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Statistics · S.MD.A.3

The geometric distribution

A caller keeps dialing until someone pledges. How likely is it that the first pledge comes on the third call, or not until after the tenth, when the number of calls is not fixed in advance?

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Why this exists

Rojin, Baran and Zhilwan volunteer at the phone bank of a community radio station. Each call they make ends in a pledge or it does not, and on their main list each call ends in a pledge with probability 0.2, independently of every other call (made-up data, as is every number in this lesson). A caller works down the list until the first pledge comes in.

The question here is not “how many pledges in 10 calls”. That count has a fixed number of trials and belongs to the binomial setting, where n is set before the process starts. The question is “on which call does the first pledge come?” Now the number of calls is decided by the outcomes themselves: it could be 1, or 3, or 25, with no largest value.

A simulation, as in Running a simulation and estimating a probability, can estimate the chance of a long wait by running many pretend calling runs. This lesson gives the exact chance, from p alone. Day 1 builds the setting and the probability that the first success lands on a given trial. Day 2 answers “more than”, “at most” and “at least” questions with a shortcut, and separates a geometric count from a binomial one.

Definition

A chance process is a geometric setting when it is a run of trials in which

  • binary outcomes: each trial ends in success or failure;
  • independent trials: the outcome of one trial does not change the chances for any other;
  • same chance of success: every trial has the same probability of success, p, with 0 < p \le 1;
  • count until the first success: the trials go on until the first success, and the variable is the number of trials it takes.

The random variable X is the number of the trial on which the first success happens, counting that successful trial. Its possible values are 1, 2, 3, \ldots with no upper limit. A random variable of this kind is a geometric random variable with parameter p, and its probability distribution is the geometric distribution.

Three of the conditions are shared with the binomial setting. The fourth replaces “a fixed number of trials n” by “keep going until the first success”, which is why a geometric model has only one parameter: there is no n.

Intuition

The event X = 3 is one single pattern: no pledge, no pledge, pledge. Calling stops at the pledge, so nothing after call 3 is part of the event, and no other pattern gives X = 3.

1 F 0.8 2 F 0.8 3 S 0.2
X = 3
Figure 1One way a calling run can go. Calling stops at the first pledge, so this run has X = 3, and its probability is the product of the chances under the boxes (made-up data).

Each extra call adds one more factor of 0.8 before the final 0.2. So the probabilities shrink by the same factor, 0.8, from each value of X to the next, and the bar graph steps down like a staircase that never quite reaches zero. The tallest bar is always the first: the single most likely call for the first pledge is call 1, even though most runs take longer.

0.200 1 0.160 2 0.128 3 0.102 4 0.082 5 0.066 6 0.052 7 0.042 8 0.034 9 0.027 10 ... 0 0.05 0.10 0.15 0.20 P(X = k) Call on which the first pledge comes, k
Figure 2The geometric distribution with p = 0.2. Each bar is 0.8 times the one before. The shaded bars, from 5 on and continuing past the edge, add to P(X > 4) = 0.8^4 = 0.4096 (made-up data).

Results

For a geometric random variable X with probability of success p on each trial, and any whole number k \ge 1,

P(X = k) = (1 - p)^{k - 1}\, p

For any whole number k \ge 0, the first k trials all fail exactly when X > k, so

\begin{gathered} P(X > k) = (1 - p)^k \\ P(X \le k) = 1 - (1 - p)^k \end{gathered}

and, since “at least k” is “more than k - 1”, P(X \ge k) = (1 - p)^{k - 1}. The probabilities of all the values add to 1.

Why it is true

The formula for one value. X = k happens exactly when trials 1 to k - 1 all fail and trial k succeeds. That is one pattern. The trials are independent, so by the multiplication rule for independent events its probability is the product of the chances, k - 1 factors of 1 - p and one factor of p:

\begin{aligned} P(X = k) &= \underbrace{(1 - p)\cdots(1 - p)}_{k - 1}\; p \\ &= (1 - p)^{k - 1}\, p \end{aligned}

The tail. X > k says no success has come by trial k, which is the event that trials 1 to k all fail. What happens after trial k does not matter, so by the same rule P(X > k) = (1 - p)^k. This is the “none occur” product of At least one problems, and P(X \le k) is its complement.

