MathematicsExam code: 0580
Conditional Probability
Probability
Counting inside the condition
- With a listed or tabulated sample space, cross out every outcome the condition excludes
- What remains is the new total, and the successes are counted from those survivors only
- Working from a completed grid makes this reliable, since the condition simply selects some of the cells
Worked example
A condition applied to a sample space
One fair spinner is numbered 1, 2, 3 and 4, and another fair spinner is numbered 2, 5 and 7. Both are spun and the two numbers are added. Given that the total is even, find the probability that the first spinner landed on 3.
Solution:
- There are 4 × 3 = 12 equally likely outcomes altogether
- The sample space shows every total, with the first spinner across the top and the second down the side
| + | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| 2 | 3 | 4 | 5 | 6 |
| 5 | 6 | 7 | 8 | 9 |
| 7 | 8 | 9 | 10 | 11 |
- The even totals are 4 and 6 in the first row, 6 and 8 in the second and 8 and 10 in the third, so 6 outcomes meet the condition
- You no longer divide by 12: the restricted total is 6 outcomes
- Of those 6 outcomes, the first spinner shows 3 in two of them, (3, 5) with total 8 and (3, 7) with total 10
- P(first spinner shows 3, given the total is even) = 2/6 = 1/3
- Dividing by 12 would answer a different question: the probability that the total is even and the first spinner shows 3, which is 2/12 = 1/6