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Consider a channel N, a user is only allowed to transmit on this channel if the user detects the channel to be vacant. If the channel is detected to be vacant and it is actually vacant, then the rate of the user is $ｒ＝a$. If the channel is detected to be vacant but it is actually occupied, then the rate of the user is $ｒ＝b$. Define $d_n=0$ as the event that the channel is detected to be vacant, $d_n=1$ otherwise. Moreover, define the event $n=0$ as the event that the channel is actually vacant, $n=1$ otherwise. What is the expected value of the user?

I have two solutions, both of them make some sense to me:

$$E[r] = Pr(n=0|d_n=0)\times a + Pr(n=1|d_n=0)\times b \tag{1}$$ Eq.(1) makes sense to me since the given condition is the detection information, and the user only transmits when $d_n=0$ where two cases exist if $d_n=0$. $$E[r] = Pr(n=0,d_n=0)\times a + Pr(n=1,d_n=0)\times b + Pr(n=0,d_n=1)\times0 + Pr(n=1,d_n=1)\times0 \tag{2}$$ Eq.(2) also makes some sensing to me since we can consider $r$ in 4 joint cases, with the last two cases transmitting rate 0 (no transmission means rate 0).

Although $(1)\times Pr(d_n=0) = (2)$, they are different answers. Which one do you think is correct? How about (3) below? I personally think (3) doesn't make sense. $$E[r] = Pr(d_n=0|n=0)\times a + Pr(d_n=0|n=1)\times b \tag{3}$$

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Your first expression is $$ \begin{align} & a \mathbb P(n = 0\mid d_n = 0) + b \mathbb P(n = 1\mid d_n = 0)\\ &= \frac{a \mathbb P(n=0, d_n = 0)}{\mathbb P(d_n=0)} + \frac{b \mathbb P(n=1, d_n = 0)}{\mathbb P(d_n=0)}\\ &= \frac{\mathbb E[r1_{\lbrace d_n = 0 \rbrace} ]}{\mathbb P(d_n = 0)}\\ &= \mathbb E[r \mid d_n=0] \end{align} $$ which, evidently, is the expectation conditioned on the event that $d_n = 0$. Your second version is the unconditional one, and I assume the one you want.