Probability Practice

GitHub Changelog - Last modified Aug 24, 2026

Suggest edit

0 of 27 completed

Create worksheet

Title

Selected problems

No problems selected.

Difficulty

Direct

1.1Count a Simple Sample Space

Exam I | Problem 1.1 |

A fair coin is flipped twice. What is the probability of getting exactly one head?

Solution

The sample space is

\{HH, HT, TH, TT\}

There are 4 equally likely outcomes, and 2 of them have exactly one head: $HT$ and $TH$.

P(\text{exactly one head}) = \frac{2}{4} = \frac{1}{2}

1.2Choose a Committee

Exam I | Problem 1.2 |

How many ways can you choose 3 students from a group of 8?

Solution

Use combinations because the order does not matter:

{8 \choose 3} = \frac{8!}{3!5!} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56

1.3Arrange Runners

Exam I | Problem 1.3 |

How many ordered outcomes are there for the gold, silver, and bronze places among 7 runners?

Solution

Use permutations because order matters:

{}_7P_3 = \frac{7!}{(7-3)!} = 7 \cdot 6 \cdot 5 = 210

1.4Count Distinct Arrangements

Exam I | Problem 1.4 |

How many distinct arrangements of the letters in BOOK are there?

Solution

The word BOOK has 4 letters, with O repeated twice. The number of distinct arrangements is

\frac{4!}{2!} = 12

1.5Use the Complement Rule

Exam I | Problem 1.5 |

Complement Rule

A fair die is rolled 3 times. What is the probability of getting at least one 6?

Solution

It is easier to compute the complement: no 6s on all 3 rolls.

P(\text{at least one 6}) = 1 - \left(\frac{5}{6}\right)^3

1 - \frac{125}{216} = \frac{91}{216}

1.6Apply the Addition Rule

Exam I | Problem 1.6 |

Suppose

P(A)=0.42,\quad P(B)=0.31,\quad P(A \cap B)=0.08.

Find $P(A \cup B)$.

Solution

Use the addition rule:

P(A \cup B)=P(A)+P(B)-P(A \cap B)

Substitute the given values:

P(A \cup B)=0.42+0.31-0.08=0.65

1.7Find a Conditional Probability

Exam I | Problem 1.7 |

Conditional Probability

Suppose

P(A \cap B)=0.12 \quad \text{and} \quad P(B)=0.3.

Find $P(A \mid B)$.

Solution

Use the definition of conditional probability:

P(A \mid B)=\frac{P(A \cap B)}{P(B)}

P(A \mid B)=\frac{0.12}{0.3}=0.4

1.8Evaluate a Bernoulli Mean and Variance

Exam I | Problem 1.8 |

If $X \sim \mathrm{Bernoulli}(0.7)$, find $E[X]$ and $\mathrm{Var}(X)$.

Solution

For a Bernoulli random variable,

E[X]=p \quad \text{and} \quad \mathrm{Var}(X)=p(1-p)

Here $p=0.7$, so

$$ E[X]=0.7 $$

and

\mathrm{Var}(X)=0.7(0.3)=0.21

1.9Compute a Binomial Probability

Exam I | Problem 1.9 |

Binomial Distribution

If $X \sim \mathrm{Binomial}(5,0.2)$, find $P(X=2)$.

Solution

Use the binomial formula:

P(X=2)={5 \choose 2}(0.2)^2(0.8)^3

Now compute:

{5 \choose 2}=10

$$ P(X=2)=10(0.04)(0.512)=0.2048 $$

1.10Standardize a Normal Random Variable

Exam I | Problem 1.10 |

If $X \sim \mathcal{N}(100,15^2)$, what is the $z$-score for $x=130$?

Solution

Standardize using

z=\frac{x-\mu}{\sigma}

Here $\mu=100$ and $\sigma=15$, so

z=\frac{130-100}{15}=\frac{30}{15}=2

Difficulty

Integrated

2.1Count Draws Without Replacement

Exam II | Problem 2.1 |

A box contains 6 good parts and 4 defective parts. Two parts are drawn without replacement. What is the probability that exactly 1 part is defective?

Solution

Use the hypergeometric model:

P(X=1)=\frac{{4 \choose 1}{6 \choose 1}}{{10 \choose 2}}

Compute the values:

\frac{4 \cdot 6}{45}=\frac{24}{45}=\frac{8}{15}

2.2Update a Belief with Bayes' Theorem

Exam II | Problem 2.2 |

A disease affects 2% of a population. A test is 95% accurate for people who have the disease, and it gives a false positive 10% of the time for people who do not have the disease. If a person tests positive, what is the probability that the person actually has the disease?

Solution

Let $D$ be the event that the person has the disease and $+$ be the event of a positive test.

