• Geometric distribution X ~ Geo(p): models the number of trials needed to get the first success.
• Four conditions must hold: indefinite trials, independence, two outcomes per trial, constant probability of success.
• PMF: P(X = x) = p(1−p)^x−1, for x = 1, 2, 3, …
• CDF: P(X ≤ x) = 1 − (1−p)^x, and P(X > x) = (1−p)^x
• Mean E(X) = 1/p, Variance Var(X) = (1−p)/p²
• The Geometric distribution has no memory — past failures don't affect future probabilities.
• The mode is always X = 1 (probabilities strictly decrease from the first trial onward).
• Negative Binomial X ~ Negative B(r, p): models the number of trials needed to get a fixed number, r, of successes.
• Same four conditions as Geometric, but now r successes are required instead of just 1.
• PMF: P(X = x) = ^x−1C_r−1 · p^r(1−p)^x−r, for x = r, r+1, r+2, …
• Mean E(X) = r/p, Variance Var(X) = r(1−p)/p²
• When r = 1, the Negative Binomial formula collapses exactly into the Geometric formula.

Picture rolling a die and waiting for your first six. You don't know if it'll take 1 roll, 5 rolls, or 50 rolls — but eventually (with probability 1) it will happen. The Geometric distribution is the tool for exactly this situation: counting trials until the very first success happens.

We write X ~ Geo(p), where:

• X = the number of trials it takes to get the first success (so X = 8 means the first success happened on the 8th go, meaning the first 7 were failures)
• p = the fixed probability of success on any single trial

A Geometric model is only valid if all four of these hold:

1. The experiment consists of an indefinite number of successive trials
2. The outcome of each trial is independent of all the others
3. Each trial has exactly two possible outcomes (success / failure)
4. The probability of success is constant, p, on every trial

A neat pattern worth noticing: the sequence of probabilities P(X=1), P(X=2), P(X=3), … is p, p(1−p), p(1−p)², … — a geometric sequence with first term p and common ratio (1−p). That's literally where the distribution gets its name. Since 0 < (1−p) < 1, this sequence is always decreasing, which means P(X=1) is always the biggest probability — so the mode of a Geometric distribution is always X = 1, no matter what p is.

This is the single most elegant idea in the whole chapter, so let's really unpack it. Suppose you've already failed 5 times in a row. What's the probability the first success happens on trial 7 (i.e. 2 more trials from now)?

Your gut might say "well, it's been failing a lot, so a success is 'due' soon" — but that's wrong, and it's a famous fallacy (the gambler's fallacy). Because each trial is independent, the die/coin has no memory of what happened before. The probability of succeeding exactly 2 trials from now is simply p(1−p) — identical to the probability of succeeding on trial 2 if you were starting completely fresh.

From this you instantly get P(X > x) = (1 − p)^x — the probability you need more than x trials, which is just "fail x times in a row." This little formula shows up constantly in exam questions, so it's worth knowing without deriving it every time.

Joshua inspects widgets from a machine until he finds a defective one. The probability any widget is defective is 0.002. Let X = number of widgets inspected until the first defective one, so X ~ Geo(0.002).

a) Find P(the 10th widget is the first defective one), i.e. P(X = 10).

Using P(X = x) = p(1−p)^x−1:

P(X = 10) = 0.002 × (0.998)⁹ = 0.00196428… ≈ 0.00196

b) Find P(X = 250):

P(X = 250) = 0.002 × (0.998)²⁴⁹ ≈ 0.00121

c) Find P(X = 250 | X > 240) — given the first 240 widgets were fine, find P the 250th is the first defective one.

This is exactly where the memoryless property saves you time. Since the process resets after every "failure" (non-defective widget), this conditional probability is the same as asking for the probability of the first defective occurring on the 10th widget starting fresh:

P(X = 250 | X > 240) = P(X = 10) ≈ 0.00196 — identical to part (a)!

d) Find P(X ≤ 250) — Joshua inspects 250 or fewer before finding a defective one.

P(X ≤ 250) = 1 − (0.998)²⁵⁰ = 0.393772… ≈ 0.394

e) Find P(X > 250) — he needs more than 250 inspections.

