Why the Success Probability Changes
When a sample is drawn without replacement, an observed item is no longer available for later draws. If the population contains a fixed number of successes, observing a success removes one of them; observing a failure removes an item that was not a success. The composition of the remaining population therefore changes, and so does the probability of success on the next draw.
This matters whenever information arrives sequentially. A calculation based on the original success proportion may be appropriate for the first draw, but not generally for later draws conditional on what has already been observed. The update is simple: count the successes and total items still available, then take their ratio.
The feasible values of \(s\) are the integers satisfying \(\max(0,r-(N-K))\le s\le \min(r,K)\). The numerator counts the successes left, and the denominator counts all items left. The same formula applies if the information is only that the first \(r\) draws contained \(s\) successes: every history consistent with that count leaves exactly \(K-s\) successes among \(N-r\) items.
Deriving the Update and Reading Its Direction
Conditional on an observed history, the remaining items are equally likely to be selected next. Of the \(N-r\) remaining items, exactly \(K-s\) are successes. Since each remaining item has the same chance of being the next draw, the probability of selecting a success is the number of remaining successes divided by the number of remaining items. This gives the update formula.
The effect of the latest observation can also be written precisely. Suppose \(m\) items remain, including \(a\) successes, and \(m>1\), so another draw will still be available after the next one. The current next-draw probability is \(a/m\). If the next observed item is a success, the following probability is \((a-1)/(m-1)\), provided \(a\ge1\). If it is a failure, the following probability is \(a/(m-1)\), provided \(m-a\ge1\).
Thus, an observed success lowers or leaves unchanged the next-draw probability; it lowers it strictly when at least one failure remains. An observed failure raises or leaves unchanged that probability; it raises it strictly when at least one success remains. If no successes remain, the probability stays zero after a failure. These qualifications matter at the boundaries of the model.
Write the population size \(N\) and the number of successes \(K\).
After \(r\) draws with \(s\) successes, \(N-r\) items and \(K-s\) successes remain.
Divide the remaining success count by the remaining population size. Confirm that \(r<N\) before referring to a next draw.
Subtract one from the remaining total, and subtract one from the success count only if the observed item was a success.
Worked Sequential Updates
Worked Example: Updating After a Mixed History
A hypothetical inspection tray contains 15 components, 6 of which have a specified surface defect. Four components are drawn uniformly without replacement. The first three inspection results are failure, success, failure. Find the probability that the fourth component has the defect. Also calculate the probability of the observed three-result history and the joint probability that this history occurs and the fourth component is defective.
Method: Update the remaining counts after the exact observed history. This avoids using the original defect proportion for a later draw.
Initially there are 15 components and 6 defective components. After a failure, a success, and a failure, 3 components have been removed, including 1 defective component. Thus 12 remain, of which 5 are defective. The fourth-draw conditional probability is
For the probability of the observed history, multiply the successive conditional probabilities. At the first draw, 9 of 15 are failures. After that failure, 6 of 14 are successes. After that success, 8 of 13 are failures. Therefore
The joint probability of the history and a defective fourth component is the history probability multiplied by the conditional probability of that fourth result:
As a check, multiplying all four sequential probabilities gives \((9/15)(6/14)(8/13)(5/12)=6/91\), the same joint probability. The common mistake is to use \(6/15\) for the fourth-draw probability; that is the original proportion, not the proportion among the 12 components that remain.
Worked Example: Conditioning on a Success Count
A hypothetical sample is drawn from 10 records, 4 of which are marked for review. Given that exactly one of the first two records drawn is marked, find the probability that the third record is marked. Compare it with the probability if the first two are both marked, if both are unmarked, and with the unconditional probability that the third record is marked.
Method: Use the observed success count to update the remaining composition. The order of the first two results is not needed when their total number of marked records is specified.
If exactly one of the first two is marked, 2 records have been removed and 1 marked record has been removed. There are 8 records left, including 3 marked records, so
If both first records are marked, 2 of the original 4 marked records have been removed. The conditional probability for the third is \(2/8=1/4\). If both are unmarked, all 4 marked records remain, and the conditional probability is \(4/8=1/2\).
Before observing any records, the third position is equally likely to contain any one of the 10 records, so its unconditional probability of being marked is \(4/10=2/5\). The conditional values \(3/8\), \(1/4\), and \(1/2\) differ from this unconditional probability because the earlier results provide information about which records remain. Each value is checked directly as remaining marked records divided by remaining total records. The common mistake is to interpret “exactly one” as a particular order: either marked-then-unmarked or unmarked-then-marked gives the same remaining counts.
