Why the Probability Changes from Draw to Draw
A bag of marbles makes it easy to see what conditional probability means: after a marble is drawn and not returned, the bag’s contents change. The probability for the next draw must be calculated from the marbles that remain, not from the original contents. The result of the first draw can therefore affect the probability of the second.
In The General Multiplication Rule, we used \(P(A\cap B)=P(A)P(B\mid A)\) to find the probability that two events both occur. A draw without replacement gives a concrete way to calculate the conditional factor \(P(B\mid A)\): restrict attention to the bag contents after event \(A\) has occurred, then count the marbles that meet event \(B\).
Suppose a bag starts with 5 green marbles and 3 blue marbles. The probability of green on the first draw is \(5/8\). If the first marble is green, only 4 green marbles remain among 7 total, so the conditional probability of green on the second draw is \(4/7\). If the first marble is blue instead, all 5 green marbles remain, and the conditional probability of green on the second draw is \(5/7\).
These are different conditional probabilities because the first draw left different contents in the bag. This is not the independent-trial model discussed in The Gambler’s Fallacy and Independent Trials. With replacement, the marble is returned and the original mix is restored. Without replacement, the next draw depends on what has already been drawn.
Keep an Inventory as You Draw
A simple inventory helps prevent errors. Record the number of each color before drawing, update the counts after each result, and use the updated counts for the next conditional probability. The denominator is always the total number of marbles currently in the bag.
| Situation | Target marbles remaining | Total marbles remaining | Conditional probability of target |
|---|---|---|---|
| Before any draw: 5 green, 3 blue | 5 | 8 | \(5/8\) |
| After drawing a green | 4 | 7 | \(4/7\) |
| After drawing a blue | 5 | 7 | \(5/7\) |
The table shows a useful pattern. Removing a target-color marble lowers the target count as well as the total. Removing a different-color marble leaves the target count unchanged while lowering the total. Either way, calculate the next probability from the current bag.
Worked Example: Two Green Marbles in a Row
Worked Example: Two Green Marbles in a Row
A bag contains 5 green marbles, 3 blue marbles, and 2 yellow marbles. Two marbles are drawn without replacement. Find the probability that both are green.
State: Let \(G_1\) mean that the first marble is green and \(G_2\) mean that the second marble is green. We want \(P(G_1\cap G_2)\).
Plan: Use the general multiplication rule, \(P(G_1\cap G_2)=P(G_1)P(G_2\mid G_1)\). There are 10 marbles initially. Given that the first marble is green, 4 green marbles remain among 9 total, so \(P(G_2\mid G_1)=4/9\). The condition probability \(P(G_1)=5/10\) is positive, so the conditional probability is defined.
Do: Substitute the initial probability and the updated conditional probability:
As a count check, there are \(5\cdot4=20\) ordered ways to draw two green marbles. There are \(10\cdot9=90\) ordered ways to draw any two marbles without replacement. Thus the count ratio is \(20/90=2/9\), the same result.
Conclude: The probability of drawing two green marbles in a row is \(2/9\), or about 0.2222. After the first green marble is removed, the probability of green on the second draw is \(4/9\), not the original \(5/10\).
When More Than One Draw Order Works
Sometimes the question names the marbles drawn but does not specify their order. “One blue and one yellow” includes blue first and yellow second, as well as yellow first and blue second. Each order is a separate path, with its own sequence of conditional probabilities.
Find the probability of each path by multiplying along that path. Then add the path probabilities because the two orders cannot both happen in the same pair of draws. This is the same approach used in The General Multiplication Rule when more than one order satisfies an event.
Worked Example: One Blue and One Yellow
Worked Example: One Blue and One Yellow
A bag contains 4 blue marbles, 3 yellow marbles, and 2 red marbles. Two marbles are drawn without replacement. Find the probability of drawing one blue and one yellow, in either order.
State: Let \(BY\) be the event that the first marble is blue and the second is yellow. Let \(YB\) be the event that the first marble is yellow and the second is blue. The requested event is \(BY\cup YB\).
Plan: There are 9 marbles at first and 8 after the first draw. Use the general multiplication rule separately for each order. The paths \(BY\) and \(YB\) are disjoint: a single ordered pair cannot have both a blue-first and a yellow-first result. Therefore, add their probabilities.
