I could really use some help solving this problem. Here is the fact pattern:
There are four (4) balls in a bag from which to choose (A, B, C, & D). The picking is 100% random, so the probability of choosing any of the balls is equal (1 out of 4, or 25%). One (1) ball will be chosen during each draw. After each draw, the chosen ball is put back. There will be a total of six (6) draws, all of which are 100% independent of one another. Your job is to find the probability of choosing each ball BEFORE each draw.
** Even though these events are independent of one another, each next draw will take into account the probabilities of the previous draw(s). HINT: Probability of rolling a die and getting a three (3) is 1/6. The probability of rolling the die and getting a three (3) again is (1/6) x (1/6) = 1/36. And the probability of rolling the die again and getting another three (3) is (1/6) x (1/6) x (1/6) = 1/216.
On the first draw, ball A was chosen. On the second draw, ball A was chosen. On the third draw, ball B was chosen. On the fourth draw, ball C was chosen. On the fifth draw, ball A was chosen. On the sixth draw, ball D was chosen.
Question 1: BEFORE the first draw, what is the probability of choosing balls A through D?
Question 2 – AFTER the first draw, but BEFORE the second draw, what is the probability of choosing balls A through D in the second draw?
Question 3 – AFTER the second draw, but BEFORE the third draw, what is the probability of choosing balls A through D in the third draw?
Question 4 – AFTER the third draw, but BEFORE the fourth draw, what is the probability of choosing balls A through D in the fourth draw?
Question 5 – AFTER the fourth draw, but BEFORE the fifth draw, what is the probability of choosing balls A through D in the fifth draw?
Question 6 – AFTER the fifth draw, but BEFORE the sixth draw, what is the probability of choosing balls A through D in the sixth draw?
Question 7 – AFTER the sixth draw, but BEFORE the next draw, what is the probability of choosing balls A through D in the next draw?
It is not clear what is asked for, and why similar things are asked for so many times. Let our sample space consist of the $4^6$ words of length $6$ over the alphabet A, B, C, D.
I would guess that (for the second draw) the following (totally unnecessary) argument is wanted.
After the first draw, the total number of possible outcomes is $(1)(4^5)$, and they are all equally likely. The number of possible outcomes in which the second ball is, say, B is $(1)(1)(4^4)$. So the probability that the second ball is B, given that the first ball was A, is $\frac{(1)(1)(4^4)}{(1)(4^5)}$. Of course this simplifies to $\frac{1}{4}$.