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 $= \binom{6}{1} \times \binom{8}{4}$.

📝 Full Solution

Step 1: Total number of 5-person committees from 14 people: $\binom{14}{5} = 2002$.

Step 2: Committees with 0 women (all men): $\binom{8}{5} = 56$.

Step 3: Committees with exactly 1 woman: $\binom{6}{1} \times \binom{8}{4} = 6 \times 70 = 420$.

Step 4: Committees with at least 2 women = Total - (0 women) - (1 woman): $$2002 - 56 - 420 = 1526$$

Answer: 1526 committees

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: $P(\text{Pos}) = P(\text{Pos}|\text{Dis})P(\text{Dis}) + P(\text{Pos}|\text{No Dis})P(\text{No Dis}) = 0.95 \times 0.01 + 0.05 \times 0.99$.

📝 Full Solution

Step 1: Define events: $D$ = has disease, $+$ = tests positive. Given: $P(+|D) = 0.95$, $P(-|\bar{D}) = 0.95$ (so $P(+|\bar{D}) = 0.05$), $P(D) = 0.01$.

Step 2: Find $P(+)$ using total probability: $$P(+) = P(+|D) \cdot P(D) + P(+|\bar{D}) \cdot P(\bar{D}) = 0.95(0.01) + 0.05(0.99) = 0.0095 + 0.0495 = 0.059$$

Step 3: Apply Bayes' Theorem: $$P(D|+) = \frac{P(+|D) \cdot P(D)}{P(+)} = \frac{0.0095}{0.059} \approx 0.161$$

Step 4: Despite the 95% accuracy, only about 16.1% of positive results are true positives. This is because the disease is rare (base rate fallacy).

Answer: $P(D|+) = \dfrac{19}{118} \approx 16.1\%$

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.

📝 Full Solution

Step 1: Both dice must be odd (1, 3, or 5) for the product to be odd. There are $3 \times 3 = 9$ such outcomes out of 36 total.

Step 2: List all odd products: $1 \times 1 = 1$, $1 \times 3 = 3$, $1 \times 5 = 5$, $3 \times 1 = 3$, $3 \times 3 = 9$, $3 \times 5 = 15$, $5 \times 1 = 5$, $5 \times 3 = 15$, $5 \times 5 = 25$.

Step 3: Sum of all products: $1 + 3 + 5 + 3 + 9 + 15 + 5 + 15 + 25 = 81$.

Step 4: Expected winnings: $E[\text{win}] = \frac{9}{36} \times \frac{81}{9} + \frac{27}{36} \times 0 = \frac{1}{4} \times 9 = \frac{81}{36} = 2.25$.

Step 5: Expected profit: $E[\text{profit}] = 2.25 - 5 = -2.75$.

Answer: Expected profit $= -\$2.75$ (you lose $2.75 on average per game)

4 Binomial Distribution Medium

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

Show Hint

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

📝 Full Solution

Step 1: This follows a binomial distribution with $n = 10$, $p = 0.5$, $k = 7$.

Step 2: $P(X = 7) = \binom{10}{7}(0.5)^7(0.5)^3 = \binom{10}{7}\left(\frac{1}{2}\right)^{10}$.

Step 3: $\binom{10}{7} = \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.

Step 4: $P(X = 7) = \frac{120}{1024} = \frac{15}{128} \approx 0.1172$.

Answer: $\dfrac{15}{128} \approx 11.72\%$

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. $P(\text{second is king} \mid \text{first is king}) = \frac{3}{51}$.

📝 Full Solution

Step 1: We want $P(\text{both kings} \mid \text{first is king})$. By the definition of conditional probability, this equals $P(\text{second is king} \mid \text{first is king})$.

Step 2: Given the first card is a king, there are 51 cards remaining, of which 3 are kings.

Step 3: $P(\text{second is king} \mid \text{first is king}) = \frac{3}{51} = \frac{1}{17}$.

Answer: $\dfrac{1}{17} \approx 5.88\%$

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: $P(X \leq 2) = P(0) + P(1) + P(2)$.

📝 Full Solution

Step 1: Poisson formula: $P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!}$ with $\lambda = 4$.

Step 2: For exactly 6 calls: $$P(X = 6) = \frac{e^{-4} \cdot 4^6}{6!} = \frac{e^{-4} \cdot 4096}{720} \approx \frac{0.01832 \times 4096}{720} \approx 0.1042$$

Step 3: For at most 2 calls: $$P(X \leq 2) = P(0) + P(1) + P(2)$$

$P(0) = e^{-4} \approx 0.01832$

$P(1) = e^{-4} \cdot 4 \approx 0.07326$

$P(2) = \frac{e^{-4} \cdot 16}{2} \approx 0.14653$

Step 4: $P(X \leq 2) \approx 0.01832 + 0.07326 + 0.14653 = 0.2381$.

Answer: $P(X=6) \approx 10.42\%$; $P(X \leq 2) \approx 23.81\%$

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 $= \frac{44}{120}$.

