A bag contains 5 coins. Four of them are fair and one has heads on

both sides. You randomly pulled one coin from the bag and tossed it 5

times, heads turned up all five times. What is the probability that

you toss next time, heads turns up. (All this time you don't know you

were tossing a fair coin or not).

The probability of getting n consecutive heads is

P(n heads) = 1/5 * 1 + 4/5 * (1/2)^n,

Thus, the probability of getting a head on the n+1st roll given that

you have gotten heads on all n previous rolls is

P(n+1 heads | n heads) = P(n+1) / P(n)

= ( 1/5 * 1 + 4/5 * (1/2)^(n+1) ) / ( 1/5 * 1 + 4/5 * (1/2)^n ).

Multiplying numerator and denominator by 5* 2^(n-1) and recognizing 4

as 2^2 gives

P(n+1 heads | n heads) = (2^(n-1) + 1) / (2^(n-1) + 2)

put n=5 in above expression probability that you toss next time, heads turns up will be 17/18

## 1 comment:

Sorry have to disagree with the solution. First five tosses have already happened and the 6th toss is an independent event.

Hence the probability of getting a 6th head is simply

1/5*1 + 4/5*1/2 = 3/5.

If the question is, what is the probability of getting 6 consequetive heads, then your solution is right.

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