The two results fit together. The first success is on trial k exactly when there was none by trial k - 1 but there is one by trial k, so

\begin{aligned} P(X = k) &= P(X > k - 1) - P(X > k) \\ &= (1 - p)^{k - 1} - (1 - p)^k \\ &= (1 - p)^{k - 1}\bigl(1 - (1 - p)\bigr) = (1 - p)^{k - 1}\, p \end{aligned}

The total is 1. The probabilities form a geometric sequence with first term p and ratio 1 - p, which is where the distribution gets its name. When 0 < p \le 1 the ratio is between 0 and 1 (it is 0 when p = 1), so by the Infinite geometric series the whole series adds to

\frac{p}{1 - (1 - p)} = \frac{p}{p} = 1

The same follows from the tail: P(X \le k) = 1 - (1 - p)^k, and (1 - p)^k shrinks toward 0 as k grows. A success comes eventually, with probability 1.

Worked examples

Example 1

Day 1: the first pledge on the third call

Rojin starts on the main list, where each call pledges with probability 0.2, independently. She calls until the first pledge. Let X be the number of calls she makes. Check that X is geometric, give its possible values, and find the probability that the first pledge comes on the first, the second and the third call.

Solution

Check the conditions in context.

  • Binary: each call ends in a pledge (success) or no pledge (failure).
  • Independent: one person’s answer does not change the next person’s chance, as given.
  • Same chance: every call on this list pledges with p = 0.2.
  • Until the first success: Rojin stops at the first pledge and X counts the calls, the pledging call included.

So X is geometric with p = 0.2, and its values are 1, 2, 3, \ldots. A failure has probability 1 - 0.2 = 0.8.

The first pledge on call 1 is the pattern “pledge”: P(X = 1) = 0.2. On call 2 it is “no pledge, pledge”, so P(X = 2) = 0.8 \times 0.2 = 0.16. On call 3 it is the pattern of Figure 1:

P(X = 3) = (0.8)^2(0.2) = 0.128

In the long run, about 128 of every 1{,}000 calling runs like Rojin’s have the first pledge land on the third call.

Example 2

Day 1: a list with a higher chance

Baran calls the station’s list of past donors, where each call pledges with probability 0.3, independently. Let X be the number of calls up to and including his first pledge. Find P(X = k) for k = 1 to 6, show how each value is related to the one before, and say which value of X is most likely.

Solution

The conditions hold as in Example 1, now with p = 0.3 and 1 - p = 0.7. The formula P(X = k) = 0.7^{k - 1}(0.3) gives

kP(X = k)
10.3
20.21
30.147
40.1029
50.0720
60.0504

with the last two rounded to four decimal places. Going from k to k + 1 adds one more failed call at the front, one more factor of 0.7, so each probability is 0.7 times the one before: 0.3 \times 0.7 = 0.21, 0.21 \times 0.7 = 0.147, and so on.

Since each value is smaller than the one before, the most likely value is X = 1, with probability 0.3. Compared with Figure 2, Baran’s bars start higher and fall faster, by a factor of 0.7 instead of 0.8: a higher chance of success makes short waits likelier and long waits rarer.

Example 3

Day 1: finding p from one probability

Zhilwan tries a new list of names collected at a street fair. From earlier weeks, the probability that her first pledge on this list comes on the second call is 0.24, and a pledge on any single call is less likely than not. Assuming a geometric setting, find p and then P(X = 3).

Solution

The formula with k = 2 says P(X = 2) = (1 - p)p, so

p(1 - p) = 0.24

Expanding and moving every term to one side gives a quadratic equation:

p^2 - p + 0.24 = 0

It factors as (p - 0.4)(p - 0.6) = 0, since 0.4 + 0.6 = 1 and 0.4 \times 0.6 = 0.24. So p = 0.4 or p = 0.6. Both really do give 0.24: 0.6 \times 0.4 and 0.4 \times 0.6 are the same product with the factors swapped. The probability of the first pledge on call 2 alone cannot tell the two lists apart. The extra fact settles it: a pledge is less likely than not, so p < 0.5 and p = 0.4. Then

P(X = 3) = 0.6^2(0.4) = 0.144
Example 4

Day 2: more than four calls, and at most four

Back on Rojin’s main list (p = 0.2), find the probability that she needs more than 4 calls for her first pledge, the probability that she needs at most 4, and the probability that she needs at least 4.

