I came across this question:
Three contestants are each given the same question. They answer independently of each other. The probability of the answer being correct is $p$. The difficulty of the question is a random variable $\sim \operatorname{UNIF} (0,1]$
Before answering the questions, I introduce a few random variables:
$X$ is the difficulty of the question: $X \sim \operatorname{UNIF} (0, 1]$
$Y$ is the number of correct answers: $Y \sim \operatorname{BIN} (3, p)$
Intuitively, I suppose $p$ depends on $X$. The actual dependence is not specified, but I assume it follows some kind of inverse relationship to $X$, as a lower difficulty will produce a higher probability of a correct answer. This means that for a given $X = x$, it follows that $p \not = x$.
I thought that a possible function that could relate $p$ to $x$ is $p = 1-x$, because at minimum difficulty $x = 0$ the probability that the answer is right should be at its maximum, $p = 1$, and at maximum difficulty $x = 1$, the probability that the answer is right should be at its minimum, $p = 0$. However, the question does not provide this.
The questions are the following:
(a) Find $\operatorname{P} (Y = 2)$.
So if $Y \sim \operatorname{BIN} (3, p)$, then
$$\operatorname{P}(Y = 2) = \binom{3}{2} p^2(1-p) = 3p^2 - 3p^3$$
For a general function $p = g(x)$:
$$\operatorname{P}(Y = 2) = \cdots = 3p^2 - 3p^3 = 3(g(x))^2 - 3(g(x))^3$$
I use the Law of Total Probability:
$$\begin{align} \operatorname{P}(Y = 2) & = \operatorname{E}( \operatorname{P} (Y = 2 \vert X = x)) \\ & = \operatorname{E}(3(g(x))^2 - 3(g(x))^3) \\ & = 3\operatorname{E}((g(x))^2) - 3\operatorname{E}((g(x))^3) \\ & = 3 \int_0^1 (g(x))^2 dx - 3 \int_0^1 (g(x))^3 dx \end{align}$$
Since I don't know $g(x)$ I can't really compute a value for $\operatorname{P}(Y=2)$. For $p = 1-x$, I get $\operatorname{P}(Y=2) = \frac{1}{4}$
(b)Find $\operatorname{E} (Y)$.
So, again, $Y \sim \operatorname{BIN} (3, p)$. Then:
$$\operatorname{E}(Y) = np = 3p$$
Again, for a general function $p = g(x)$ I use the Law of Total Expectation:
$$\begin{align} \operatorname{E}(Y) & = \operatorname{E}(\operatorname{E}(Y \vert X = x)) \\ & = \operatorname{E}(3g(x)) \\ & = 3 \operatorname{E}(g(x))\\ & = 3 \int_0^1 g(x) dx \\ & = 3(G(1) - G(0))\\ \end{align}$$
Again, I can't really derive a numerical value for $\operatorname{E}(Y)$ because I don't know $g(x)$ or $G(x)$ (note that $\frac{d(G(x))}{dx} = g(x)$).
If I use $p = 1-x$, I get $\operatorname{E}(Y) = \frac{3}{2}$
I want to think that the question is vague and that I need $g(x)$ to be able to solve it, but maybe there's another way to derive $\operatorname{P}(Y=2)$ and $\operatorname{E}(Y)$.
Is there?