Probability With and Without Replacement

The phrase “with replacement” describes what happens to an item after it is selected. If the item goes back into the pool before the next selection, the pool resets. If it stays out, the next selection comes from a changed pool. This small procedural detail changes the denominator and often determines whether multiplication uses the same probability at each step.

With replacement: the pool resets

Suppose a bag contains three red counters and two blue counters. Draw one counter, record its color, return it, and mix the bag before drawing again. The probability of red on either draw remains 3/5. The first draw does not alter the second draw’s probability, so the events are independent. The chance of red twice is (3/5)(3/5) = 9/25.

With replacement, ordered outcomes can repeat: red then red is possible because the first red counter returns. If the question asks for one red and one blue in either order, count both sequences: red-blue or blue-red. Each has probability (3/5)(2/5) = 6/25, so the total is 12/25. Clear event definitions help prevent a common error of counting only one order.

Without replacement: the pool changes

Now draw from the same bag but keep the first counter out. If the first counter is red, only two red counters remain among four total. Therefore P(second red given first red) = 2/4, not 3/5. The joint chance of two reds is (3/5)(2/4) = 3/10. The first outcome changed the second probability, so these draws are dependent.

Without replacement, the chance of a blue counter on the second draw depends on the first color. After a red, three blue counters? No: there are still two blue counters, now among four items, so the chance is 2/4. After a blue, one blue remains among four, so the chance is 1/4. This conditional reasoning is more reliable than trying to memorize a single second-draw probability.

Order, combinations, and what the question counts

Some problems care about sequence. “First red, then blue” is one ordered event. Other questions say “one red and one blue,” where either order qualifies. When using sequential probabilities, add the valid order-specific paths. When using combinations, unordered groups are counted once. Both approaches work when used consistently, but mixing them can double-count or omit outcomes.

For example, a box has four green and three yellow marbles. Draw two without replacement. The probability of one of each color is P(green then yellow) + P(yellow then green). That equals (4/7)(3/6) + (3/7)(4/6) = 24/42 = 4/7. If only “green then yellow” is calculated, the answer misses half the valid sequence structure.

A general formula

If a pool has N total objects and K are of the target type, then for n draws with replacement the chance all n are target items is (K/N)ⁿ. This assumes the draws are made under the same conditions and returned items are indistinguishable in the probability model. For draws without replacement, the chance all n are target items is (K/N)((K−1)/(N−1))((K−2)/(N−2)) and so on, stopping after n factors. Each successful draw reduces both the target count and the total pool.

For a concrete example, a deck has four aces among 52 cards. The probability of two aces with replacement is (4/52)(4/52) = 1/169. Without replacement, it is (4/52)(3/51) = 1/221. Replacing the first ace keeps four aces available; keeping it out leaves three. The second probability is the only conceptual change, but it changes the result.

Replacement and independence are related, but read the setup

Replacement commonly makes draws independent because the composition returns to its original state. Without replacement, draws from a small finite population are dependent because each selection changes the remaining pool. However, independence should be justified from the process, not from a vocabulary shortcut. If the population is extremely large relative to the sample, the dependence may be small, but a test problem that explicitly says no replacement usually expects conditional probabilities.

There are processes where an item can be returned yet the trials are still not independent—for example, if the selection method changes or the chance of choosing each item is altered. Conversely, some events defined from a no-replacement process can be independent in special cases. The basic exam model assumes a fixed, uniformly mixed pool and asks whether the draw changes its composition.

Worked comparison

A class has 12 students, of whom five are left-handed. Two students are selected at random. If the first student is returned to the selection pool before the second draw, the probability both are left-handed is (5/12)(5/12) = 25/144. If the first is not returned, it is (5/12)(4/11) = 5/33. The second fraction changes because one left-handed student has already been removed. The no-replacement probability is lower because success on the first draw slightly reduces the available number of successful outcomes.

A step-by-step method

1. Write the starting number of total items and the number that satisfy the event.
2. Translate the process: does the selected item go back before the next draw?
3. For each step, update the numerator and denominator if the draw is without replacement.
4. Multiply along a sequence to find an ‘and’ probability.
5. Add different valid sequences when the prompt accepts more than one order.
6. Check that the final probability is between 0 and 1 and that each conditional fraction matches the pool at that step.

Common mistakes

• Keeping the original denominator after a no-replacement draw. The pool is smaller on the next step.
• Reducing the target count after a draw of the wrong type. Update the pool according to what actually happened.
• Assuming with replacement means the selected item must be selected again. Replacement makes it possible, not certain.
• Forgetting to add the reverse order when the prompt asks for a mixed pair in either order.
• Calling trials independent without checking whether the process changed.
• Using combinations for an ordered sequence or multiplying sequential probabilities while counting unordered outcomes inconsistently.

On the exam, look for the procedure word: replaced, returned, kept out, selected without replacement. That tells you whether the next probability resets or must be recalculated from the remaining pool. Then decide whether the question needs one sequence or several valid orders.