Keep repeating a trial until it succeeds for the first time, and count the trials. That waiting time follows a geometric distribution, with simple formulas for exact and cumulative probabilities and a mean of \(\tfrac{1}{p}\).
If \(X \sim \text{Geo}(p)\), \(X\) is the number of the trial on which the first success occurs: \(\text{P}(X = r) = p(1 - p)^{r - 1}\) for \(r = 1, 2, 3, \dots\). Also \(\text{P}(X > r) = (1 - p)^{r}\) (the first \(r\) trials all fail) and \(\text{E}(X) = \dfrac{1}{p}\).
Fig. 1: X ~ Geo(0.4) drawn to scale for r = 1 to 8. The most likely value is always r = 1, and the probabilities shrink by the factor 1 − p each time
Section 01
The first success.
The conditions are the same as for the binomial (independent trials, two outcomes, constant \(p\)) except that there is NO fixed number of trials: you stop at the first success. For the first success to happen on trial \(r\), the first \(r - 1\) trials must fail and trial \(r\) must succeed.
Worked example An exact probability
A fair die is thrown until a six appears. Find the probability that the first six occurs on the 4th throw.
\(X \sim \text{Geo}\left(\tfrac{1}{6}\right)\).
Three non-sixes then a six: \(\left(\tfrac{5}{6}\right)^3 \times \tfrac{1}{6}\).
\(= \dfrac{125}{1296}\).
Answer: \(\dfrac{125}{1296} = 0.0965\) (3 s.f.)
Worked example Which model?
For each situation, state whether a binomial or a geometric model is suitable: (a) the number of heads in 10 tosses of a coin, (b) the number of tosses needed to get the first head.
(a) The number of trials is fixed at 10 and we count successes: binomial, \(\text{B}(10, 0.5)\).
(b) The number of trials is not fixed; we count until the first success: geometric, \(\text{Geo}(0.5)\).
Answer: (a) binomial (b) geometric
Examiner trap
Using \((1 - p)^{r}\,p\). If the first success is on trial \(r\), only \(r - 1\) trials fail before it, so the power is \(r - 1\).
Section 02
Cumulative probabilities.
No long sums are needed. 'More than \(r\) trials are needed' means the first \(r\) trials all fail, which has probability \((1 - p)^{r}\). Everything else follows from that.
Event
Probability
\(X > r\)
\(q^{r}\), where \(q = 1 - p\)
\(X \ge r\) (same as \(X > r - 1\))
\(q^{r-1}\)
\(X \le r\)
\(1 - q^{r}\)
\(a < X \le b\)
\(q^{a} - q^{b}\)
Worked example More than, at most
A fair die is thrown until a six appears. Find the probability that (a) more than 5 throws are needed, (b) at most 3 throws are needed.
(a) The first 5 throws are not sixes: \(\left(\tfrac{5}{6}\right)^5 = \dfrac{3125}{7776}\).
\(X \sim \text{Geo}(p)\). What is \(\text{P}(X = 1)\)?
Quick check+10 XP
\(X \sim \text{Geo}(p)\) with \(q = 1 - p\). What is \(\text{P}(X > 5)\)?
Quick check+10 XP
Which variable is best modelled by a geometric distribution?
Section 03
Expectation and the least n.
On average you wait \(\tfrac{1}{p}\) trials for the first success: 6 throws for a six, 2 tosses for a head. Questions about the least number of trials use \(\text{P}(X \le n) = 1 - q^{n}\) and trial values of \(n\).
Worked example Expectation both ways
(a) \(X \sim \text{Geo}\left(\tfrac{1}{6}\right)\): find \(\text{E}(X)\). (b) \(Y \sim \text{Geo}(p)\) with \(\text{E}(Y) = 2.5\): find \(p\) and \(\text{P}(Y = 2)\).