Questions on geometric and negative binomial distributions

Question 1

State how the geometric distribution differs from the binomial in what it counts.

Accepted 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.

Question 2

X follows a geometric distribution with p = 0.3. Find P(X = 4) and the mean.

Accepted 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.

Question 3

X follows a geometric distribution with p = 0.2. Find the probability that the first success occurs within the first three trials.

Accepted Answer

The complement is that all three trials fail, so the probability is 1 − 0.8³ = 1 − 0.512 = 0.488. M1 for 1 − 0.8³, A1 for 0.488. Adding 0.2 + 0.16 + 0.128 gives the same value; the complement is one power instead of three terms.

Question 4

X follows a negative binomial distribution with r = 3 and p = 0.6. State the smallest value X can take, and find the variance of X.

Accepted Answer

Three successes need at least three trials, so the smallest value is 3. Var(X) = r(1 − p)/p² = 3 × 0.4/0.36 = 3.33. B1 for 3, M1 for the variance formula with r = 3, A1 for 3.33. The distribution starts at x = r, not at x = 1, and P(X = 2) = 0.

Question 5

X follows a geometric distribution with p = 0.3. Find P(X ≥ 5) and explain why no summation is required.

Accepted 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.

Question 6

A component passes inspection with probability 0.25. Find the probability that the second pass occurs on the sixth inspection, and the mean and variance of the number of inspections needed for two passes.

Accepted 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.

Question 7

Explain why the combination in the negative binomial formula is ^x−1C_r−1 rather than ^xC_r.

Accepted 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 ^xC_r 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.

Question 8

Show that the negative binomial distribution with r = 1 reduces to the geometric distribution, in probability, mean and variance.

Accepted 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. B1 for the probability, B1 for the mean, B1 for the variance.

Question 9

X follows a geometric distribution with parameter p. Show that P(X = x + 1)/P(X = x) = 1 − p for every x, and explain what this says about the mode of X. Given that P(X = 1) = 4 P(X = 3), find p.

Accepted Answer

P(X = x + 1)/P(X = x) = p(1 − p)^x/[p(1 − p)^x−1] = 1 − p, a constant less than 1, so each probability is smaller than the one before and the mode is 1 whatever the value of p.
P(X = 1) = 4 P(X = 3) gives p = 4p(1 − p)², so (1 − p)² = 1/4 and 1 − p = ½ (the root 1 − p = −½ would need p = 3/2, which is not a probability). So p = ½. B1 for the ratio, B1 for the mode being 1 because the ratio is below 1, M1 for dividing through by p and taking the square root, A1 for p = ½. Dividing by p is allowed because p is not zero; the rejected root should be named, not silently dropped.

Question 10

A fair die is rolled repeatedly. For each of the following state the distribution, with its parameters: (i) the number of sixes in ten rolls; (ii) the number of the roll on which the first six appears; (iii) the number of the roll on which the third six appears. Find the probability that the third six appears on the eighth roll.

Accepted Answer

(i) Binomial B(10, 1/6): a fixed number of trials, counting successes. (ii) Geometric with p = 1/6: counting trials to the first success. (iii) Negative binomial with r = 3 and p = 1/6: counting trials to the third success. P(X = 8) = ⁷C₂ × (1/6)³ × (5/6)⁵ = 21 × (1/216) × 0.4019 = 0.0391. B1 B1 B1 for the three models with parameters, M1 for the negative binomial form with ⁷C₂, A1 for 0.0391. The three means are 10/6, 6 and 18.

Question 11

X follows a geometric distribution with p = 0.1. Find the smallest integer n for which P(X > n) < 0.5.

Accepted Answer

P(X > n) = 0.9ⁿ, so the condition is 0.9ⁿ < 0.5. Taking logarithms, n log 0.9 < log 0.5, and since log 0.9 is negative the inequality reverses: n > log 0.5/log 0.9 = 6.58. So n = 7. Check: 0.9⁶ = 0.5314 and 0.9⁷ = 0.4783. M1 for 0.9ⁿ < 0.5, M1 for taking logarithms and reversing the inequality, A1 for n = 7. Fewer than half of all waits exceed seven trials, even though the mean wait is ten: the mean sits well above the middle of a distribution with such a long upper tail.

