PROBABILITY Challenge!?

Question

An urn contains 3 red and 17 white balls. Balls are drawn without replacement. Find the probability that a red ball was drawn first, given that exactly one red ball was drawn on the 3rd, and 4th draws.

Update:

I have been trying to set the solution up using the Bayes' theorem, but not sure how it should be done.

MathMan TG · Accepted Answer

"given that exactly one red ball was drawn on the 3rd, and 4th draws" ...
I guess that means that just one of them was red.
(My first solution was based on both being red, but then it wouldn't need to say "exactly".
If that's the correct interpretation, it's a simpler calculation, with fewer cases, but the
basic method is the same.)

The relevant sequences are:

W W (followed by R W or W R)
W R (followed by R W or W R)
R R (followed by R W or W R)
R W (followed by R W or W R)

Let D = 20 * 19 * 18 * 17 = denominator of probability of all 4-ball sequences.
Since we looking for conditional probability, we won't need this shared denominator,
because we just care about the proportions of the numerators.

W W R W = 17 * 16 * 3 * 15 = 12240
W W W R = same, 12240
W R R W = 17 * 3 * 16 * 2 = 1632
W R W R = same 1632
R R W R = 3 * 2 * 17 * 1 = 102
R R R W = same 102
R W R W = 3 * 17 * 2 * 16 = 1632 
R W W R = same 1632

Total of all those is
12240 * 2 + 1632 * 4 + 102 * 2 = 31212

The ones with R first total 2 * 102 + 2 * 1632 = 3468

So given that exactly one of draws 3 and 4 is red, the probability that #1 was also red is
3468 / 31212 = 1/9.

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PROBABILITY Challenge!?

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