Question

Consider an unfair coin where P(tail)=1/3. The coin is flipped over and over until the tail is observed a second time. What is the probability that the tail is observed a second time on the 5th flip

Answer 1

P(Tail) = 1/3, Therefore P(Head) = 1 - 1/3 = 2/3

To get a second tails in the last flip means in our first 4 flips, 3 heads should appear and 1 tail should appear.

The possible options are HHHTT, or HHTHT or HTHHT or THHHT

= 4 * (2/3)³ * (1/3)² = 4 * (8/27) * (1/9) = 32 / 243

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This can also be done by the Binomial distribution. where nCr = (p)^r * (q) ^{n - r}.

Here we need in 4 flips, 3 heads and 1 tails. (The 5th is fixed as tails).

Therefore n = 4, r = 3, p = 2/3 and q = 1/3

Therefore The probability = 4C3 * (2/3)3 * (1/3) = 4 * 8/81 = 32/81, which is for four flips.

Therefor for 5 flips = 32/81 * 1/3 = 32 / 243

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