A fair coin is flipped repeatedly. After 10 consecutive tails, what is the probability that the next flip is heads?
AGreater than 1/2 — the coin is 'due' for heads after so many tails.
BLess than 1/2 — a long tails streak suggests the coin may be biased.
CExactly 1/2 — each flip is independent, so past results carry no information.
DExactly (1/2)^11 — the probability of heads after 10 tails is the probability of that whole sequence.
This is the memorylessness property in action. Each flip is an independent Bernoulli trial with P(heads) = 1/2. The geometric distribution's memorylessness states P(X > m+1 | X > m) = P(X > 1) = 1/2 — the conditional probability of needing one more trial is identical to the unconditional probability. Option A is the gambler's fallacy: random processes do not 'correct' for streaks. Option D confuses the probability of a specific sequence of 11 flips with the conditional probability of the next flip given the previous 10.
Question 2 Multiple Choice
A quality inspector checks items from an assembly line where each item independently has a 5% defect rate. What is the expected number of items she must inspect to find the first defect?
A5 items — because 5% of 100 is 5 defects per 100 items.
B20 items — because E[X] = 1/p = 1/0.05 = 20.
C0.05 items — because the probability of a defect on any one item is 0.05.
D95 items — because the probability of a non-defect is 0.95.
For a geometric distribution with success probability p, E[X] = 1/p. With p = 0.05, E[X] = 20. Intuitively: if each item has a 1-in-20 chance of being defective, you expect to check about 20 items before finding one. Option A confuses the rate (5%) with a count per batch; option C takes p literally as a number of items; option D has no basis in the formula. The mean 1/p also implies that rarer events require more trials on average — lower p means longer expected wait.
Question 3 True / False
The memorylessness property of the geometric distribution is a direct consequence of the independence of Bernoulli trials.
TTrue
FFalse
Answer: True
Memorylessness — P(X > m+n | X > m) = P(X > n) — follows algebraically from independence. P(X > m) = (1−p)^m is just the probability that m independent trials all fail. Conditional probability gives P(X > m+n | X > m) = (1−p)^(m+n) / (1−p)^m = (1−p)^n = P(X > n). Every step in this derivation relies on independence: the probability of a sequence of failures multiplies because trials are independent. If outcomes were correlated, the distribution would not be memoryless.
Question 4 True / False
Having already failed 10 times in a geometric trial sequence, the expected number of additional trials needed before the first success is less than 1/p, because you have 'used up' some of your expected waiting time.
TTrue
FFalse
Answer: False
This is the gambler's fallacy expressed in expectation form. The geometric distribution is memoryless: given that you have failed 10 times, the expected number of additional trials is still exactly 1/p — as if you were starting from scratch. The failures are irrelevant because each trial is independent. The mechanism (a coin, a manufacturing defect rate) has not changed. Past outcomes contain no information about future outcomes, so the remaining expected wait is always 1/p regardless of accumulated failures.
Question 5 Short Answer
Explain in plain language what the memorylessness property means for the geometric distribution, and why it holds.
Think about your answer, then reveal below.
Model answer: Memorylessness means the distribution of remaining wait time is the same regardless of how long you have already waited. If you flip a coin until heads and have seen 20 tails, the expected number of additional flips is still 1/p — identical to starting fresh. This holds because each flip is an independent Bernoulli trial: the coin has no memory of past results, and past results provide no information about future outcomes. The underlying success probability p is unchanged by any number of failures.
Formally: P(X > m+n | X > m) = P(X > n), which follows directly from independence. This is why the geometric distribution is the discrete analogue of the exponential distribution — the only continuous distribution with the same property. Memorylessness has practical implications: if you are waiting for a bus that arrives with probability p each minute, knowing you've waited 20 minutes does not change the expected additional wait. The wait distribution resets with each passing minute.