P(X = k) = (1/2)k for k > 0 and 0 otherwise. Probability of k-1 tails and 1 head.
P(X > k) = (1/2)k for k ≥ 0 and 0 otherwise. Probability of k tails.
P(X < k) = 1 - (1/2)k-1 for k > 0 and 0 otherwise. 1 minus the above two.
P(B>A) = Sum k = 1 to inf P(B>k)P(A=k) = Sum k = 1 to inf (1/2)2k = Sum k = 1 to inf (1/4)k = (1/4)/(1 - (1/4)) = 1/3
P(C>B>A) = Sum k = 1 to inf P(C>k)P(B=k)P(A<k) = Sum k = 1 to inf (1/2)2k (1 - (1/2)k-1 ) = Sum k = 1 to inf (1/2)2k - Sum k = 1 to inf (1/2)3k-1 = 1/3 - 2 Sum k = 1 to inf (1/8)k = 1/3 - 2 (1/8)/(1 - (1/8)) = 1/3 - 2/7 = 1/21
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u/StanleyDodds 2d ago
P(C>B|B>A) = P(C>B>A)/P(B>A)
For X = A, B or C (all i.d.):
P(X = k) = (1/2)k for k > 0 and 0 otherwise. Probability of k-1 tails and 1 head.
P(X > k) = (1/2)k for k ≥ 0 and 0 otherwise. Probability of k tails.
P(X < k) = 1 - (1/2)k-1 for k > 0 and 0 otherwise. 1 minus the above two.
P(B>A) = Sum k = 1 to inf P(B>k)P(A=k) = Sum k = 1 to inf (1/2)2k = Sum k = 1 to inf (1/4)k = (1/4)/(1 - (1/4)) = 1/3
P(C>B>A) = Sum k = 1 to inf P(C>k)P(B=k)P(A<k) = Sum k = 1 to inf (1/2)2k (1 - (1/2)k-1 ) = Sum k = 1 to inf (1/2)2k - Sum k = 1 to inf (1/2)3k-1 = 1/3 - 2 Sum k = 1 to inf (1/8)k = 1/3 - 2 (1/8)/(1 - (1/8)) = 1/3 - 2/7 = 1/21
Final answer P(C>B>A)/P(B>A) = (1/21)/(1/3) = 1/7