Solution

“More than 4 calls” means the first 4 calls all end without a pledge. That is a statement about four calls only, so by independence

P(X > 4) = 0.8^4 = 0.4096

These are the shaded bars of Figure 2, an endless list of them, added in one step. “At most 4” is the complement:

P(X \le 4) = 1 - 0.4096 = 0.5904

Check by adding the four bars from the formula: 0.2 + 0.16 + 0.128 + 0.1024 = 0.5904. “At least 4” means more than 3, so the first 3 calls fail: P(X \ge 4) = 0.8^3 = 0.512. About half the time Rojin is still waiting after three calls.

Example 5

Day 2: between three and six calls

On the same list, find the probability that Rojin’s first pledge comes on call 3, 4, 5 or 6.

Solution

The event 3 \le X \le 6 is “no pledge by call 2” minus the runs that have still had no pledge by call 6. Subtracting one tail from another, the way Cumulative probability subtracts cumulative probabilities,

\begin{aligned} P(3 \le X \le 6) &= P(X > 2) - P(X > 6) \\ &= 0.8^2 - 0.8^6 \\ &= 0.64 - 0.262144 \approx 0.3779 \end{aligned}

Check by adding the four values:

\begin{aligned} &0.128 + 0.1024 + 0.08192 + 0.065536 \\ &= 0.377856 \end{aligned}

The ends need care: X \ge 3 is X > 2, and X \le 6 leaves out X > 6, so both 3 and 6 are inside.

Example 6

Day 2: geometric, binomial, or neither

Each call on the main list pledges with probability 0.2, independently. For each variable, say whether it is geometric, binomial, or neither, and give its parameters.

(a) B, the number of pledges Baran gets in exactly 10 calls. (b) X, the number of calls Rojin makes until her first pledge. (c) T, the number of calls Zhilwan makes until her third pledge.

Then find P(B = 0) and P(X > 10), and explain why they are equal.

Solution

(a) The number of calls is fixed at 10 and B counts successes, so all four binomial conditions hold: B is binomial with n = 10 and p = 0.2, with values 0 to 10.

(b) The calls go on until the first success and X counts calls, so X is geometric with p = 0.2, with values 1, 2, 3, \ldots

(c) The number of calls is not fixed, so T is not binomial. It does not stop at the first success, so it is not geometric either. Its smallest value is 3. It is neither.

P(B = 0) is the chance that all 10 calls fail, 0.8^{10} \approx 0.1074. P(X > 10) is the chance that the first 10 calls fail, also 0.8^{10} \approx 0.1074. They are the same event seen two ways: “no pledge in 10 calls” counted as a number of successes, or as a first pledge that has not yet come. What separates the two models is the question asked, not the calls.

Example 7

Day 2: two callers at once

Rojin calls on the main list (p = 0.2) while Baran calls past donors (p = 0.3). They dial at the same time, one call each per round, all calls independent. Let R be the number of the first round in which at least one of them gets a pledge. Show that R is geometric and find P(R = 2) and P(R > 3).

Solution

Treat each round as one trial, a success when at least one of the two calls pledges. A round fails only if both calls fail, which by independence has probability

0.8 \times 0.7 = 0.56

so a round succeeds with probability 1 - 0.56 = 0.44. Every round has this same chance, rounds are independent because all the calls are, and R counts rounds until the first success. So R is geometric with p = 0.44. Then

\begin{gathered} P(R = 2) = 0.56 \times 0.44 = 0.2464 \\ P(R > 3) = 0.56^3 \approx 0.1756 \end{gathered}

Choosing what counts as one trial turned two processes into one geometric setting.

Boundaries

  • The same chance on every trial. Calling down a short list where the chance changes from name to name, or drawing names without replacement from a small group, breaks the setting, because the probabilities no longer come from one p.
  • Independence. A run of calls that share a cause (a storm keeps everyone at home, a news story makes everyone generous) are not independent, and the product rule no longer gives a pattern’s probability.
  • p must be more than 0. With p = 0 a success never comes and X has no value. With p = 1 every run ends on trial 1, so P(X = 1) = 1.
  • Which count. Here X counts trials up to and including the first success, so X \ge 1 and P(X = 0) = 0. Some books count the failures before the first success instead. That variable is Y = X - 1, with values 0, 1, 2, \ldots and P(Y = j) = (1 - p)^j p. Read which one a problem means before using a formula.
  • No largest value. The model allows any number of trials. A real calling shift ends, so the model is an idealization; it stays useful because the far tail is tiny: with p = 0.2, P(X > 40) = 0.8^{40} \approx 0.00013.
  • Only the first success. A count of trials until the second or third success is neither geometric nor binomial (Example 6).