Do: For blue followed by yellow, the first-draw probability is \(4/9\). After a blue is drawn, all 3 yellow marbles remain among 8:
For yellow followed by blue, the first-draw probability is \(3/9\). After a yellow is drawn, all 4 blue marbles remain among 8:
Add the two disjoint path probabilities. As a check, the favorable ordered draws total \(4\cdot3+3\cdot4=24\), out of \(9\cdot8=72\) possible ordered draws:
Conclude: The probability of drawing one blue and one yellow marble in either order is \(1/3\), or about 0.3333. Including both orders matters; counting only blue followed by yellow would give just half the desired probability.
Updating Probabilities Through Three Draws
For three or more draws, update the bag after every result. A conditional probability can depend on the entire sequence so far, not just the immediately preceding draw. For example, after drawing two marbles, the third-draw probability must reflect both of them.
A tree diagram can display this process: each branch gives a possible draw and its probability given the branches already taken. Multiply along one path to find the probability of that particular sequence. If different paths satisfy the event, add their probabilities. The branch probabilities generally change as the contents change.
Worked Example: Exactly Two Orange Marbles in Three Draws
Worked Example: Exactly Two Orange Marbles in Three Draws
A bag contains 3 orange marbles, 4 purple marbles, and 2 white marbles. Three marbles are drawn without replacement. Find the probability of drawing exactly two orange marbles.
State: The event is that exactly two of the three draws are orange. The possible color sequences are orange-orange-not orange (OON), orange-not orange-orange (ONO), and not orange-orange-orange (NOO), where “not orange” includes either purple or white.
Plan: Calculate each path using the changing counts, then add the path probabilities. The three paths are disjoint because a sequence cannot have two different color orders at once. There are 9 marbles initially, then 8, then 7.
Do: For OON, two orange marbles are drawn first, leaving 6 non-orange marbles among 7:
For ONO, after the first orange, 6 non-orange marbles remain among 8. After a non-orange marble is drawn, 2 orange marbles remain among 7:
For NOO, there are 6 non-orange marbles initially. After one is drawn, the bag has 3 orange marbles among 8; after an orange is drawn, 2 orange marbles remain among 7:
Now add the three disjoint paths. The count check gives \(3\cdot2\cdot6=36\) ordered outcomes for each path, so there are \(108\) favorable ordered outcomes out of \(9\cdot8\cdot7=504\):
Conclude: The probability of drawing exactly two orange marbles in three draws is \(3/14\), or about 0.2143. Each sequence has the same probability here, but its branch probabilities still need to be updated after each draw.
Common Mistakes and AP Exam Tips
- Reusing the original fraction. After a marble is removed, the total in the denominator decreases. If the removed marble matches the target color, the target count decreases too. Show the updated numerator and denominator for each conditional factor.
- Assuming draws are independent. Without replacement, the first result changes the bag, so the conditional probability for a later draw may differ from its original probability. A full-credit explanation says which marbles remain after the earlier draw.
- Multiplying when the question allows different orders, but counting only one. For “one blue and one yellow,” include both blue-then-yellow and yellow-then-blue. Multiply within each path, then add the disjoint path probabilities.
- Adding the probabilities of stages in a single sequence. “First orange and then purple” requires both results, so multiply the branch probabilities. Add probabilities only for separate paths that each meet the event.
- Using a conditional probability with the wrong reference group. In \(P(B\mid A)\), the condition \(A\) determines which outcomes are considered. State what has already happened before identifying the remaining marbles and calculating the next probability.
A clear solution names the event, shows how the bag changes, writes the conditional factors, and interprets the answer in context. Avoid writing only a decimal: the setup is what shows that you understood which draws the probability describes.
Check Your Understanding
For each question, show how the bag contents change and give the probability in context.
- A bag contains 4 red and 5 green marbles. Two marbles are drawn without replacement. What is the probability both are red?
- A bag contains 3 blue and 2 yellow marbles. What is the probability of drawing one of each in either order?
- A bag contains 5 white and 3 black marbles. Given that the first marble is black, what is the probability that the second marble is white?
- A bag contains 2 purple and 3 orange marbles. Three marbles are drawn without replacement. List the possible color sequences for exactly two purple marbles.
- In your own words, explain why the probability of a color on the next draw can change after a marble is removed without replacement.