University of Washington
Department of Mathematics Midterm Examination Solutions

MATH 394: Probability I

Summer 2026 — Instructor: Arman Jahangiri

Exam date: July 22, 2026

Answering time: 60 minutes

Total: 60 points

Problem 1. 5 points

A student organization has 7 mathematics graduate students, 5 statistics graduate students, and 6 undergraduate students. A committee of 5 students is selected uniformly at random from all committees of 5.

Find the probability that the committee contains exactly 2 mathematics graduate students, exactly 1 statistics graduate student, and exactly 2 undergraduate students.

Give an exact answer. A correct expression involving binomial coefficients is acceptable.

Solution

There are \[ \binom {18}{5} \] possible committees of 5 students, and all are equally likely.

To form a committee of the required type, choose:

Thus the number of favorable committees is \[ \binom {7}{2}\binom {5}{1}\binom {6}{2}. \] Therefore, \[ P(\text {required composition}) = \frac {\binom {7}{2}\binom {5}{1}\binom {6}{2}}{\binom {18}{5}}. \] Numerically, \[ \binom {7}{2}\binom {5}{1}\binom {6}{2}=21\cdot 5\cdot 15=1575, \qquad \binom {18}{5}=8568, \] so \[ \frac {1575}{8568}=\frac {525}{2856}=\frac {175}{952}. \]

Problem 2. 10 points

A spam filter is designed by looking at commonly occurring phrases in spam. Suppose that \(80\%\) of email is spam. In \(10\%\) of the spam emails, the phrase “free money” is used, whereas this phrase is only used in \(1\%\) of non-spam emails. A new email has just arrived, which does mention “free money.” What is the probability that it is spam?

Solution

(a) Let \(S\) be the event that the email is spam, and let \(F\) be the event that the email contains the phrase “free money.” We are given \[ P(S)=0.80,\qquad P(F\mid S)=0.10,\qquad P(F\mid S^c)=0.01. \] Bayes’ rule gives \[ P(S\mid F) = \frac {P(F\mid S)P(S)}{P(F\mid S)P(S)+P(F\mid S^c)P(S^c)}. \] Substituting the given values, \[ P(S\mid F) = \frac {0.10\cdot 0.80}{0.10\cdot 0.80+0.01\cdot 0.20} = \frac {0.08}{0.082} = \frac {40}{41}. \]

Problem 3. 10 points

Let \(A\) and \(B\) be events.

(a)
Let \(S\) be a sample space and \(A_1,\dots , A_n\in \mathcal {F}\) be events. Explain what it means for \(A_i's\) to for a partition of \(S\), (3 points)
(b)
Suppose \(P(A) = 0.6\) and \(P(B)=0.5\). Can \(A\) and \(B\) be disjoint? (3 points)
(c)
Suppose that \(A\) is independent of itself. Use the definition of independence to conclude that \(P(A)=0\) or \(P(A)=1\). (4 points)

Solution

(a) The events \(A_1,\ldots ,A_n\) form a partition of the sample space \(S\) if they are pairwise disjoint, \[ A_i\cap A_j=\emptyset ,\qquad i\neq j, \] and \[ \bigcup _{i=1}^n A_i=S. \]

(b) If \(A\) and \(B\) were disjoint, then \[ 1\geq P(A\cup B)=P(A)+P(B)=0.6+0.5=1.1, \] which is impossible since probabilities cannot exceed \(1\). Hence \(A\) and \(B\) cannot be disjoint.

(c) If \(A\) is independent of itself, then \[ P(A\cap A)=P(A)P(A). \] Since \(A\cap A=A\), \[ P(A)=P(A)^2. \] Let \(p=P(A)\). Then \[ p(1-p)=0, \] so \(p=0\) or \(p=1\).

Final Answer

(a) \(\{A_i\}_{i=1}^n\) is a partition iff the events are pairwise disjoint and their union is \(S\).

(b) No.

(c) \[ {P(A)=0\quad \text {or}\quad P(A)=1.} \]

Problem 4. 15 points

Suppose that \(X\) has probability density function \[ f_X(x)= \begin {cases} c\,x(2-x), & 0<x<2,\\ 0, & \text {otherwise}, \end {cases} \] where \(c\) is a constant.

(a)
Find \(c\). (3 points)
(b)
Find the cumulative distribution function \(F_X(x)\) for all real \(x\). (5 points)
(c)
Find \(P(X>3/2)\). (2 points)
(d)
Find \(\mathbb {E}[X]\) and \(\operatorname {Var}(X)\). (5 points)

Solution

(a) A density must integrate to \(1\), so \begin {align*} 1 &=c\int _0^2 x(2-x)\,dx\\ &=c\int _0^2(2x-x^2)\,dx\\ &=c\left [x^2-\frac {x^3}{3}\right ]_0^2 =c\left (4-\frac 83\right ) =c\frac 43. \end {align*}

Hence \[ c=\frac 34. \]

(b) For \(0<x<2\), \begin {align*} F_X(x) &=\int _0^x \frac 34t(2-t)\,dt\\ &=\frac 34\left [t^2-\frac {t^3}{3}\right ]_0^x\\ &=\frac {3x^2}{4}-\frac {x^3}{4}. \end {align*}

Therefore, \[ F_X(x)= \begin {cases} 0, & x\leq 0,\\[2mm] \displaystyle \frac {3x^2-x^3}{4}, & 0<x<2,\\[2mm] 1, & x\geq 2. \end {cases} \]

