A juggler practices juggling. His results vary quite a bit. Sometimes he drops the balls quickly, sometimes he can hang in there for a while. Here is a distribution of his juggling durations after 36 attempts:
| Duration (s) | No of attempts |
|---|---|
| 5 | 9 |
| 10 | 9 |
| 15 | 6 |
| 20 | 5 |
| 25 | 3 |
| 30 | 4 |
The question is: how likely is he going to continue past a certain duration once he reaches it?
My idea would be e.g. for 5 seconds: He stops after 5 seconds 9 times, but he achieves > 5 seconds 36-9=27 times Hence his likelihood of reaching more than 5 seconds once he is at the 5 s mark is 1-(9/27) =66%
Or, once he reaches 10 seconds: 9 times he reaches 10 seconds (attempts to 5 dont count because they dont reach 10 seconds), 18 times he goes past 10, so the likelihood of passing 10s once reached is 50% or:p= 1-(9/18).
Is this right?
Note: The table gives the frequency of stopping at each duration.
However, you seek the conditional probability for continuing when given that a duration is reached.
This uses Bayes' Rule. When $X$ is the duration, then $\{X>a\}$ is the event of exceeding duration $a$, and $\{X\geq a\}$ is the event of reaching duration $a$. So the probability we seek is: $$\mathsf P(X>a\mid X\geq a)=\dfrac{\mathsf P(X > a)}{\mathsf P(X\geq a)}$$
Thusly the probability for exceeding $10$s when given that at least $10$s has been reached is: $$\mathsf P(X> 10\mid X\geq 10)=\dfrac{~~~~~~~6+5+3+4}{9+6+5+3+4}=\dfrac{18}{27}$$