Where this goes next

The next question about a wait is how long it is on average. Mean and standard deviation of a geometric random variable finds the mean number of trials and the standard deviation of a geometric random variable from p alone. Exact answers here sit beside simulated ones: Simulating a waiting time estimates waits by repeating a chance process, which is how a wait is studied when the trials do not meet these conditions. For counts in a fixed number of trials, The binomial probability formula is the formula on the binomial side, and both models belong to the study of Random variables.

Exercises

All data are made up. Round probabilities to four decimal places.

Day 1: the setting and the formula

  1. On Zhilwan’s list of lapsed donors, each call pledges with probability 0.1, independently. She calls until her first pledge; let X be the number of calls. Is X geometric? Give its possible values, P(X = 1) and P(X = 4).
  2. In exercise 1, find the probability that Zhilwan’s first pledge comes on call 10.
  3. On a weekend list each call pledges with probability 0.25. Find the probability that the first pledge comes on call 2 and on call 3, and the factor that takes one to the other.
  4. Is each variable geometric? If so, give p. (a) Rojin calls until someone picks up the phone, when each call is answered with probability 0.4, independently. (b) Baran rolls a fair six-sided number cube until it shows a 6 and counts the rolls. (c) Zhilwan draws cards one at a time, without putting any back, from a box of 5 cards of which 2 say “matching gift”, and counts the draws until the first “matching gift”.
  5. In exercise 4(b), find the probability that the first 6 comes on the third roll.
  6. For a geometric variable, P(X = 1) = 0.35. Find P(X = 3).
  7. A geometric variable has P(X = 4) = 0.0864 and P(X = 5) = 0.05184. Find p.

Day 2: cumulative questions and choosing the model

  1. On the lapsed list of exercise 1 (p = 0.1), find P(X > 5) and P(X \le 5).
  2. On the weekend list (p = 0.25), find the probability that the first pledge takes (a) at least 4 calls, (b) fewer than 4 calls.
  3. On the main list (p = 0.2), find the probability that the first pledge comes on one of calls 2 to 5.
  4. Zhilwan’s shift on the lapsed list (p = 0.1) is 15 calls long. Find the probability that she gets no pledge in the whole shift, and the probability that her first pledge comes on one of calls 11 to 15.
  5. On the lapsed list (p = 0.1), find the smallest number of calls k for which P(X \le k) \ge 0.5.
  6. Say whether each variable is geometric, binomial, or neither, and give its parameters. Each call pledges with probability 0.2 and each call ends in a hang-up with probability 0.3, independently. (a) The number of pledges in 25 calls. (b) The number of calls until the first hang-up. (c) The number of calls until the second pledge.
  7. Baran makes exactly 7 calls on the main list (p = 0.2). Find the probability that none of them pledges, and name the geometric event with the same probability.
  8. On the lapsed list (p = 0.1), find the probability that the first pledge comes on call 3, and the probability of exactly one pledge in the first 3 calls. Explain why the second is 3 times the first.
  9. Zhilwan alternates lists, past donors (p = 0.3) on odd calls and lapsed donors (p = 0.1) on even calls, starting with a past donor. (a) Is the number of calls until her first pledge geometric? (b) Treating each pair of calls as one trial, find the probability that a pair has no pledge and the probability of no pledge in her first 6 calls.
  10. On the main list (p = 0.2), Baran has made 4 calls with no pledge. Find the probability that his first pledge comes within his next 2 calls, and compare it with P(X \le 2).