By Bayes' theorem,

P(D \mid +)=\frac{P(+ \mid D)P(D)}{P(+)}

First find the total probability of a positive test:

P(+)=P(+ \mid D)P(D)+P(+ \mid D^c)P(D^c)

$$ P(+)=0.95(0.02)+0.10(0.98)=0.019+0.098=0.117 $$

Now compute the posterior:

P(D \mid +)=\frac{0.95(0.02)}{0.117}=\frac{0.019}{0.117}=\frac{19}{117}

So the probability is about $0.162$.

2.3Use the Geometric Memoryless Property

Exam II | Problem 2.3 |

If $X \sim \mathrm{Geometric}(0.25)$, find $P(X>7 \mid X>3)$.

Solution

Use the memoryless property:

P(X>m+n \mid X>m)=P(X>n)

Here we want

P(X>7 \mid X>3)=P(X>4)

For a geometric random variable with success probability $0.25$,

P(X>4)=(1-0.25)^4=\left(\frac{3}{4}\right)^4=\frac{81}{256}

So the probability is about $0.316$.

2.4Solve a Negative Binomial Count

Exam II | Problem 2.4 |

Negative Binomial Distribution

A basketball player makes each free throw with probability $0.6$, independently. What is the probability that the third make occurs on the fifth attempt?

Solution

This is a negative binomial situation with $r=3$ and $k=5$.

P(X=5)={5-1 \choose 3-1}(0.6)^3(0.4)^{5-3}

P(X=5)={4 \choose 2}(0.6)^3(0.4)^2

$$ =6(0.216)(0.16)=0.20736 $$

2.5Compute a Poisson Count Probability

Exam II | Problem 2.5 |

Poisson Distribution

A support line receives 4 calls per hour on average. What is the probability of exactly 2 calls in a half hour?

Solution

In half an hour, the mean count is

\lambda=4 \cdot \frac{1}{2}=2

Use the Poisson formula:

P(X=2)=e^{-2}\frac{2^2}{2!}

P(X=2)=2e^{-2}

which is about $0.271$.

2.6Compute an Exponential Waiting-Time Probability

Exam II | Problem 2.6 |

Exponential Distribution

The waiting time to the next event has an exponential distribution with rate 3 per hour. What is the probability of waiting more than 20 minutes?

Solution

Convert 20 minutes to hours:

20 \text{ min}=\frac{1}{3} \text{ hour}

For an exponential random variable,

P(X>t)=e^{-\lambda t}

P\left(X>\frac{1}{3}\right)=e^{-3(1/3)}=e^{-1}

This is about $0.368$.

2.7Use Inclusion-Exclusion for Three Events

Exam II | Problem 2.7 |

Inclusion-Exclusion

Suppose

P(A)=0.5,\quad P(B)=0.4,\quad P(C)=0.3

and

P(A \cap B)=0.2,\quad P(A \cap C)=0.1,\quad P(B \cap C)=0.08,\quad P(A \cap B \cap C)=0.05.

Find $P(A \cup B \cup C)$.

Solution

Use inclusion-exclusion:

P(A \cup B \cup C) =P(A)+P(B)+P(C)-P(A \cap B)-P(A \cap C)-P(B \cap C)+P(A \cap B \cap C)

Substitute the values:

$$ 0.5+0.4+0.3-0.2-0.1-0.08+0.05=0.87 $$

2.8Check Independence from a Joint Table

Exam II | Problem 2.8 |

A joint pmf is given by

\begin{array}{c|cc} & Y=0 & Y=1 \\ \hline X=0 & 0.12 & 0.18 \\ X=1 & 0.28 & 0.42 \end{array}

Find the marginal distributions of $X$ and $Y$, and determine whether $X$ and $Y$ are independent.

Solution

First find the marginals by summing rows and columns.

For $X$:

$$ P(X=0)=0.12+0.18=0.30 $$

$$ P(X=1)=0.28+0.42=0.70 $$

For $Y$:

$$ P(Y=0)=0.12+0.28=0.40 $$

$$ P(Y=1)=0.18+0.42=0.60 $$

Check one joint probability:

$$ P(X=0)P(Y=0)=0.30(0.40)=0.12 $$

The same factorization works for the other entries as well, so the joint distribution factors into the product of the marginals.

Therefore, $X$ and $Y$ are independent.

Difficulty

Applied

3.1Find an Expected Value from a Discrete Distribution

Final | Problem 3.1 |

Let $X$ take the values $1$, $2$, and $4$ with probabilities $0.2$, $0.5$, and $0.3$, respectively. If the payoff is $X^2$, what is the expected payoff?

Solution

Use the formula for the expected value of a function of $X$:

E[X^2]=\sum_x x^2 P(X=x)

$$ E[X^2]=1^2(0.2)+2^2(0.5)+4^2(0.3) $$

$$ =0.2+2+4.8=7 $$

The expected payoff is $7$.

