Question

A certain country has a careless mint. One coin in every 801 manufactured has heads on both sides. The remaining coins have heads on one side, tails on the other, as usual. A randomly selected coin made at the mint is tossed five times and falls heads each time. What is the probability that this same coin will fall heads again if it is tossed for a sixth time?

Answer 1

solution:-

Let

F = fair coin

D = double headed coin

H = comes up heads

Thus,

P(5 consecutive H) = P(D) P(5 consecutive H/D) + P(F) P(5 consecutive H/F)

= (1/801)*(1)^5 + (800/801)*(1/2)^5

= 0.032459426

Now,

P(D/5 consecutive H) = P(D) P(5 consecutive H/D) / P(5 consecutive H) = (1/801)*(1)^5 / 0.032459426 = 0.038461538

Thus,

P(F/5 consecutive H) = 1 - P(D/5 consecutive H) = 1 - 0.038461538 = 0.961538462

Thus,

P(6th is H) = P(D/5 consecutive H) p(H/D) + P(F/5 consecutive H) P(P(H/F)

= 0.038461538*1 + 0.961538462*(1/2)

= 0.519230769 [ANS]