Quantify uncertainty, model randomness, and draw powerful conclusions from data — the mathematics that governs our world.
A club has 8 men and 6 women. How many 5-person committees can be formed that contain at least 2 women?
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
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?
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: \(P(D|+) = \dfrac{19}{118} \approx 16.1\%\)
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?
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)
A fair coin is tossed 10 times. What is the probability of getting exactly 7 heads?
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: \(\dfrac{15}{128} \approx 11.72\%\)
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?
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: \(\dfrac{1}{17} \approx 5.88\%\)
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?
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: \(P(X=6) \approx 10.42\%\); \(P(X \leq 2) \approx 23.81\%\)
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?
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: \(\dfrac{11}{30} \approx 36.67\%\)
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%?
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%.
A stick of length 1 is broken at two random points. What is the probability that the three pieces can form a triangle?
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: \(\dfrac{1}{4} = 25\%\)
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?
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 (\(z = -2\), \(p \approx 0.023\)).
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.
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!
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 \(-3\)?
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: \(\dfrac{1}{8} = 12.5\%\)