Question 12

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.

Accepted 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.
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.

Question 13

X follows a geometric distribution with parameter p. Prove that P(X > m + n | X > m) = P(X > n) for all positive integers m and n, and explain in words what this says about a run of failures. A machine is tested repeatedly and passes each test independently with probability 0.3. Given that the first four tests were failed, find the probability that the first pass occurs on the seventh test.

Accepted Answer

P(X > k) = (1 − p)^k, since X exceeds k exactly when the first k trials all fail. Then P(X > m + n | X > m) = P(X > m + n)/P(X > m), because X > m + n implies X > m, and this is (1 − p)^m+n/(1 − p)^m = (1 − p)ⁿ = P(X > n).
In words: a run of m failures has no effect on how much longer the wait will be. The trials are independent, so the process does not remember its failures.
Given the first four tests failed, the first pass being on the seventh test needs tests five and six to fail and test seven to pass: 0.7² × 0.3 = 0.147. Equivalently P(X = 7)/P(X > 4) = 0.3 × 0.7⁶/0.7⁴ = 0.147. M1 for P(X > k) = (1 − p)^k, M1 for writing the conditional probability as a quotient with the intersection simplified, A1 for (1 − p)ⁿ, B1 for the interpretation, A1 for 0.147. Multiplying 0.147 by 0.7⁴ again would count the four known failures twice: the condition has already fixed them.

Question 14

The number of trials needed for the rth success, X, can be written as X = X₁ + X₂ + … + X_r, where X_i is the number of trials from the (i − 1)th success to the ith success. Explain why each X_i has the geometric distribution with parameter p and why the X_i are independent, and hence derive the mean and variance of the negative binomial distribution from those of the geometric. Trials succeed with probability 0.2. Find the probability that the third success occurs within the first six trials.

Accepted Answer

After the (i − 1)th success the process starts again with the same p and independent trials, so the wait for the next success is a fresh geometric wait: each X_i is geometric with parameter p, and the waits are independent because they involve disjoint sets of trials.
Means add for any random variables: E(X) = r × 1/p = r/p. Variances add for independent ones: Var(X) = r × (1 − p)/p² = r(1 − p)/p². With r = 3 and p = 0.2 these are 15 and 60.
P(X = x) = ^x−1C₂ × 0.2³ × 0.8^x−3, so the probability that X is at most 6 is 0.2³(1 + 3 × 0.8 + 6 × 0.8² + 10 × 0.8³) = 0.008 × (1 + 2.4 + 3.84 + 5.12) = 0.008 × 12.36 = 0.0989. B1 for each wait being geometric with the same p, B1 for independence from disjoint trials, A1 for r/p, A1 for r(1 − p)/p² with independence named as the reason variances add, M1 for the four terms from x = 3 to x = 6, A1 for 0.0989. The sum starts at x = 3, because the third success cannot arrive before the third trial.

Question 15

A sales agent makes calls, each of which results in a sale independently with probability 0.1. The agent's target for a shift of 30 calls is at least three sales. Explain why the event 'at least three sales in 30 calls' is the same event as 'the third sale occurs on or before the 30th call', decide which distribution to use to find its probability, and find the probability that the target is met.

Accepted Answer

If the third sale has arrived by call 30 then the 30 calls contain at least three sales, and if the 30 calls contain at least three sales then the third of them came on some call up to the 30th. The two descriptions pick out the same outcomes, so the negative binomial variable X (the call of the third sale) and the binomial variable Y ~ B(30, 0.1) (sales in 30 calls) satisfy P(X is at most 30) = P(Y ≥ 3).
The binomial is the one to use: the shift fixes the number of trials at 30, and P(Y ≥ 3) needs only three terms to subtract, where the negative binomial would need the 28 terms from x = 3 to x = 30.
P(Y ≥ 3) = 1 − 0.9³⁰ − 30 × 0.1 × 0.9²⁹ − 435 × 0.1² × 0.9²⁸ = 1 − 0.0424 − 0.1413 − 0.2277 = 0.589. B1 for the two descriptions picking out the same outcomes, B1 for choosing the binomial because the number of calls is fixed, M1 for 1 − P(Y is at most 2), A1 for the three terms, A1 for 0.589.