P(X > 250) = (0.998)²⁵⁰ = 0.606227… ≈ 0.606

Now imagine you're not stopping at the first six — you're rolling until you get your third six. That's the Negative Binomial distribution: it counts the number of trials needed to reach a fixed number of successes, r.

We write X ~ Negative B(r, p) (also sometimes written NB(r, p)), where:

• X = number of trials needed to reach a total of r successes
• r = the fixed target number of successes
• p = the fixed probability of success on any one trial

Same four conditions apply, just with "r successes" instead of "1 success":

1. Indefinite number of successive trials
2. Independent trials
3. Two outcomes per trial (success/failure)
4. Constant probability of success, p

The same criticism applies too: if someone improves with practice while chasing their r-th success, independence and constant p both come into question.

This is worth really understanding rather than memorising blind, because it shows up in the reasoning of harder exam questions. Say you want the 3rd six to land exactly on the 12th roll of a fair die. Two things must both happen:

1. Exactly 2 sixes occur somewhere in the first 11 rolls (it doesn't matter which 2 of the 11) — this is a plain Binomial probability, P(Y = 2) for Y ~ B(11, 1/6)
2. AND a six occurs on the 12th roll itself — probability p

Multiply these together (independence again) and you get the general Negative Binomial formula. This is exactly why the formula contains a binomial coefficient — you're really doing a Binomial calculation for the first (x−1) trials, then bolting on one guaranteed success at the very end.

Set r = 1 in the Negative Binomial formula and watch what happens:

Emanuel plays chess in a tournament where his probability of winning any one game is 0.55. This single set-up is a great illustration of how Binomial, Geometric, and Negative Binomial can all appear in the same question depending on exactly what's being asked.

a) His first win is in the third game he plays.

This is asking for the first success on trial 3 — use X ~ Geo(0.55), or equivalently NB(1, 0.55):

P(X = 3) = (0.55)(0.45)² = 0.111375 ≈ 0.111

b) He wins exactly 4 of his first 7 games.

Careful — this is not a "trials until success" question at all. The number of games (7) is fixed and we're counting wins, so this is a regular Binomial: X ~ B(7, 0.55).

P(X = 4) = ⁷C₄(0.55)⁴(0.45)³ = 0.291847… ≈ 0.292

c) He wins for the fourth time in his seventh game.

Now we're fixing the number of successes (4) and asking about the trial count — Negative Binomial. X ~ Negative B(4, 0.55):

P(X = 7) = ⁶C₃(0.55)⁴(0.45)³ = 0.166770… ≈ 0.167

d) He wins for the fourth time in his seventh game, given that he won his first game.

If game 1 was already a win, he now needs 3 more wins within the remaining 6 games, with the 4th win landing exactly on game 6 of those remaining games. So use X ~ Negative B(3, 0.55):

P(X = 6) = ⁵C₂(0.55)³(0.45)³ = 0.151609… ≈ 0.152

e) His fourth win occurs in or before his seventh game.

This is subtly different again — "in or before game 7" means at least 4 wins within a fixed 7 games, which flips it back into a Binomial cumulative question: X ~ B(7, 0.55), find P(X ≥ 4).

P(X ≥ 4) = 1 − P(X ≤ 3) = 0.608287… ≈ 0.608

• I can state what X and p represent for X ~ Geo(p)
• I can state and apply all four conditions for a Geometric/Negative Binomial model
• I can criticise a Geometric/Negative Binomial model in context (independence, constant p)
• I can write down and use P(X=x) = p(1−p)^x−1
• I can explain why the mode of a Geometric distribution is always 1
• I can use the cumulative formula P(X≤x) = 1−(1−p)^x and its complement P(X>x)
• I can explain the memoryless property and apply it to conditional probability questions
• I can recall and apply E(X) = 1/p and Var(X) = (1−p)/p² for Geometric
• I understand what X, r and p represent for X ~ Negative B(r, p)
• I can write down and use P(X=x) = ^x−1C_r−1p^r(1−p)^x−r
• I understand where the Negative Binomial formula comes from (Binomial × final success)
• I can explain why Geometric is the r=1 special case of Negative Binomial
• I can recall and apply E(X) = r/p and Var(X) = r(1−p)/p² for Negative Binomial
• I can distinguish Binomial vs Geometric vs Negative Binomial from question wording
• I can solve for p given the mean and/or variance, rejecting invalid (negative) solutions