Probability & Statistics

Key Formulas & Concepts

Bayes' Theorem: $$P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}$$
Combinations: $$\binom{n}{k} = \frac{n!}{k!\,(n-k)!}$$
Permutations: $$P(n, k) = \frac{n!}{(n-k)!}$$
Expected Value: $$E[X] = \sum x \cdot P(X = x)$$
Variance: $$\operatorname{Var}(X) = E[X^2] - (E[X])^2$$
Binomial Distribution: $$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$
Poisson Distribution: $$P(X = k) = \frac{e^{-\lambda}\,\lambda^k}{k!}$$
Normal (68-95-99.7 Rule): ~68% within $1\sigma$, ~95% within $2\sigma$, ~99.7% within $3\sigma$
Conditional Probability: $$P(A|B) = \frac{P(A \cap B)}{P(B)}$$

1 Combinatorics — Committees Medium

A club has 8 men and 6 women. How many 5-person committees can be formed that contain at least 2 women?

Show Hint

Use complement: Total committees $-$ (committees with 0 women) $-$ (committees with exactly 1 woman). Total $= \binom{14}{5}$. Zero women $= \binom{8}{5}$. One woman $= \binom{6}{1} \times \binom{8}{4}$.

2 Bayes' Theorem Hard

A medical test is 95% accurate (true positive rate = 95%, true negative rate = 95%). A disease affects 1% of the population. If a person tests positive, what is the probability they actually have the disease?

Show Hint

$P(\text{Disease}|\text{Positive}) = \frac{P(\text{Pos}|\text{Dis}) \times P(\text{Dis})}{P(\text{Pos})}$. Use total probability: $P(\text{Pos}) = P(\text{Pos}|\text{Dis})P(\text{Dis}) + P(\text{Pos}|\text{No Dis})P(\text{No Dis}) = 0.95 \times 0.01 + 0.05 \times 0.99$.

3 Expected Value — Dice Game Medium

You pay $5 to roll two fair dice. You win the dollar amount equal to the product of the two dice if the product is odd; otherwise you win nothing. What is your expected profit?

Show Hint

The product is odd only if both dice show odd numbers. There are 9 such outcomes out of 36. List all possible odd products and compute their average, then subtract the $5 cost.

4 Binomial Distribution Medium

A fair coin is tossed 10 times. What is the probability of getting exactly 7 heads?

Show Hint

Use $B(10, 0.5)$: $P(X=7) = \binom{10}{7} \times (0.5)^7 \times (0.5)^3 = \frac{\binom{10}{7}}{2^{10}}$.

5 Conditional Probability — Cards Hard

Two cards are drawn without replacement from a standard 52-card deck. Given that the first card is a king, what is the probability that both cards are kings?

Show Hint

Given the first card is a king (already happened), there are 3 kings remaining out of 51 cards. $P(\text{second is king} \mid \text{first is king}) = \frac{3}{51}$.

6 Poisson Distribution Hard

A call center receives an average of 4 calls per minute. What is the probability of receiving exactly 6 calls in a given minute? What is the probability of receiving at most 2 calls?

Show Hint

Use Poisson with $\lambda = 4$. $P(X=6) = \frac{e^{-4} \cdot 4^6}{6!}$. For at most 2: $P(X \leq 2) = P(0) + P(1) + P(2)$.

7 Derangements Hard

Five people check their hats at a restaurant. If the hats are returned randomly, what is the probability that no one gets their own hat back?

Show Hint

This is the derangement problem. $D(n) = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!}$. For $n = 5$: $D(5) = 5!\left(1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}\right) = 44$. Probability $= \frac{44}{120}$.

8 Normal Distribution Medium

Exam scores are normally distributed with mean $\mu = 72$ and standard deviation $\sigma = 8$. What percentage of students scored between 64 and 88? What score separates the top 5%?

Show Hint

$64 = 72 - 8$ ($1\sigma$ below), $88 = 72 + 2 \times 8$ ($2\sigma$ above). Use the empirical rule or z-table. For top 5%, $z \approx 1.645$, so score $\approx 72 + 1.645 \times 8$.

9 Geometric Probability Hard

A stick of length 1 is broken at two random points. What is the probability that the three pieces can form a triangle?

Show Hint

Let the break points be at positions $x$ and $y$ (uniformly distributed). The three pieces form a triangle iff each piece $< \frac{1}{2}$. Map this to a geometric probability problem on the unit square. The answer is a simple fraction.

10 Hypothesis Testing Hard

A company claims their light bulbs last 1000 hours on average. A sample of 36 bulbs has a mean life of 980 hours with a standard deviation of 60 hours. At the 5% significance level, is there evidence the true mean is less than 1000?

Show Hint

$H_0: \mu = 1000$, $H_1: \mu < 1000$ (one-tailed). Test statistic $z = \frac{980 - 1000}{60/\sqrt{36}} = \frac{-20}{10} = -2$. Compare to critical value $-1.645$.

11 Birthday Paradox Olympiad

In a group of 30 people, what is the probability that at least two share a birthday? (Assume 365 equally likely birthdays.) Compute the exact expression and approximate numerically.

Show Hint

$P(\text{all different}) = \frac{365}{365} \cdot \frac{364}{365} \cdot \frac{363}{365} \cdots \frac{336}{365} = \frac{365!}{335! \times 365^{30}}$. Then $P(\text{at least one match}) = 1 - P(\text{all different}) \approx 0.706$.

12 Random Walk Olympiad

A particle starts at the origin on a number line. Each step, it moves $+1$ with probability $\frac{2}{3}$ or $-1$ with probability $\frac{1}{3}$. What is the probability it ever reaches position $-3$?

Show Hint

For a biased random walk with $P(+1) = p$ and $P(-1) = q = 1-p$, the probability of reaching $-n$ starting from 0 is $\left(\frac{q}{p}\right)^n$ when $p > q$. Here it's $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$.