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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

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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\ $.

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