On the lecture I am taking, a example was given to do hypothesis test on population with unknown variance. And the solution is first test the samples' deviation based on F-test and favor the Null hypotheis of equal variance and perform t-test for equal variance.
But (I assume) the the null hypothesis of the followed t-test would be $$ \mathcal{H_0} : \{\mu_{diff} = 0 \mid accepting\ hypothesis\ that\ variances\ are\ equal\}$$
And then the Type I error would be the probability of : $$ P(Type\ I\ error\mid equal\ variances) = \\P(reject\ \mathcal{H}_0\mid \mathcal{H}_0:\{\mu_{diff} = 0 \mid accepting\ hypothesis\ that\ variances\ are\ equal\}) $$
Which is not the same as the common hypothesis test which probability of Type I error given null hypothesis is true is happen to be $\alpha$ that we set.
Given the consideration of above nonsense, the should the total probability of Type I error of this experiment be $$ P(Type\ I\ error) =\\ P(Type\ I\ error\mid equal\ variances)\times\ P(samples\ variances\ are\ equal)\\ +P(Type\ I\ error\mid unequal\ variances)\times\ P(samples\ variances\ are\ unequal) $$
If correct, what are the $P(samples\ variances\ are\ equal)\ and\ P(samples\ variances\ are\ unequal)$?