Find the a posteriori probability? (Ch-4,Exercise-21, Probability, Random Variables and Stochastic Processes-Papoulis) The probability of heads of a random coin is a random variable p uniform in the interval (0, 1). (a) Find P{O.3 <= P <= O.7}. (b) The coin is tossed 10 times and heads shows 6 times. Find the a posteriori probability that p is between 0.3 and 0.7. a) Got P{O.3 <= P <= O.7} = 0.4 b) For the second part, the prob. of getting 6 heads in 10 tosses acc. to me should be (10 6) (1/2)^6 (1/2)^4, and suppose that is event B. P(A/B) = P(AB)/P(B). Here what would be P(AB)(A is event of interest). Am i doing something wrong, is my approach correct

For part b) you've incorrectly taken $\ p=\frac{1}{2}\ $. If $\ A\ $ is the event $\ \left\\{ 0.3\le p\le0.7\right\\}\ $, and $\ B\ $ the event that $6$ out $10$ tosses come up heads, then \begin{align} P\left(A\cap B\right) &= \int_{0.3}^{0.7} P\left(B\left\vert\, p=x\right.\right)dx\ ,\ \text{and}\\\ P\left(B\right) &= \int_0^1 P\left(B\left\vert\, p=x\right.\right)dx\ , \end{align} where $\ P\left(B\left\vert\, p=x\right.\right)={10\choose 6}x^6(1-x)^4\ $.