(c) Using the CDF, \begin {align*} P\left (X>\frac 32\right ) &=1-F_X\left (\frac 32\right )\\ &=1-\frac {3(3/2)^2-(3/2)^3}{4}\\ &=1-\frac {27}{32} =\frac 5{32}. \end {align*}

(d) First, \begin {align*} \mathbb {E}[X] &=\int _0^2 x\frac 34x(2-x)\,dx\\ &=\frac 34\int _0^2(2x^2-x^3)\,dx\\ &=\frac 34\left [\frac {2x^3}{3}-\frac {x^4}{4}\right ]_0^2 =1. \end {align*}

Also, \begin {align*} \mathbb {E}[X^2] &=\int _0^2 x^2\frac 34x(2-x)\,dx\\ &=\frac 34\int _0^2(2x^3-x^4)\,dx\\ &=\frac 34\left [\frac {x^4}{2}-\frac {x^5}{5}\right ]_0^2\\ &=\frac 65. \end {align*}

Thus \[ \operatorname {Var}(X)=\mathbb {E}[X^2]-\bigl (\mathbb {E}[X]\bigr )^2 =\frac 65-1=\frac 15. \]

Final Answer

\[ {c=\frac 34},\qquad {P(X>3/2)=\frac 5{32}},\qquad {\mathbb {E}[X]=1},\qquad {\operatorname {Var}(X)=\frac 15}. \] The CDF is \[ { F_X(x)= \begin {cases} 0, & x\leq 0,\\ (3x^2-x^3)/4, & 0<x<2,\\ 1, & x\geq 2. \end {cases}} \]

Problem 5. 15 points

A company has found that \(40\%\) of customers who visit its website make a purchase. Suppose that \(200\) customers visit the website independently on a particular day.

Let \(X\) denote the number of customers who make a purchase.

1.
Determine whether a Poisson approximation or a normal approximation is more appropriate by verifying the required conditions. Then clearly state the approximating distribution; that is, provide \(\lambda \) if a Poisson approximation is appropriate, or provide \(\mu \) and \(\sigma ^2\) if a normal approximation is appropriate. (5 points)
2.
Use the approximation from part (1) to estimate each of the following: (10 points)
(a)
\[ \mathbb {P}(70<X<90) \] (3 points)
(b)
\[ \mathbb {P}(X>90) \] (3 points)
(c)
\[ \mathbb {P}(X=80) \] (4 points)

Solution

Since each customer independently makes a purchase with probability \(p=0.40\), we have \[ X\sim \operatorname {Bin}(200,0.40). \]

(1) For a Poisson approximation to be appropriate, \(n\) should be large and \(p\) should be small. Although \[ n=200 \] is large, \(p=0.40\) is not small. Therefore, a Poisson approximation is not appropriate.

For a normal approximation, we verify that \[ np=200(0.40)=80 \] and \[ n(1-p)=200(0.60)=120. \]

Since both quantities are greater than \(10\), the normal approximation is appropriate.

The mean and variance of \(X\) are \[ \mu =np=80 \] and \[ \sigma ^2=np(1-p)=200(0.40)(0.60)=48. \]

Therefore, we approximate \(X\) by \[ Y\sim N(80,48), \] where \[ \sigma =\sqrt {48}\approx 6.93. \]

(2)

(a) Using the continuity correction, \[ \mathbb {P}(70<X<90) \approx \mathbb {P}(70.5<Y<89.5). \]

Standardizing, \[ \begin {aligned} \mathbb {P}(70.5<Y<89.5) &= \mathbb {P}\left ( \frac {70.5-80}{\sqrt {48}} < Z < \frac {89.5-80}{\sqrt {48}} \right )\\ &\approx \mathbb {P}(-1.37<Z<1.37). \end {aligned} \]

Using the standard normal table, \[ \Phi (1.37)=0.9147 \] and \[ \Phi (-1.37)=1-\Phi (1.37)=0.0853. \]

Therefore, \[ \mathbb {P}(70<X<90) \approx 0.9147-0.0853 = 0.8294. \]

(b) Using the continuity correction, \[ \mathbb {P}(X>90) \approx \mathbb {P}(Y>90.5). \]

Standardizing, \[ \begin {aligned} \mathbb {P}(Y>90.5) &= \mathbb {P}\left ( Z> \frac {90.5-80}{\sqrt {48}} \right )\\ &\approx \mathbb {P}(Z>1.52). \end {aligned} \]

Using the standard normal table, \[ \Phi (1.52)=0.9357. \]

Thus, \[ \mathbb {P}(X>90) \approx 1-\Phi (1.52) = 1-0.9357 = 0.0643. \]

(c) Using the continuity correction, \[ \mathbb {P}(X=80) \approx \mathbb {P}(79.5<Y<80.5). \]

Standardizing, \[ \begin {aligned} \mathbb {P}(79.5<Y<80.5) &= \mathbb {P}\left ( \frac {79.5-80}{\sqrt {48}} < Z < \frac {80.5-80}{\sqrt {48}} \right )\\ &\approx \mathbb {P}(-0.07<Z<0.07). \end {aligned} \]

Using the standard normal table, \[ \Phi (0.07)=0.5279 \] and \[ \Phi (-0.07)=0.4721. \]

Therefore, \[ \mathbb {P}(X=80) \approx 0.5279-0.4721 = 0.0558. \]