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Geometric and negative binomial distributions questions
Stop counting successes in a fixed number of trials and start counting trials until a fixed number of successes. The question turns round, and so does the distribution.
7 original questions · 24 marks · the geometric and negative binomial distributions notes · Further Statistics 1
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State how the geometric distribution differs from the binomial in what it counts.
Worked answer
The binomial fixes the number of trials and counts successes; the geometric fixes the number of successes at one and counts the trials needed to get it. B1 B1 for the two descriptions. The roles of the two quantities swap over.X follows a geometric distribution with p = 0.3. Find P(X = 4) and the mean.
Worked answer
P(X = 4) = 0.3 × 0.7³ = 0.1029: three failures then a success. The mean is 1/p = 3.33 trials. M1 for the product, A1 for 0.1029, B1 for the mean. The mode is 1 even though the average wait is longer than three, because the probabilities only ever decrease.X follows a geometric distribution with p = 0.3. Find P(X ≥ 5) and explain why no summation is required.
Worked answer
Needing at least five trials means the first four all failed, so P(X ≥ 5) = 0.7⁴ = 0.2401. The whole upper tail collapses to a single power, because 'at least x trials' and 'the first x − 1 all failed' describe the same event. M1 for the first four failing, A1 for 0.2401, B1 for the reason no summation is needed. Summing 0.3(0.7)x−1 from x = 5 upwards gives the same number and takes far longer.A component passes inspection with probability 0.25. Find the probability that the second pass occurs on the sixth inspection, and the mean number of inspections needed for two passes.
Worked answer
This is negative binomial with r = 2 and p = 0.25, so P(X = 6) = ⁵C₁ × 0.25² × 0.75⁴ = 5 × 0.0625 × 0.3164 = 0.0989. The mean is r/p = 8 inspections, with variance r(1 − p)/p² = 2(0.75)/0.0625 = 24. M1 for the negative binomial form, A1 for 0.0989, B1 for the mean, B1 for the variance.Explain why the combination in the negative binomial formula is x−1Cr−1 rather than xCr.
Worked answer
The final trial is pinned as a success by definition, so it is not free to move. Only the earlier r − 1 successes can be arranged, and only among the first x − 1 trials. Using xCr would count arrangements in which the rth success arrives early, which describes a different event. B1 for the last trial being fixed, B1 for arranging the earlier successes only, B1 for what the wrong combination would count.Show that the negative binomial distribution with r = 1 reduces to the geometric distribution, in probability, mean and variance.
Worked answer
With r = 1 the combination is x−1C0 = 1, so P(X = x) = p(1 − p)x−1, the geometric formula. The mean r/p becomes 1/p and the variance r(1 − p)/p² becomes (1 − p)/p². Waiting for one success is the special case of waiting for r of them. B1 for the probability, B1 for the mean, B1 for the variance.The random variable X follows a negative binomial distribution with parameters r and p, and has mean 12 and variance 24. Find the values of p and r. Hence find P(X = 5) and the probability that at least 6 trials are needed, each as an exact fraction.
Worked answer
Mean r/p = 12 and variance r(1 − p)/p² = 24. Dividing the second by the first removes r altogether: (1 − p)/p = 24/12 = 2.
So 1 − p = 2p and p = 1/3. Then r = 12p = 4. Dividing the two moments is the move worth finding; solving the pair by substitution works but takes twice the algebra.
P(X = 5) = ⁴C₃ (1/3)⁴ (2/3) = 4 × (1/81) × (2/3) = 8/243, about 0.0329.
For the tail, note that X cannot be less than r = 4, so only two values sit below 6. P(X = 4) = ³C₃ (1/3)⁴ = 1/81 = 3/243, with no arrangement to make, since all four trials are successes.
P(X ≥ 6) = 1 − 3/243 − 8/243 = 232/243, about 0.955. Starting the tail at X = 1 rather than X = r is the slip here, and it produces probabilities that do not sum to one.
M1 for dividing the variance by the mean, A1 for p = 1/3, A1 for r = 4, M1 A1 for 8/243, A1 for 232/243.
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