A cookie jar has 3 red marbles and 1 white marble. A shoebox has 1 red marble and 1 white marble. Three marbles are chosen at random without replacement from the cookie jar and placed in the shoebox. Then 2 marbles are chosen at random and without replacement from the shoebox. What is the probability that both marbles chosen from the shoebox are red?

The 3/4 refers to the idea there are 3 red out of 4 total. Then 2/3 is 2 red leftover out of 3 total remaining, and so on. The numerators and denominators count down (3,2,1 up top and 4,3,2 down below). This countdown happens because we do not replace the marbles selected.

Since case B is the opposite of case A, it means

P(B) = 1 - P(A) = 1 - (1/4) = 3/4

Or we can compute it like this

P(B) = 3*(3/4)*(2/3)*(1/2) = 3/4

The 3 out front is the number of ways to arrange the 2 red and 1 white where we cannot distinguish between the 2 reds. There are 3 spots to pick for the white marble, and the red marbles will go in the remaining blank spots.

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Let C represent the event "getting 2 red marbles from the shoebox after either case A or case B happens".

If case A happens, then the shoebox will have 4 red and 1 white.

P(C given A) = (4/5)*(3/4) = 3/5

If case B happens, then the shoebox will have 3 red and 2 white.

P(C given B) = (3/5)*(2/4) = 3/10

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To recap so far we have the following probabilities:

P(A) = 1/4
P(B) = 3/4
P(C given A) = 3/5
P(C given B) = 3/10

The ultimate goal is to calculate P(C)

Use the law of total probability to say the following:

P(C) = P(C and A) + P(C and B)

P(C) = P(C given A)*P(A) + P(C given B)*P(B)

P(C) = (3/5)*(1/4) + (3/10)*(3/4)

P(C) = 3/8