Why Sequence Probabilities Need More Than One Probability
When an experiment has several stages, the chance of a particular sequence depends on what happens at each stage. For example, drawing a green item and then a yellow item is not the same ordered event as drawing a yellow item and then a green item. And if the first item is not returned before the second draw, the first result can change what is available next. The multiplication rule accounts for these changes directly.
The rule does not require the stages to have the same probabilities. It uses the probability of the first event, then the probability of the next event given that the first event occurred. That conditional probability is what lets the calculation reflect the actual sequence.
Why the Rule Works
The rule follows from the definition of conditional probability. When \(P(A)>0\), conditional probability is the fraction of the probability of \(A\) that also belongs to \(B\). In notation, this is \(P(B\mid A)=P(A\text{ and }B)/P(A)\). Multiplying both sides by \(P(A)\) gives the multiplication rule.
This justification also clarifies why the conditional term matters: the rule calculates the joint event using the probability of the second event within the circumstances created by the first. If \(P(A)=0\), the conditional probability \(P(B\mid A)\) is not defined by this ratio. In that case, the joint event has probability zero because it requires \(A\) to occur.
For three stages, apply the same rule first to the event “\(A\) and \(B\)” and then to \(C\). Provided the relevant conditional probabilities are defined, this gives the sequential form
Each factor matches a stage: the chance of the first event, the chance of the second given the first, and the chance of the third given the first two. The same reasoning extends to longer sequences by conditioning each stage on the results that came before it.
Draws With Replacement
In a draw with replacement, the selected item is returned before the next draw. If the draw is made in the same way each time, replacement restores the original set of items and its category counts. To calculate a sequence, specify each stage’s event and use the probabilities that match this sampling process.
Worked Example: Green Then Yellow With Replacement
A bag contains 5 green tokens and 3 yellow tokens. A token is drawn at random, recorded, returned to the bag, and then another token is drawn at random. Find the probability of drawing green first and yellow second.
Model and assumptions. Each draw selects one of the 8 tokens with equal probability. The first token is returned before the second draw, so the bag again contains 5 green and 3 yellow tokens at the second draw. Let \(G_1\) be the event that the first draw is green and \(Y_2\) the event that the second draw is yellow.
The first-stage probability is \(P(G_1)=5/8\). After the token is returned, the conditional probability of yellow on the second draw is \(P(Y_2\mid G_1)=3/8\). Thus
The probability of the specified order, green and then yellow, is \(15/64\), or about 23.44%. The denominator remains 8 for both stages because the token is returned before the second draw.
Draws Without Replacement
In a draw without replacement, the selected item is not returned. The pool for the next stage has one fewer item, and its category counts may have changed. Therefore the second probability must be calculated for the pool that remains after the first specified result, not automatically copied from the first stage.
Worked Example: Green Then Yellow Without Replacement
Use the same bag of 5 green and 3 yellow tokens, but now keep the first token out of the bag. Find the probability of drawing green first and yellow second.
Model and assumptions. Initially, each of the 8 tokens is equally likely to be selected. After a green token is drawn and kept out, 7 tokens remain: 4 green and 3 yellow. Let \(G_1\) and \(Y_2\) have the same meanings as in the previous example.
The first-stage probability is \(P(G_1)=5/8\). Given that the first token was green and was not replaced, the probability of yellow on the second draw is \(P(Y_2\mid G_1)=3/7\). Applying the multiplication rule,
The probability is about 26.79%. Although both stages call for a particular color, the denominator for the second stage is 7, not 8, because only 7 tokens remain. The answer differs from the with-replacement calculation because the sampling process is different.
Tracking Several Stages
For a longer sequence, calculate each stage’s probability after applying the conditions from all earlier stages. Do not skip a stage or use the original counts after items have been removed. The multiplication rule then combines the stage probabilities into the chance of the entire ordered sequence.
Worked Example: Three Specified Colors Without Replacement
A container holds 4 red, 3 blue, and 2 green balls. Three balls are drawn at random without replacement. Find the probability that the colors, in order, are red, then blue, then green.
Model and assumptions. There are 9 balls initially, and each remaining ball is equally likely to be drawn at each stage. No ball is returned. Let \(R_1\), \(B_2\), and \(G_3\) represent the specified colors at the first, second, and third draws.
The probability of red first is \(4/9\). Given a red first, 8 balls remain, including 3 blue, so the chance of blue second is \(3/8\). Given red first and blue second, 7 balls remain, including 2 green, so the chance of green third is \(2/7\). Therefore
The probability of this particular order is about 4.762%. Each denominator records the number of balls available at that stage, and each numerator records how many balls meet the required color condition at that stage.
Order and Event Definitions
A probability statement must describe the event whose probability is being calculated. “Red then blue” specifies both colors and their order. “One red and one blue in two draws” does not specify an order; it includes red then blue and blue then red. Those are different sequences, so a calculation for one order alone does not give the probability of the unordered event.
For example, with the 4 red and 3 blue balls among 9 in the previous example, the probability of red then blue is
The probability of blue then red is
Because these two orders cannot both happen in the same two draws, the probability of one red and one blue in either order is \(1/6+1/6=1/3\). This addition is only for combining the two distinct orders; the multiplication rule calculates the probability within each specified sequence.
A Reliable Calculation Method
Before multiplying, describe the random process and write down what is known at each stage. A compact check is to move through the sequence in order:
State what must happen at each stage, such as “green first, then yellow.”
Check whether selection is at random and whether items are replaced or kept out.
For every later stage, use the pool remaining after the earlier specified events.
Multiply the stage probabilities, then report the chance of the full sequence in context.
The order of the factors can be changed arithmetically, but their meanings should still be clear: each conditional probability must be based on the preceding events in the sequence. Keeping the order visible helps prevent using the wrong denominator or describing a different event from the one asked for.
Common Mistake: Reusing the First Probability
A frequent error is to calculate the probability of the first event and then use the same probability for the second draw without checking what has changed. For instance, in the without-replacement example, one might write \((5/8)(3/8)\) for green then yellow merely because there are 3 yellow tokens out of 8 initially. That reasoning uses the initial pool for the second draw even though one green token has been removed.
The correction is to calculate the second probability conditionally. Given a green first draw without replacement, 7 tokens remain and 3 are yellow, so the correct second factor is \(3/7\). With replacement, the second factor is \(3/8\) because the original composition is restored. Always match each factor to the conditions stated after the previous stages.
Check Your Understanding
For each question, identify the ordered event and use the probabilities appropriate to the stated sampling process.
- A box contains 6 white and 4 black counters. One counter is drawn, returned, and another is drawn. What is the probability of black first and white second?
- In a bag with 5 red and 2 blue marbles, two marbles are drawn without replacement. What is the probability of blue first and red second?
- Explain why the second factor in a without-replacement calculation may differ from the unconditional probability of the same event on the first draw.
- A jar contains 3 orange and 5 purple beads. Three beads are drawn without replacement. Write the multiplication-rule expression for orange, then purple, then purple, and evaluate it.
- Why is the probability of red then blue not, by itself, the probability of drawing one red and one blue in either order?