JEEOlympiad

A Polya urn starts with 2 red and 3 blue balls. A ball is drawn…

Question

A Polya urn starts with 2 red and 3 blue balls. A ball is drawn uniformly at random; its colour is noted, and then it is returned to the urn together with one additional ball of the SAME colour. This procedure is repeated for a total of 4 draws. Let the probability that exactly 2 of the 4 drawn balls are red be m/n in lowest terms. Find m+n.

✓ Verified answer: 44checked by our engine — not a guess

Step-by-step solution

Each ordered colour sequence of 4 draws with exactly 2 reds has its probability computed by multiplying sequential conditional probabilities, where after drawing a colour its count rises by 1 and the total rises by 1.

A key property of the Polya urn: for a fixed number of reds, every order of the same multiset of colours gives the SAME probability (the numerator factors of red counts {2,3} and blue counts {3,4} appear in some order, and the denominators are always 5,6,7,8).

So compute one order and multiply by the number of arrangements.

Take order R,R,B,B: P = (2/5)(3/6)(3/7)(4/8) = (2*3*3*4)/(5*6*7*8) = 72/1680 = 3/70.
Number of ways to place 2 reds among 4 positions = C(4,2) = 6.
P(exactly 2 red) = 6 * 3/70 = 18/70 = 9/35.
Thus m/n = 9/35, m+n = 44.

Final answer44

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