Ballot boxes probability problem collection I tag these questions as homework, as they are older exam questions and every year
Can you try to solve/explain how to solve with a method some of these?
If something is answered in an old post please post the link.

1. A ballot box contains 2 white and three black balls. Two players are pulling one ball out at a time (first player pulls one then the second pulls one etc) without putting the ball back. The winner is whoever pulls first a white ball. Find the probability for the first player to win.

2. We have three ballot boxes: the 1st contains two white balls, the 2nd contains two black balls and the 3rd contains one black and one white ball. We pick randomly one ball from a random ballot box. If the ball that we pulled is white, what's the probability that we chose the first ballot box?

To community: I'm familiar with many terms of probability theory so with a good analysis I think I might understand your answers.
Reminder:Their diffictulty may seem easy to medium but not everyone is great at this.
 A: Problem 1
There are two outcomes when first player wins. One is when he draw white ball from the ballot box and the second is when both players draw black ball in their respective first draw.
So the probability for the first outcome is $\frac 25$.
The probability for the second outcome is: $\frac 35 \cdot \frac 24 \cdot \frac 23 = \frac {12}{60} = \frac{1}{5}$
So the total probability will be the sum of the probabilities of the two outcomes. Which means:
$$\frac 25 + \frac 15 = \frac 35$$
Problem 2
If we know that we pick white ball that means that it's either the first or the third ballot box. But since there are 2 white balls in the first and 1 white ball in the third. The probability is:
$$\frac 23$$
A: ANS 1
only 2 cases will lead to 1st person's win namely :-  {W} & {B,B,W}    (these sets display the order in which the balls were drawn)
so the Probability would be [$\frac{2}{5} (case 1) + \frac{3}{5} \cdot \frac{2}{4} \cdot \frac {2}{3} (case 2)$]  
ANS 2
f -> event of choosing 1st box
w -> event of getting a white ball
P(f) =1/3
p(w)= 1/3*1 + 1/3*1/2
P(f/w)= P(w/f)P(f)/P(w)
  = 1*(1/3)/[(1/3)*1 +1/3*(1/2)]

  = 2/3