Worked Example: A Conditional Probability and a Joint Probability
A hypothetical shipment has 12 packages, 5 of which require additional screening. Four packages are selected uniformly without replacement. Find the probability that exactly two of the first three selected packages require screening. Then find the probability that exactly two of the first three require screening and the fourth also requires screening.
Method: First calculate the probability of the condition, then multiply by the updated fourth-draw probability. This makes clear which probability is conditional and which is joint.
Among the first three packages, choosing exactly two that require screening means choosing 2 from the 5 such packages and 1 from the 7 others. There are \(\binom{12}{3}\) equally likely sets of three packages, so
Given this count, 3 of the 5 screening-required packages have been selected, so 2 remain among the 9 packages left. The probability that the fourth package requires screening is \(2/9\). Therefore
The calculation is consistent with counting ordered status patterns directly: there are 3 positions for the one non-screening package among the first three. Each such four-draw pattern has probability \((5\cdot4\cdot3\cdot7)/(12\cdot11\cdot10\cdot9)=7/198\), and \(3(7/198)=7/66\). This direct check reveals a mismatch: it does not equal \(7/99\). The reason is that the previous conditional step used the wrong remaining count: after exactly two successes in the first three, \(5-2=3\) screening-required packages remain, not 2. The correct conditional probability is \(3/9=1/3\), and the correct joint probability is
The direct pattern count and the corrected conditional calculation agree. This check illustrates the central rule: subtract the number of observed successes from the original success total, not from an intermediate count already reduced for another purpose.
Why Draws Without Replacement Are Dependent
Let \(I_j\) equal 1 if draw \(j\) is a success and 0 otherwise. Each draw has marginal success probability \(P(I_j=1)=K/N\), because every population item is equally likely to occupy any fixed draw position. But conditioning on the first draw changes the probability for the second: after a success it is \((K-1)/(N-1)\), while after a failure it is \(K/(N-1)\), when the relevant outcome is possible. These conditional probabilities generally differ from the unconditional probability \(K/N\), so the draw indicators are generally dependent.
For \(N>1\), the dependence can also be shown through covariance. For two distinct draw positions, the probability both are successes is \(K(K-1)/(N(N-1))\). Since \(E[I_j]=K/N\) and \(I_1I_2\) equals 1 exactly when both draws are successes,
When \(0<K<N\), this covariance is negative: learning that one draw is a success tends to reduce the chance of success on another draw. If all items have the same status, \(K=0\) or \(K=N\), the outcomes are constant and the covariance is zero. The conditional update and the covariance describe the same finite-population effect from two different perspectives.
Common Mistakes and Checks
- Using the starting proportion after observations. Recount the remaining successes and total items; do not automatically reuse \(K/N\).
- Forgetting that a next draw must exist. The formula for draw \(r+1\) requires \(r<N\), as well as a feasible history with positive probability.
- Updating the wrong count. A failure reduces the total remaining but does not reduce the number of remaining successes; a success reduces both.
- Claiming every update is strict. After a failure the probability increases only if at least one success remains. If none remains, it stays zero. Likewise, a success does not lower the probability if no failures remain.
- Confusing conditional and joint probabilities. A probability given a history is not the probability that the history and the next outcome both occur; multiply by the history probability to obtain the joint probability.
Check Your Understanding
For each problem, update the remaining counts before calculating a conditional probability.
- A box contains 14 sensors, 5 of which fail a calibration test. After two failures and one pass are drawn without replacement, what is the probability the next sensor fails? (Answer: \(4/11\))
- A population has 9 items, 3 successes, and 6 failures. Given that the first two draws contain exactly one success, find the probability of success on draw 3. Compare with the probability after two failures. (Answer: \(2/7\) and \(3/7\), respectively)
- A tray contains 11 parts, 4 defective. Find the probability the first two draws are both defective, then find the conditional probability the third is defective given that history. (Answer: \(4/11\cdot3/10=6/55\); conditional probability \(2/9\))
- A population contains 8 items, 3 successes. Given exactly one success among the first three draws, what is the probability the fourth draw is a success? (Answer: \(2/5\))
- There are 10 cards, 4 red and 6 black, drawn without replacement. Calculate \(P(\text{second red}\mid\text{first red})\), \(P(\text{second red}\mid\text{first black})\), and the unconditional probability the second card is red. (Answer: \(1/3,\ 4/9,\ 2/5\); the differing conditional probabilities show dependence)