3.2Model an Expected Count with Indicators

Final | Problem 3.2 |

A fair die is rolled 5 times. Let $X$ be the number of adjacent pairs that match. Find $E[X]$.

Solution

There are 4 adjacent pairs: $(1,2)$, $(2,3)$, $(3,4)$, and $(4,5)$.

Let $I_i$ be the indicator for the event that the $i$th pair matches. Then

$$ X=I_1+I_2+I_3+I_4 $$

Each pair matches with probability $1/6$, so

E[I_i]=\frac{1}{6}

By linearity of expectation,

E[X]=4\cdot \frac{1}{6}=\frac{2}{3}

3.3Model a Sampling Situation

Final | Problem 3.3 |

A box has 8 good components and 4 defective components. Three components are drawn without replacement. What is the probability that at least one component is defective?

Solution

Use the complement event: no defective components are drawn.

There are 8 good components, so

P(\text{no defective})=\frac{{8 \choose 3}}{{12 \choose 3}}

Thus

P(\text{at least one defective})=1-\frac{{8 \choose 3}}{{12 \choose 3}}

Compute the combinations:

1-\frac{56}{220}=1-\frac{14}{55}=\frac{41}{55}

3.4Waiting Time to the Third Event

Final | Problem 3.4 |

Calls arrive at a rate of 2 per hour. Under the gamma model, what is the mean waiting time until the third call?

Solution

The waiting time to the third event is modeled by a gamma distribution with shape $3$ and rate $2$.

For $X \sim \mathrm{Gamma}(\alpha,\lambda)$,

E[X]=\frac{\alpha}{\lambda}

So here

E[X]=\frac{3}{2}

The mean waiting time is $1.5$ hours.

3.5Infer a Beta Model for a Proportion

Final | Problem 3.5 |

A parameter $p$ represents a conversion rate, so it must stay between 0 and 1. Which distribution from the note is a natural choice for modeling $p$?

Solution

The beta distribution is defined on the interval $[0,1]$, so it is a natural model for probabilities and proportions.

Therefore, the appropriate choice is the beta distribution.

Difficulty

Challenge

4.1Combine Total Probability and Bayes

Final | Problem 4.1 |

Machine A makes 60% of the items and has a defect rate of 1%. Machine B makes the other 40% and has a defect rate of 4%. If an item is defective, what is the probability that it came from Machine A?

Solution

Let $A$ be the event that the item came from Machine A and let $D$ be the event that the item is defective.

First use the law of total probability:

P(D)=P(D \mid A)P(A)+P(D \mid B)P(B)

$$ P(D)=0.01(0.6)+0.04(0.4)=0.006+0.016=0.022 $$

Now apply Bayes' theorem:

P(A \mid D)=\frac{P(D \mid A)P(A)}{P(D)}

P(A \mid D)=\frac{0.01(0.6)}{0.022}=\frac{0.006}{0.022}=\frac{3}{11}

So the probability is about $0.273$.

4.2Approximate a Binomial Count with a Normal Model

Final | Problem 4.2 |

If $X \sim \mathrm{Binomial}(100,0.2)$, approximate $P(16 \le X \le 24)$ using a normal model.

Solution

For a binomial random variable,

\mu=np=100(0.2)=20

and

\sigma=\sqrt{np(1-p)}=\sqrt{100(0.2)(0.8)}=\sqrt{16}=4

Use the continuity correction:

P(16 \le X \le 24) \approx P(15.5 \le Y \le 24.5)

where $Y \sim \mathcal{N}(20,16)$.

Standardize:

z_1=\frac{15.5-20}{4}=-1.125

z_2=\frac{24.5-20}{4}=1.125

P(16 \le X \le 24) \approx P(-1.125 \le Z \le 1.125) \approx 0.739

4.3Compare Sample Sizes with Standard Error

Final | Problem 4.3 |

A population has mean 50 and standard deviation 12. Compare the standard error of the sample mean for samples of size 36 and 144. Which sample mean should be more stable?

Solution

The standard error of the sample mean is

\sigma_{\bar X}=\frac{\sigma}{\sqrt{n}}

For $n=36$:

\sigma_{\bar X}=\frac{12}{\sqrt{36}}=\frac{12}{6}=2

For $n=144$:

\sigma_{\bar X}=\frac{12}{\sqrt{144}}=\frac{12}{12}=1

The sample mean from 144 observations has the smaller standard error, so it should be more stable and closer to the true mean. That is the kind of behavior predicted by the law of large numbers.

4.4Use the Union Bound

Final | Problem 4.4 |

Three independent backup checks have failure probabilities 0.03, 0.05, and 0.02. Give an upper bound on the probability that at least one check fails.

Solution

Use the union bound:

P(A_1 \cup A_2 \cup A_3) \le P(A_1)+P(A_2)+P(A_3)

P(\text{at least one fails}) \le 0.03+0.05+0.02=0.10

The upper bound is $0.10$.