Parking lot problem I'm facing some difficulties in this question:
The probability of a person arrive to a parking lot before 8:00 am is 0.4, the probability to find a place to park given that she arrived before 8:00am is 0.6. If she get after 8:00 am, the probability to find a place to park is 0.3.
Given that the person, in 3 days, arrived exactly 2 days before 8:00 am, what is the probability that she successfully parked at least 1 day.
I tried different approaches, but in each of those I got a different result.
I tried to define: X: Number of days that she successfully parkedP(X>=1) = $1-P(X=0)$ = and I got $0.66$, but when I try to do P(X=1)+P(X=2)+P(X=3) i got $0.875$ . I don't know what I'm missing.
I tried:$1-P(X=0) = 1 - \binom{3}{1}$ *P(not park in a day | arrived to the parking lot before 8:00) * P(not park in a day | arrived to the parking lot before 8:00) * P(not park in a day | arrived to the parking lot after 8:00)
= $1 - 3*0.4*0.4*0.7 = 1 - 0.336 = 0.664$
The second approach was thinking that if I know that the person arrives 2 days before 8:00 and 1 day after 8:00, so now the probability of arrive to a parking lot in 3 days before 8:00 is $\frac{2}{3}$ and $\frac{1}{3}$ before 8:00. And I did $1 - P(Not Park)^3 = 1 - .5^3 = 0.875$
 A: Without loss of generality: the person arrived before 8am on the 1st two days
and after 8am on the 3rd day.
When the person arrives before 8am, p(failure) = 0.4. 
When the person arrives after 8am, p(failure) = 0.7.
Therefore, chance of 3 consecutive failures if $(0.4)^2 \times 0.7$.
Then chance of success is $1 - $ chance of 3 consecutive failures.

The probability is given that the chance of a person arriving  before 8am is $(0.4)$.  Because you are given that the person arrived before 8am on exactly 2 of the 3 days, the probability equaling $(0.4)$ may reasonably be
construed to be a red herring.
If you had not been given the information that the person arrived before 8am on exactly 2 of the 3 days, then the problem would have been significantly more complicated.
You would then have had to reason that the chance of there being no parking space on 1 specific day is
$$[(0.4) \times (0.4)] + [(0.6) \times (0.7)].$$
A: Here is how you should look at it:
If $X$ is the event of she finding parking space,
$P (X = 0) =  (1- 0.6)^2 \times (1-0.3)$
So the probability that she gets parking at least one of the days
$P(X \geq 1) = 1 - P(X=0)$
Even if you want to apply conditional probability formula, here is how you may want to look at it.
Even B: Probability that she reaches exactly two days before 8 AM
Event A: She does not find parking space any of the days
$P(B) = \displaystyle {3 \choose 2} \times 0.4^2 \times (1-0.4)$
$P(A \cap B) = \displaystyle {3 \choose 2} \times (0.4 \times 0.4)^2 \times (0.6 \times 0.7)$
So $P(A|B) = 0.4^2 \times 0.7$
What you are interested in is $ \ 1 - P(A|B)$
