We are doing a Bernoulli Experiment with $P=0.4$ unlimited times.
What is the probability that we will get two successes before two failures?
(I need to solve it via conditional probability).
What I got is $p=\frac{4}{9}$, I'm right?
I say:
$E$ = We get two successes in sequence.
$F$ = We get success at the first time.
$G$ = We got success at the second time.
Now:
$$P(E)=P(E|F)\cdot P(F)+P(E|F^C)\cdot P(F^C)\\ =P(E|F) \cdot \frac25+P(E|F^C)\cdot \frac35$$
and then: $$P(E|F)=P(E|FG)\cdot P(G|F)+P(E|FG^C)\cdot P(G^C|F) \\ =1\cdot \frac25 + P(E)\cdot \frac35 \\-----------------------\\ P(E|F^C)=P(E|F^CG)\cdot P(G|F^C)+P(E|F^CG^C)\cdot P(G^C|F^C) \\ =P(E)\cdot \frac25+0 $$
Then we mark: $p=P(E)$ - $$p=(\frac25+\frac35p)\frac25 +\frac25p\Rightarrow p=\frac49$$
Thank you!
Hint: We need to assume that the results of the experiment are independent.
The time may be unlimited, but the experiment will not take long.
Write S for success, and F for failure. The probability of S on any trial is $0.4$, and therefore the probability of F is $0.6$.
The only ways that we can have $2$ successes before $2$ failures are SS, SFS, and FSS.
Calculate. We get $\frac{2}{5}\cdot \frac{2}{5}+2\cdot\frac{2}{5}\cdot\frac{3}{5}\cdot\frac{2}{5}=\frac{44}{125}$.
Added: We interpreted "two successes before two failures" to mean "a total of two successes before a total of two failures." But conceivably what is intended (and should have been said) is "What is the probability of two consecutive successes before two consecutive failures?"
Let $a$ be the probability we ultimately win, given that we have not yet won, but the last result we have is a success. Let $b$ be the probability that we ultimately win, given that we had not yet lost, but the last result was a failure.
By conditional probabilities, we have $$a=\frac{2}{5}+\frac{3}{5}b$$ and $$b=\frac{2}{5}a.$$ So $a=\frac{10}{19}$ and $b=\frac{20}{95}$.
The probability we ultimately win is $$\frac{2}{5}\cdot\frac{10}{19}+\frac{3}{5}\cdot \frac{20}{95}.$$