Dependent Events: Probability Without Replacement for Cards, Marbles, and Quality Checks
October 29, 2026
Two events are dependent when the outcome of the first changes the probability of the second. This happens whenever you sample without replacement: after drawing an item, there are fewer items left, and the mix changes. The probability that both occur is P(A and B) = P(A) × P(B given A).
A bag contains 5 red and 3 blue marbles. The probability that the first marble is red is 5/8. If it is red, there remain 4 red among 7 marbles, so the probability that the second is also red is 4/7. Both red: 5/8 × 4/7 = 20/56 = 35.7%.
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Two aces from a standard deck: 4/52 × 3/51 = 12/2,652 = 0.45%. Three hearts in a row: 13/52 × 12/51 × 11/50 = 1,716/132,600 = 1.29%. Drawing two cards of the same color: after one card, 25 of the remaining 51 cards match, so the probability is 25/51 = 49.0%, slightly less than a half. Compare with drawing with replacement, where two reds would be 5/8 × 5/8 = 39.1%: removing the first red makes the second red less likely.
| Event | Calculation | Probability |
|---|---|---|
| Two reds from 5 red, 3 blue | 5/8 x 4/7 | 35.7% |
| Two aces from a deck | 4/52 x 3/51 | 0.45% |
| Three hearts in a row | 13/52 x 12/51 x 11/50 | 1.29% |
| Same color, two cards | 26/52 x 25/51 | 24.5% for red-red, 49.0% either color |
| Two reds with replacement | 5/8 x 5/8 | 39.1% |
The Complement for 'At Least One'
A box of 20 parts contains 2 defective ones. You inspect 3 at random. The probability that none are defective is 18/20 × 17/19 × 16/18 = 4,896/6,840 = 71.6%. So the probability of finding at least one defective part is 1 − 0.716 = 28.4%. Inspecting more parts increases the odds: with a sample of 6 from the same box, none defective is C(18,6) ÷ C(20,6) = 18,564 ÷ 38,760 = 47.9%, so at least one is found 52.1% of the time. Sampling plans in quality control use these calculations to decide how many items to test.
Combinations Approach
The same results come from counting combinations: the chance of drawing 2 reds from 5 red and 3 blue is C(5,2) ÷ C(8,2) = 10 ÷ 28 = 35.7%, which equals the multiplication result. The combination method is convenient when the order does not matter and when many items are drawn: the probability of being dealt exactly 2 aces in a 5-card hand uses C(4,2) × C(48,3) ÷ C(52,5).
Common Mistakes
Treating draws as independent when they are not (using the same probability for the second draw) is the typical error. Assuming past outcomes change future odds for independent events like coin flips (the gambler's fallacy) is the opposite error. Check whether the population is large relative to the sample: if you draw 5 people from a city of a million, replacement hardly matters, and the independent approximation works well.
Frequently Asked Questions
What are dependent events?
Events where the outcome of one changes the probability of the other.
What is the multiplication rule for dependent events?
P(A and B) = P(A) × P(B given A).
What is the chance of two aces from a deck?
About 0.45%.
How do I find at least one defective in a sample?
Compute the probability of none defective and subtract from 1.
When can I treat draws as independent?
When the sample is tiny compared with the population.