When an item is taken out and not replaced, the contents change, so the second probability must use the new counts. Write the first fraction, then recount what is left before writing the second.
This is the dependent case that the independence lesson warned about, and it is drawn on the same kind of two-stage tree.
How do you update the second stage?
- Count the start. Note how many of each kind and the total.
- Write the first-stage fractions from those counts.
- On each branch, remove the item that was taken. Reduce that kind by 1 and the total by 1.
- Write the second-stage fractions from the new counts, branch by branch.
- Multiply along paths, add the paths you need, and check that all paths total 1.
Worked example
A bag holds 5 red and 3 green counters. Two are taken one after the other without replacement. Find (a) P(two red), (b) P(two different colours).
Step 1, first stage. P(R) = 5/8 and P(G) = 3/8.
Step 2, second stage.
- After R: 4 red and 3 green remain, total 7. P(R) = 4/7, P(G) = 3/7.
- After G: 5 red and 2 green remain, total 7. P(R) = 5/7, P(G) = 2/7.
Step 3, paths.
| Path | Working | Probability |
|---|---|---|
| R then R | 5/8 × 4/7 | 20/56 |
| R then G | 5/8 × 3/7 | 15/56 |
| G then R | 3/8 × 5/7 | 15/56 |
| G then G | 3/8 × 2/7 | 6/56 |
Step 4, check. 20 + 15 + 15 + 6 = 56, so the total is 56/56 = 1.
Step 5, (a). P(two red) = 20/56 = 5/14.
Step 6, (b). Different colours means R then G or G then R: 15/56 + 15/56 = 30/56 = 15/28.
Check (b). P(same colour) = 20/56 + 6/56 = 26/56, and 1 − 26/56 = 30/56. Both routes agree.
The mistake to watch for
The second-stage fractions are copied from the first stage.
Mistaken working: P(two red) = 5/8 × 5/8 = 25/64.
This uses 5 red out of 8 counters twice, as if the first counter had been put back.
Correction. One red counter has left the bag, so 4 red remain out of 7 counters. The answer is 5/8 × 4/7 = 20/56 = 5/14. Notice that 25/64 is close to 0.39 while 5/14 is about 0.36. The two values are not far apart, so the error is easy to miss unless you recount.
A quick sanity check: the second denominator is always one less than the first, and the numerator of the same colour is also one less. If the denominators are the same on both stages, you have probably forgotten the removal.
Check yourself
Try these, then open each answer.
1. A box has 4 milk and 6 dark chocolates. Two are taken without replacement. Find P(two dark).
Show answer
First dark: 6/10. Then 5 dark remain out of 9. P = 6/10 × 5/9 = 30/90 = 1/3.
2. Using the same box, find P(one milk and one dark, in either order).
Show answer
Milk then dark: 4/10 × 6/9 = 24/90. Dark then milk: 6/10 × 4/9 = 24/90. Add: 48/90 = 8/15. Check: two milk is 4/10 × 3/9 = 12/90, and 12 + 30 + 48 = 90, so the total is 1.
3. A bag has 3 red and 2 blue counters. Two are taken without replacement. Find P(at least one blue).
Show answer
The opposite is two red: 3/5 × 2/4 = 6/20 = 3/10. So P(at least one blue) = 1 − 3/10 = 7/10. Check by paths: RB 6/20 + BR 6/20 + BB 2/20 = 14/20 = 7/10.
Where this leads next
The last step in any tree is to confirm the outcomes add up, which is the focus of checking that an outcome model totals one. The probability tree and counting board lets you switch between replacement and no replacement and watch the second-stage fractions change. The non-calculator working trainer helps keep the fractions exact.
If you understand the idea but keep slipping on the counts, online one-to-one Mathematics tuition gives a teacher the chance to watch where the recount step goes missing.