Three coins are tossed. If all three show tails, $H_0$ is rejected, otherwise, $H_0$ is not rejected. Using this procedure, what is the probability of committing a Type II error?
I have that $1 - [P(\text{get tail 1st toss}|H_1\text{ true})*P(\text{get tail 2nd toss}|H_1\text{ true})*P(\text{get tail 3rd toss}|H_1\text{ true})]$
But I am not sure of this $1 - ((1/8)(1/8)(1/8))$
Something is missing in the text. Assuming that the null hypothesis is
$$H_0: \theta_0\leq\frac{1}{2}$$
Against
$$H_1: \theta_1 >\frac{1}{2}$$
Where $\theta$ is the probability to get tail, type II error is
$$\beta=1-\mathbb{P}[\text{TTT}|\theta_1]=1-\theta_1^3$$
Observe that $\beta$ decreases when $\theta_1$ increases in the Interval $\left(\frac{1}{2};1\right]$ as it must be.