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We perform Bernoulli processes of respective probabilities p and q. What is the probability of getting n consecutive successes before getting m consecutive failures?
Sol: We define the following events:
E: getting n consecutive successes before getting m consecutive failures.
F: The first process is a success.
G: The n-1 processes following the first are successes.
H: The m-1 processes following the first are failures.
We condition on the first process:
P(E)=P(E|F)P(F)+P(E|F^c)P(F^c)
We condition P(E|F) on the event G:
P(E|F)=P(E|FG)P(G|F)+P(E|FG^c)P(G^c|F)
The solution then says that P(E|FG)=1 and P(G|F)=p^{n-1}. On that I agree. But it also says that P(E|FG^c)=P(E|F^c) and P(G^c|F)=q^{n-1}. I can't make any sense of the first one, but the second is obviously false, because G^c means "the n-1 processes following the first are not all successes", while q^{n-1} is the probability for the event "the n-1 processes following the first are all failures".
But "not all sucesses" does not imply "all failures". Am I crazy?
Sol: We define the following events:
E: getting n consecutive successes before getting m consecutive failures.
F: The first process is a success.
G: The n-1 processes following the first are successes.
H: The m-1 processes following the first are failures.
We condition on the first process:
P(E)=P(E|F)P(F)+P(E|F^c)P(F^c)
We condition P(E|F) on the event G:
P(E|F)=P(E|FG)P(G|F)+P(E|FG^c)P(G^c|F)
The solution then says that P(E|FG)=1 and P(G|F)=p^{n-1}. On that I agree. But it also says that P(E|FG^c)=P(E|F^c) and P(G^c|F)=q^{n-1}. I can't make any sense of the first one, but the second is obviously false, because G^c means "the n-1 processes following the first are not all successes", while q^{n-1} is the probability for the event "the n-1 processes following the first are all failures".
But "not all sucesses" does not imply "all failures". Am I crazy?
