The Probability of a Biased Coin: n Flips, m Heads

[markosheehan]

the probability of getting a head on flipping a biased coin is p. the coin is flipped n times producing a sequence containing m heads and (n-m) tails what is the probability of obtaining this sequence from n flips.
i can't understand the wording

 

[MarkFL]

I've moved this thread since our advanced forum is for calculus based stats.

A few things we need to observe:

The probability of getting heads is:

$$P(H)=p$$

Now, we know that it is certain that we will either get heads or tails, so we may state:

$$P(H)+P(T)=1\implies P(T)=1-P(H)=1-p$$

So, the probability of getting $m$ heads is:

$$P\left(H_m\right)=p^m$$

And the probability of getting $n-m$ tails is:

$$P\left(T_{n-m}\right)=(1-p)^{n-m}$$

Next we need to look at the number $N$ of ways to choose $m$ from $n$:

$$N={n \choose m}$$

Can you put all this together to find the requested probability?

 

[markosheehan]

when i put this all together i get (n ncr m)*p*(1-p)^n-m however at the back of the book it says the answer is p^m(1-p)^n-m

 

[MarkFL]

What I get is:

$$P(X)={n \choose m}p^m(1-p)^{n-m}$$

And this agrees with the binomial probability formula. :D

This is the probability of getting any sequence with $m$ heads, for any particular such sequence, then it would be:

$$P(X)=p^m(1-p)^{n-m}$$

 

[HallsofIvy]

when i put this all together i get (n ncr m)*p*(1-p)^n-m however at the back of the book it says the answer is p^m(1-p)^n-m

Was it possible that the problem asked for the probability of m heads in a row followed by n-m tails in a row? As MarkFl said, that probability if for any particular such sequence- "m heads in a row followed by n- m tails in a row" or "n- m tails in a row followed by m heads in a row" or "A head, then a tail, then a head, followed by m- 2 heads in a row, followed by n- m- 1 tails in a row", etc.