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Suppose that you toss a coin until, for example, 5 heads occur. What would be the probability that it takes 8 tosses?

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closed as off-topic by Thomas Shelby, NCh, Robert Z, Cesareo, Gibbs Feb 15 at 10:33

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    $\begingroup$ You want to read about the negative binomial distiribution. $\endgroup$ – Arthur Feb 11 at 11:29
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    $\begingroup$ You need the $8^{th}$ toss to be $H$ and you need exactly $4$ of the previous $7$ to be Heads as well, so... $\endgroup$ – lulu Feb 11 at 11:36
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[H or T ][H or T][H or T][H or T][H or T][H ot T][H or T][H]

First assume that the coin is fair so P(H) or P(T) = 1/2

The last tosses need to be H as we want it to occur in 8 tosses

So now we find all the combinations of 4 heads and 3 tails in the tosses

n choose k

(7 choose 4 ) 7!/4!3! = 35

All possible tosses combinations = 2^8 =256

P( H = 5 in 8 tosses ) = 35/256 = 0.1367

But if the coin is not fair YEAH as arthur mentioned https://en.wikipedia.org/wiki/Negative_binomial_distribution

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  • $\begingroup$ That's if you know you're doing 8 tosses. The way i read the question, we're instead tossing until we get the fifth head. That's a different question. Doesn't matter whether the coin is fair. $\endgroup$ – Arthur Feb 11 at 11:48
  • $\begingroup$ You are right!! I am totally wrong $\endgroup$ – EconBoy Feb 11 at 14:39
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The last toss must result in a head.

Further of the former $7$ tosses exactly $4$ must result in heads.

There are $\binom74$ possibilities and if the coin is fair then each of them has probability $2^{-8}$ to occur.

So the probability that exactly $8$ tosses are needed equals:$$\binom742^{-8}$$

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  • $\begingroup$ That's if you know you're doing 8 tosses. The way i read the question, we're instead tossing until we get the fifth head. That's a different question. $\endgroup$ – Arthur Feb 11 at 11:47

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