} \int_0^2 x^6\; dx\\
&= \frac{15}{128} \cdot \frac{128}{7}\\
&= \frac{15}{7}.
\end{aligned}
\]
Thus,
\[
\begin{aligned}
\operatorname{Cov}(X, Y) &= \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y] \\
&= \frac{15}{7} - \frac{5}{3} \cdot \frac{5}{4} \\
&= \frac{15}{7} - \frac{25}{12} \\
&= \frac{5}{84} \approx 0.0595.
\end{aligned}
\]
Question 19
Let \(X\) and \(Y\) have joint PMF:
| \(X = 0\) |
0.2 |
0.3 |
0.1 |
| \(X = 1\) |
0.1 |
0.2 |
0.1 |
Let \(Z = 2X + Y\).
- Find the MGF of \(Z = 2X + Y\) directly from the joint PMF.
We compute \(\mathbb{E}[e^{tZ}] = \mathbb{E}[e^{t(2X + Y)}]\) by summing \(e^{t(2x+y)} \cdot p(x,y)\) over all cells:
| \((0, -1)\) |
\(-1\) |
0.2 |
\(0.2\, e^{-t}\) |
| \((0,\ 0)\) |
\(0\) |
0.3 |
\(0.3\) |
| \((0,\ 1)\) |
\(1\) |
0.1 |
\(0.1\, e^{t}\) |
| \((1, -1)\) |
\(1\) |
0.1 |
\(0.1\, e^{t}\) |
| \((1,\ 0)\) |
\(2\) |
0.2 |
\(0.2\, e^{2t}\) |
| \((1,\ 1)\) |
\(3\) |
0.1 |
\(0.1\, e^{3t}\) |
\[
m_Z(t) = 0.2e^{-t} + 0.3 + 0.2e^{t} + 0.2e^{2t} + 0.1e^{3t}.
\]
- Use the MGF to find \(\mathbb{E}[Z]\).
Differentiate and evaluate at \(t = 0\):
\[
m_Z'(t) = -0.2e^{-t} + 0.2e^{t} + 0.4e^{2t} + 0.3e^{3t}.
\]
\[
\mathbb{E}[Z] = m_Z'(0) = -0.2 + 0.2 + 0.4 + 0.3 = 0.7.
\]
We can also verify this result: \(\mathbb{E}[Z] = \\E[2X+Y] = 2\mathbb{E}[X] + \mathbb{E}[Y]\).
\(\mathbb{E}[X] = 0(0.6) + 1(0.4) = 0.4\). \(\mathbb{E}[Y] = (-1)(0.3) + 0(0.5) + 1(0.2) = -0.1\).
\(2(0.4) + (-0.1) = 0.7\).
Question 20
A hospital emergency department receives an average of 4 patients per hour during the overnight shift. Assume patient arrivals follow a Poisson distribution. Let \(X\) be the number of patients that arrive in a given hour.
- Find the MGF of \(X\) and use it to find \(E[X]\).
\(X \sim {\mathrm{Poiss}}(\lambda)\).
The MGF of a \({\mathrm{Poiss}}(\lambda)\) random variable is derived from the definition \(m_X(t) = E[e^{tX}]\):
\[
m_X(t) = \sum_{x=0}^{\infty} e^{tx} \cdot \frac{e^{-\lambda}\lambda^x}{x!}
= e^{-\lambda} \sum_{x=0}^{\infty} \frac{({\lambda e^t})^x}{x!}
= e^{-\lambda} \cdot e^{\lambda e^t} = e^{\lambda(e^t - 1)}
\]
With \(\lambda = 4\):
\[
m_X(t) = e^{4(e^t - 1)}
\]
To find \(E[X]\), differentiate \(m_X(t)\) and evaluate at \(t = 0\):
\[
m_X'(t) = e^{4(e^t - 1)} \cdot 4e^t
\]
\[
E[X] = m_X'(0) = e^{4(1-1)} \cdot 4e^0 = 1 \cdot 4 = 4.
\]
As expected, the mean equals the rate parameter \(\lambda = 4\).
