Probability & Distributions

A permutation counts ordered arrangements; a combination counts unordered selections. Taking k items from n gives n!/(n-k)! permutations and n!/(k!(n-k)!) combinations. The extra k! divides out the orderings of each selected group.

A sample space is the set of all possible outcomes of a random experiment, listed so exactly one occurs. An event is any subset of it. Two dice give 36 outcomes; "the sum is 7" is a 6-outcome event.

Answer two yes/no questions: does order matter, and may items repeat? Ordered with repeats is n^k; ordered without repeats is n!/(n-k)!; unordered without repeats is n choose k; unordered with repeats is C(n+k-1, k).

Adding P(A) and P(B) counts outcomes in both events twice. The addition rule subtracts the overlap: P(A or B) = P(A) + P(B) - P(A and B). For one card, P(red or face) = 26/52 + 12/52 - 6/52 = 8/13.

Adding 1/6 four times double-counts rolls containing more than one six, and would give a probability above 1 for seven rolls. Use the complement: P(no six) = (5/6)^4, so P(at least one six) is about 0.518.

Your first pick wins only one time in three, so two times in three the car is behind another door. The host, who knows where it is, opens a losing door and concentrates that 2/3 onto the single door left.

Mutually exclusive means the two events cannot both happen, so P(A and B) = 0. Independence requires P(A and B) = P(A) times P(B), which is strictly positive when both probabilities are. Exclusivity therefore forces maximal dependence, not independence.

The overall defect rate is 3.2%. The law of total probability weights each line's defect rate by that line's share of production: 0.60 * 0.02 + 0.40 * 0.05 = 0.032.

About 17%. In 10,000 people, 100 have the disease and 99 of them test positive, while 495 of the 9,900 healthy people also test positive. Rare conditions make positives mostly false positives.

P(landed) times P(started given landed) times P(finished given landed and started). The chain rule turns a joint probability into a product of conditionals, each measured on the survivors of the previous step, no independence needed.

Shannon entropy is a distribution's average uncertainty: H = -sum p log2 p, the expected number of yes/no questions needed to pin down one draw. It is largest for equally likely outcomes and zero when one outcome is certain.

Multiply each outcome by its probability and add the pieces up. On a 38-pocket wheel a $1 straight-up bet nets +$35 with probability 1/38 and -$1 with probability 37/38, giving about -$0.053 per dollar staked.

Exactly zero. Under a continuous model, any single point has zero width and therefore zero area under the density, so only intervals carry probability. Ask instead for P(199.5 < X <= 200.5).

KL(P||Q) = sum P(x) log(P(x)/Q(x)) is the cost of describing draws from P as if they came from Q. It is never negative and zero only when P equals Q, but it is asymmetric, so it is a divergence, not a metric.

Linearity of expectation is proved by regrouping a sum over the joint distribution, never by multiplying probabilities. E[X + Y] = E[X] + E[Y] holds for any variables with finite means. Independence is needed for products, not sums.

About 2.5%. A score of 130 is two standard deviations above the mean, so its z-score is 2; the rule puts roughly 95% of values within two SDs, leaving about 5% split evenly between the two tails.

A binomial count over n independent trials with success probability p has mean np and variance np(1-p). It is the sum of n Bernoulli(p) indicators, each contributing mean p and variance p(1-p).

The Poisson mass function is P(X = k) = e^(-λ) λ^k / k!. With λ = 3 tickets per hour, P(X = 0) = e^(-3), about 0.0498, so roughly a 5 percent chance of a completely quiet hour.

Counts in a fixed window, outages per week, are Poisson. The gaps between consecutive outages, hours between them, are exponential. Both describe one event stream: one view counts events, the other measures waiting times.

The exponential distribution is memoryless: P(T > s + t | T > s) = P(T > t). Time already spent waiting carries no information about what remains, so the expected remaining wait is still the full 10 minutes.

The Central Limit Theorem says that averaging many independent draws from almost any distribution with finite variance produces an average whose distribution is approximately normal, even when the individual observations are not remotely normal.

No. Spins are independent, so the chance of black is exactly what it was before the streak. The law of large numbers dilutes early results under a huge volume of later ones; it never reaches back to correct them.

The law of large numbers says that for independent draws from one distribution with a finite mean, the average of the observations converges to that distribution's true expected value as the number of draws grows.

The Markov property says the next state depends only on the current state, not on the path taken to reach it. A user in the paid state has the same transition probabilities regardless of how long ago they upgraded.

Draw many independent points uniformly in the unit square and count the fraction that land inside the quarter circle of radius 1. That fraction estimates pi/4, so multiply it by 4. Accuracy improves like 1/sqrt(n).