Reasoning

  1. Explain why P(X > k) = (1 - p)^k without adding any of the values P(X = 1), P(X = 2), \ldots
  2. Explain why, in every geometric distribution, X = 1 is the most likely value.
  3. Explain why a geometric variable has no largest value and yet its probabilities add to 1.
  4. On the main list (p = 0.2), Zhilwan says: “Every call pledges with probability 0.2, so the probability that my first pledge comes on call 10 is 0.2.” Explain what is wrong, and find the actual probability.
Answers
  1. Yes: binary, independent, same p = 0.1, calling until the first pledge. Values 1, 2, 3, \ldots P(X = 1) = 0.1 and P(X = 4) = 0.9^3 \times 0.1 = 0.0729.
  2. P(X = 10) = 0.9^9 \times 0.1 \approx 0.0387.
  3. P(X = 2) = 0.75 \times 0.25 = 0.1875 and P(X = 3) = 0.75^2 \times 0.25 \approx 0.1406. Each is 0.75 times the one before.
  4. (a) Geometric, p = 0.4. (b) Geometric, p = \tfrac{1}{6}. (c) Not geometric: without replacement the chance of a match on each draw depends on what was drawn before (after one blank card it is \tfrac{2}{4}, not \tfrac{2}{5}), so the draws are not independent. The count also cannot pass 4, since only 3 cards are blank.
  5. \left(\tfrac{5}{6}\right)^2 \times \tfrac{1}{6} = \tfrac{25}{216} \approx 0.1157.
  6. P(X = 1) = p, so p = 0.35 and P(X = 3) = 0.65^2 \times 0.35 \approx 0.1479.
  7. Each value is 1 - p times the one before, so 1 - p = \tfrac{0.05184}{0.0864} = 0.6 and p = 0.4. Check: 0.6^3 \times 0.4 = 0.0864.
  8. P(X > 5) = 0.9^5 \approx 0.5905 and P(X \le 5) = 1 - 0.9^5 \approx 0.4095.
  9. (a) At least 4 means the first 3 fail: 0.75^3 \approx 0.4219. (b) The complement, 1 - 0.75^3 \approx 0.5781.
  10. P(2 \le X \le 5) is P(X > 1) - P(X > 5), which is 0.8 - 0.8^5 \approx 0.4723.
  11. No pledge in 15 calls: 0.9^{15} \approx 0.2059. First pledge on calls 11 to 15: 0.9^{10} - 0.9^{15} \approx 0.1428.
  12. Need 1 - 0.9^k \ge 0.5, so 0.9^k \le 0.5. Since 0.9^6 \approx 0.5314 and 0.9^7 \approx 0.4783, the smallest is k = 7.
  13. (a) Binomial, n = 25, p = 0.2. (b) Geometric, p = 0.3. (c) Neither: the number of calls is not fixed, and it does not stop at the first success.
  14. 0.8^7 \approx 0.2097. The same event is a first pledge after call 7, P(X > 7).
  15. P(X = 3) = 0.9^2 \times 0.1 = 0.081. Exactly one pledge in 3 calls has three patterns (the pledge on call 1, 2 or 3), each with probability 0.081, so 3 \times 0.081 = 0.243. The geometric event is only the pattern with the pledge last.
  16. (a) No: the chance changes between odd and even calls. (b) A pair fails with probability 0.7 \times 0.9 = 0.63, so pairs form a geometric setting with p = 0.37. No pledge in 6 calls is 3 failed pairs: 0.63^3 \approx 0.2500.
  17. Of the runs with no pledge in 4 calls, those still without one after 6 have chance \tfrac{0.8^6}{0.8^4} = 0.64 (Finding a conditional probability), so the answer is 1 - 0.64 = 0.36. This equals P(X \le 2) = 1 - 0.8^2 = 0.36: the calls already made do not change the chances for the next ones.
  18. X > k means no success in the first k trials, that is, all k fail. By independence that has probability (1 - p)^k, whatever happens afterward.
  19. P(X = k + 1) = (1 - p) \times P(X = k) and 1 - p < 1 (it is 0 when p = 1), so the probabilities shrink from the first value on; the largest is P(X = 1) = p.
  20. Any number of failures in a row is possible, since (1 - p)^k > 0 when p < 1. Yet P(X \le k) = 1 - (1 - p)^k gets as close to 1 as we like as k grows, and the series p + (1 - p)p + \cdots sums to \tfrac{p}{p} = 1.
  21. 0.2 is the chance that call 10 pledges, whatever happened before. A first pledge on call 10 also needs calls 1 to 9 to fail, so P(X = 10) = 0.8^9 \times 0.2 \approx 0.0268, the last bar of Figure 2.
kg1128 · updated 2026-10-20 · CC-BY-SA-4.0