Later success on a given trial
A question may ask for the second or third success on a specified trial. Here . Only success number 1 gives a geometric distribution. For later successes, use a binomial count in the earlier block and then force the final success.
Split the experiment before the final trial
Second success on trial 8
trial 1
F
q
trial 2
F
q
trial 3
S
p
trial 4
F
q
trial 5
F
q
trial 6
F
q
trial 7
F
q
trial 8
S
p
first 7: exactly 1 success
trial 8: success
P = ⁷C₁p²q⁶
For success number to occur on trial :
- the first trials contain exactly successes;
- trial is a success.
The first part is a binomial count. The final success is then multiplied on.
Symbols
- = probability that success number k occurs on trial r (none)
- = required success number (none)
- = trial on which it occurs (trials)
- = positions of the earlier k - 1 successes (none)
Count the earlier block, then force the final success
Worked example
Second success on trial 10
With , the first 9 trials must contain exactly one success, and trial 10 must succeed. (9709/52/M/J/24 Q1(c))
Worked example
Third success on trial 7
The first 6 trials need exactly two successes, followed by a success on trial 7. (9709/52/F/M/24 Q2(c))
Do not confuse three different events
Swipe left or right to read every column →
| Event | What must happen |
|---|---|
| exactly k - 1 successes before trial r, then success |
Common mistake
Independent trials have success probability 0.25. Find the probability that the third success occurs on trial 6.
Show worked answer
The first 5 trials need exactly two successes, then trial 6 must succeed.