📝 Full Solution

Step 1: A derangement is a permutation where no element appears in its original position. The number of derangements of $n$ elements is: $$D(n) = n!\sum_{k=0}^{n}\frac{(-1)^k}{k!}$$

Step 2: For $n = 5$: $$D(5) = 5!\left(1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}\right)$$

Step 3: Compute inside the parentheses: $\frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} = \frac{60 - 20 + 5 - 1}{120} = \frac{44}{120}$.

Step 4: $D(5) = 120 \times \frac{44}{120} = 44$.

Step 5: Probability = $\frac{D(5)}{5!} = \frac{44}{120} = \frac{11}{30}$.

Answer: $\dfrac{11}{30} \approx 36.67\%$

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 $\approx 72 + 1.645 \times 8$.

📝 Full Solution

Step 1: Convert to z-scores: $z_{64} = \frac{64 - 72}{8} = -1$ and $z_{88} = \frac{88 - 72}{8} = 2$.

Step 2: $P(64 \leq X \leq 88) = P(-1 \leq Z \leq 2)$. From z-tables: $P(Z \leq 2) = 0.9772$ and $P(Z \leq -1) = 0.1587$.

Step 3: $P(-1 \leq Z \leq 2) = 0.9772 - 0.1587 = 0.8185$, or about 81.85%.

Step 4: For the top 5%: the z-score is $z = 1.645$. The cutoff score: $$x = \mu + z\sigma = 72 + 1.645 \times 8 = 72 + 13.16 = 85.16$$

Answer: About 81.85% scored between 64 and 88. A score of approximately 85.2 separates the top 5%.

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.

📝 Full Solution

Step 1: Let the two break points be at positions $x$ and $y$ chosen uniformly on $[0, 1]$. Without loss of generality, assume $x < y$. The three pieces have lengths $x$, $y - x$, and $1 - y$.

Step 2: The triangle inequality requires each piece to be less than $\frac{1}{2}$: $$x < \frac{1}{2}, \quad y - x < \frac{1}{2}, \quad 1 - y < \frac{1}{2}$$

Step 3: This simplifies to: $x < \frac{1}{2}$, $y > \frac{1}{2}$, and $y - x < \frac{1}{2}$.

Step 4: On the unit square (sample space for the ordered pair $(x, y)$ with $x < y$), the favorable region has area $\frac{1}{4}$ out of total area $\frac{1}{2}$ (the triangle $x < y$).

Step 5: Probability $= \frac{1/4}{1/2} = \frac{1}{4}$. Alternatively, by symmetry on the full unit square, the favorable area is $\frac{1}{4}$ of the total.

Answer: $\dfrac{1}{4} = 25\%$

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 $-1.645$.

📝 Full Solution

Step 1: Set up hypotheses: $H_0: \mu = 1000$ (null), $H_1: \mu < 1000$ (alternative, one-tailed left).

Step 2: Compute the test statistic: $$z = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{980 - 1000}{60 / \sqrt{36}} = \frac{-20}{10} = -2$$

Step 3: For a one-tailed test at the 5% significance level, the critical value is $z_{\alpha} = -1.645$.

Step 4: Since $z = -2 < -1.645$, we reject $H_0$. The p-value is $P(Z < -2) \approx 0.0228$, which is less than 0.05.

Answer: Yes, there is significant evidence at the 5% level that the true mean is less than 1000 hours ($z = -2$, $p \approx 0.023$).

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 $P(\text{at least one match}) = 1 - P(\text{all different}) \approx 0.706$.

📝 Full Solution

Step 1: Use the complement: $P(\text{at least one match}) = 1 - P(\text{all different})$.

Step 2: $P(\text{all different})$: The first person can have any birthday ($\frac{365}{365}$). The second must differ ($\frac{364}{365}$). The third ($\frac{363}{365}$), and so on.

Step 3: $$P(\text{all different}) = \frac{365}{365} \cdot \frac{364}{365} \cdot \frac{363}{365} \cdots \frac{336}{365} = \prod_{i=0}^{29}\frac{365 - i}{365}$$

Step 4: This can be written as $\frac{365!}{335! \cdot 365^{30}}$. Numerically: $P(\text{all different}) \approx 0.2937$.

Step 5: $P(\text{at least one match}) = 1 - 0.2937 \approx 0.7063$.

Answer: $P \approx 70.6\%$ -- surprisingly high for just 30 people!

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 $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$.

📝 Full Solution

Step 1: Let $r$ be the probability of ever reaching position $-1$ starting from 0. By the Gambler's Ruin theory, for a biased walk with $p > q$, this probability is $\frac{q}{p}$.

Step 2: Here $p = \frac{2}{3}$ and $q = \frac{1}{3}$, so: $$r = \frac{q}{p} = \frac{1/3}{2/3} = \frac{1}{2}$$

Step 3: To reach $-3$, the particle must reach $-1$ from 0, then $-2$ from $-1$, then $-3$ from $-2$. By the Markov property, each step has probability $r = \frac{1}{2}$.

Step 4: $$P(\text{reach } {-3}) = r^3 = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$$

Answer: $\dfrac{1}{8} = 12.5\%$