A coin with uniform probability

Question

Assume that we have a coin that the probability of head in this coin is $p^2$, which $p$ itself is generated by a uniform distribution on $[0,1]$.

If we flip the coin and see head, then what is the probability of $p<\frac{1}{2}$?

I personally could compute the probability of head, which is:

$\int_{0}^{1} P(\text{head comes}\mid P=p).f_{P} (p ) d_{p}= \int_{0}^{1} p^2 d_{p} = \frac{1}{3}$

But I can not find the probability of the question...

Answer 1

Let me try. We have: $P(H / p=p) = p^2 $

By Bayes Theorem: $$P(p=p / H) = \frac{P(H/p=p) P(p=p)}{P(H)}$$ Considering the probability P(p=p) as uniform, we have: $$P(p=p / H) = \frac{p^2 . 1}{\int{P(H / p=p)P(p=p)dp}}$$

$$P(p=p / H) = \frac{p^2 . 1}{\int{p^2dp}}$$

$$P(p=p / H) = 3p^2$$

So, if I'm not mistaken: $$P(p \le \frac{1}{2}) = \int_0^\frac{1}{2}{3p^2dp} $$

$$P(p \le \frac{1}{2}) = \frac{1}{8} $$

Answer 2

let

H: be event that coin is head

L : event that $p<\frac{1}{2}$

This we will use later to compute conditional probability $P(p=t|L) = \frac{P(p=t,L)}{P(L)} = \frac{P(p=t)}{P(L)} = 2 $ for $t \in [0, 1/2]$.

what you want is following

$$P(L|H) = \frac{P(H|L)P(L)}{p(H)}$$

so calculate $$P(H|L) = \int_{0}^{1/2} P(H|L,p =t) . p(t|L) dt = \int_0^{1/2} t^2.2.dt = 2. (t^3/3)|_{t=0}^{t = 1/2} = \frac{1}{12}$$

and I am using from what you have computed $p(H) = 1/3$ so answer is following.

$$ = \frac{1/12 . 1/2}{1/3} = \frac{1}{8}$$