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

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The discussion centers on calculating the probability of obtaining a specific sequence of heads and tails when flipping a biased coin. The probability of getting heads is denoted as p, while tails is 1-p. The probability of observing m heads in n flips is derived using the binomial probability formula: P(X) = {n choose m} * p^m * (1-p)^(n-m). There is clarification on whether the problem specifically asks for a sequence of m heads followed by n-m tails, which is not the case, as the formula accounts for any arrangement of heads and tails. Ultimately, the correct interpretation aligns with the binomial distribution, confirming the probability calculation.
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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
 
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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?
 
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
 
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:

$$P(X)=p^m(1-p)^{n-m}$$
 
markosheehan said:
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.
 
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