# Selecting a coin from from collection of unfair coins toss n times mean and probability of no tails

We have a collection of unfair coins, for which the probability of getting head after tossing them is uniformly distributed in the range (0, 1). We randomly select a coin from this collection and toss it n times.

(a) What is the expected number of heads?

(b) Compute the probability of having no tails in these trials.

For part a) since its binomial I think it should be np but if we gonna use expectation of uniform on (0,1) which is 0.5 I don't know how to put it in

For b) i think it should be $1-{n \choose n}p^nq^{n-n}=1-p^n$

You have a free variable $p$ in your answer, but there's no such variable in the problem, so something must be wrong.

For (a), note that by symmetry the expected number of heads is equal to the expected number of tails, and these must add up to $n$, so they're both $n/2$.

For (b), you need to integrate over the distribution. If $p\in(0,1)$ is the probability of tossing heads, the probability of no tails (and thus all heads) is $p^n$. Integrating over the uniform distribution yields the probability

$$\int_0^1p^n\mathrm dp=\frac1{n+1}\;.$$

Note that this makes sense for $n=1$, where the probability must be $\frac12$ (again by symmetry).

• These are unfair coins why is the expected number of heads is equal to the expected number of tails? Is it because of it says that "for which the probability of getting head after tossing them is uniformly distributed in the range (0, 1). " i know that expecation of such uniform is 1/2. I did not understand part b) at all. Can you please explain it in more details? – Mathkid113 Feb 22 '16 at 1:11
• @edipo: For (a), there's no need to form the expectation. All you need is the symmetry that the uniform distribution has the same density at $p$ as at $1-p$, so heads and tails play symmetric roles. For (b), I'm not sure where to begin the explanation -- are you generally aware of how to compute probabilities from parametrized families of distributions given a distribution of the parameters? – joriki Feb 22 '16 at 1:54
• No im anot aware of how to compute probabilities from parametrized families of distributions given a distribution of the parameters. – Mathkid113 Feb 22 '16 at 3:05
• @Mathkid113: Then this is probably not a suitable exercise for you, since that's what's required in part (b). You need to calculate the desired probability as a function of the distribution parameter $p$ and then take the expected value with respect to its given distribution. In the present case, taking the expected value with respect to the uniform distribution over $(0,1)$ is done by integrating over $(0,1)$. – joriki Feb 22 '16 at 7:27