# Given confidence interalvals $p_u, p_o$ and a random variable $S_2$ find $P_p(p < P_u(S_2)), P_p(p > P_o(S_2))$ [closed]

I am given a random variable distributed binomially: $$S_2$$ ~ $$Bin(2, p)$$ and the following intervals:

• $$[p_u(0), p_o(0)] = [0, 1 / 4]$$
• $$[p_u(1), p_o(1)] = [1 / 4, 3 / 4]$$
• $$[p_u(2), p_o(2)] = [1 / 2, 1]$$

How can I find $$P_p(p < P_u(S_2)), P_p(p > P_o(S_2))$$ if

1. $$0 < p < 1 / 4$$
2. $$1 / 4 < p < 1 / 2$$
3. $$1 / 2 < p < 3 / 4$$
4. $$3 / 4 < p < 1$$

Any suggestions are highly appreciated!

• What do things like $p_u(2)$ and $P_p(p < P_u(S_2))$ mean? Sep 14 at 23:26

$$P_u(S_2)$$ is a discrete random variable with atoms $$\{0,1/4,1/2\}$$ and probabilities $$[(1-p)^2,2p(1-p),p^2]$$. Then
• $$0) $$\qquad\mathbf{P}_p\left(P_u(S_2)>p\right)=\mathbf{P}_p\left(S_2\geq1\right)=2p(1-p)+p^2$$
• $$1/4) $$\qquad\mathbf{P}_p\left(P_u(S_2)>p\right)=\mathbf{P}_p\left(S_2=2\right)=p^2$$
• $$1/2) $$\qquad\mathbf{P}_p\left(P_u(S_2)>p\right)=0$$.
$$P_o(S_2)$$ is a discrete RV with atoms $$\{1/4,3/4,1\}$$ and probabilities $$[(1-p)^2,2p(1-p),p^2]$$. Then
• $$0) $$\qquad\mathbf{P}_p\left(P_o(S_2)
• $$1/4) $$\qquad\mathbf{P}_p\left(P_o(S_2)
• $$3/4) $$\qquad\mathbf{P}_p\left(P_o(S_2).