A cookie jar, a shoebox and a lot of time on our hands-Probability?
A cookie jar has 3 red marbles and 1 white marble.
A shoebox has 1 red marble and 1 white marble.
3 marbles are chosen at random without replacement from the cookie jar and placed in the shoebox.
THEN, 2 marbles are chosen at random without replacement from the shoebox.
What is the probability that both marbles chosen from the shoebox are red?
Answer: 3/8
How to solve?
Condition on how many red marbles were chosen from the cookie jar. This could be 0, 1, 2 or 3.
Define some events, for instance
J0 = 0 red marbles picked from the cookie jar
J1 = 1 red marble picked from the cookie jar
and similarly for J2 and J3.
Let A be the event you're interested in, that both marbles chosen from the shoebox are red. You want P(A). So as I said, condition on J0, J1, J2 and J3.
P(A) = P(A|J0) P(J0) + P(A|J1) P(J1) + P(A|J2) P(J2) + P(A|J3) P(J3)
(Hint: I've always found that defining the events and then writing down the notation for what probabilities you're given is 90% of the problem. The rest is almost mechanical).
Now you just have to work out each of those things on the right. First of all, if you didn't get at least 2 red marbles on the first draw (J2 or J3), then you can't possibly get 2 red marbles on the second draw. So P(A|J0) = P(A|J1) = 0.
What's P(J2)? The probability of getting exactly 2 red marbles out of 3 marbles chosen. That's (3C2)(1C1)/(4C3) because there are 3C2 pairs of red marbles, 1C1 way to choose the one marble, and 4C3 ways of choosing 3 marbles out of the cookie jar.
What's P(A|J2)? This is the probability of drawing both red marbles when the shoebox contains 2 red marbles and 1 white marble. So that's (2C2)/(3C2). Do you see why? 2C2 pairs of red marbles (only 1 of course) and 3C2 pairs altogether.
Similar reasoning for P(J3) and P(A|J3).
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