Question 21
Let \(X \sim {\mathrm{Gam}}(\alpha, \lambda)\) where \(\alpha>0\) and \(\lambda > 0\). Find the MGF \(m_X(t)\).
\[
\begin{aligned}
\mathbb{E}[\exp(tX)] &= \int_0^\infty \exp(tx) \frac{\lambda^\alpha}{\Gamma(\alpha)} x^{\alpha - 1} e^{-\lambda x} \mathsf{d}x \\
&= \frac{\lambda^\alpha}{\Gamma(\alpha)} \int_0^\infty x^{\alpha - 1} e^{tx - \lambda x} \mathsf{d}x \\
&= \frac{\lambda^\alpha}{\Gamma(\alpha)} \int_0^\infty x^{\alpha - 1} e^{-(\lambda - t)x} \mathsf{d}x \\
&= \frac{\lambda^\alpha}{\Gamma(\alpha)} \cdot \frac{\Gamma(\alpha)}{(\lambda - t)^\alpha}
\quad\quad\text{because } \int_0^\infty x^{\alpha - 1} e^{-(\lambda - t)x} \mathsf{d}x =
\frac{\Gamma(\alpha)}{(\lambda - t)^\alpha},\\
&= \left(\frac{\lambda}{\lambda - t}\right)^\alpha
= \left(1 - \frac{t}{\lambda}\right)^{-\alpha}.
\end{aligned}
\]
Question 22
A public health nurse has a group of 20 patients waiting to receive their test results, of which 8 have tested positive. Due to limited appointment slots, the nurse randomly selects 6 patients to be seen first. Let \(X\) be the number of positive patients in the group of 6.
- What is the probability that exactly 3 of the selected patients have tested positive?
\(X\) follows a Hypergeometric distribution with population size \(N = 20\), number of positive patients \(K = 8\), and sample size \(n = 6\):
\[
P(X = 3) = \frac{\dbinom{K}{x}\dbinom{N-K}{n-x}}{\dbinom{N}{n}}
= \frac{\dbinom{8}{3}\dbinom{12}{6-3}}{\dbinom{20}{6}}= \frac{56 \times 220}{38760}
= \frac{12320}{38760}
\]
- What is the probability that at most 1 of the selected patients has tested positive?
\(P(X \leq 1) = P(X = 0) + P(X = 1)\)
\[
P(X = 0) = \frac{\dbinom{8}{0}\dbinom{12}{6}}{\dbinom{20}{6}}
= \frac{1 \times 924}{38760}
= \frac{924}{38760}
\]
\[
P(X = 1) = \frac{\dbinom{8}{1}\dbinom{12}{5}}{\dbinom{20}{6}}
= \frac{8 \times 792}{38760}
= \frac{6336}{38760}
\]
Thus,
\[
P(X \leq 1) = \frac{924 + 6336}{38760} = \frac{7260}{38760} = 0.1873
\]
Question 23
A forensic laboratory uses a DNA-based screening test to detect the presence of a particular genetic marker associated with an inherited blood clotting disorder. The marker is present in 3% of the population.
Among all patients who were correctly diagnosed in a clinical study (whether they carry the marker or not), the test gave the correct result 92% of the time.
Let \(M\) be the event that a patient carries the marker, and \(T\) be the event that the screening test returns positive. Researchers also found that the probability of a \(\mathbb{P}(T \ \vert\ M^c\)) is known to be four times \(\mathbb{P}(T^c \ \vert\ M)\).
- Find \(P(T \mid M^c)\) and \(P(T^c \mid M)\). Hint: how do these relate to \(P(\text{correct result})\)?
Let \(P(T \mid M^c) = p\) and \(P(T^c \mid M) = q\) We are told:
\[p = 4q\]
A correct diagnosis occurs when a patient with the marker tests positive, or a patient without the marker tests negative. The overall accuracy is given by:
\[
\begin{aligned}
P(\text{correct}) &= P(T \mid M)P(M) + P(T^c \mid M^c)P(M^c) \\
&= (1 - q)(0.03) + (1 - p)(0.97) = 0.92
\end{aligned}
\]
Substituting \(p = 4q\):
\[
(1 - q)(0.03) + (1 - 4q)(0.97) = 0.92
\]
\[
0.03 - 0.03q + 0.97 - 3.88q = 0.92
\]
\[
1 - 3.91q = 0.92 \implies q = \frac{0.08}{3.91} \approx 0.02
\]
Thus:
\[
P(T^c \mid M) = 0.02, \qquad P(T \mid M^c) = 0.08
\]
(b) Calculate the probability that a randomly selected patient tests positive.
Using the results from (a), \(P(T \mid M) = 1 - 0.02 = 0.98\), and applying the law of total probability:
\[
\begin{aligned}
P(T) &= P(T \mid M)P(M) + P(T \mid M^c)P(M^c)\\
&= (0.98)(0.03) + (0.08)(0.97)\\
&= 0.0294 + 0.0776\\
&= 0.107
\end{aligned}
\]
(c) Given that a patient tests positive, what is the probability they actually carry the marker?
By Bayes’ Theorem:
\[
\begin{aligned}
P(M \mid T) &= \frac{P(T \mid M)\,P(M)}{P(T)}\\
&= \frac{(0.98)(0.03)}{0.107}\\
&= \frac{0.0294}{0.107}\\
&= 0.2748
\end{aligned}
\]