Conditional Probability in a Survey anonymized using another random experiment. Suppose a survey is being carried out to check the number of people accessing pirated movies at the workplace. Since some employees might not be comfortable sharing this data, employees roll a fair die and if the outcome is 1, reply "yes", "no" if the outcome is 6 and reply truthfully if the outcome is anything else.
The probability that an employee answers yes in this experiment is 3/4. Then,
(a) what is the true probability that an employee accesses pirated movies
(b) how likely is it that an employee who answered yes accesses pirated movies?
(c) how likely is it that an employee who answered no accesses pirated movies?
I am thinking that 
P(Yes) = 2/3 = P(Yes|1)P(1) + P(yes|6)p(6) + P(yes|2..5)P(2..5) = 1\6 + 0 +  P(Yes|2..5).(4/6)
So (a) P(yes| 2..5) = 3/4.
for (b) and (c) I am not sure how to define the conditional probabilities though. 
 A: Let $Y$ and $A$ be binary variables ($Y,A \in \{0,1\})$ representing "Yes" and "Accessed". For example, $Y=1$ means the yes was given, while $A = 0$ means it was not accessed. Let $D$ represent the role of the die which we model as taking values $\{1,6,\omega\}$ where $\omega$ is any value besides $1$ and $6$. We have
\begin{align}
Y = \begin{cases}
1 & D=1 \\
0 & D=6 \\
A & D=\omega
\end{cases}
\end{align}
or more compactly using indicators
$$Y = 1\cdot 1_{\{D=1\}} + 0\cdot 1_{\{D=6\}} + A \cdot 1_{\{D=\omega\}}$$
that is,
$$ Y =  1_{\{D=1\}} + A \cdot 1_{\{D=\omega\}} $$
We also know that $D$ and $A$ are independent. We have
$$ E[Y] = E[1_{\{D=1\}}] + E[A] \cdot E[1_{\{D=\omega\}}] $$
that is,
$$ \frac34 = E[Y] = P(D=1) + E[A] P(D = \omega) = \frac16 + E[A] \frac{2}{3} $$
which gives $E[A] = 7/8$, the probability of accessing pirated movies.
We use properties of conditional expectation to compute $E(A|Y)$. This is a function of $Y$ and since $Y$ is binary, a general function of $Y$ can be written as
$$ E[A\mid Y] = \alpha Y + \beta.$$ 
By smoothing property $E[E[A\mid Y]] = E[A]$. Then, we have
$$
E[A] = \alpha + \beta E[Y] \implies \frac78 = \frac34  \alpha + \beta. \quad (*)
$$
We also have $E[Y E[A\mid Y]] = E[Y A]$. Since $Y E[A \mid Y] = Y(\alpha Y + \beta) = (\alpha + \beta ) Y$ (recall that $Y$ is binary), we obtain
$$
E[AY] = (\alpha + \beta) E[Y].
$$
But, using $A^2 = A$, we have $A Y = A (1_{\{D=1\}} + 1_{\{D=\omega\}}) = A(1-1_{\{D = 0\}})$ and 
\begin{align}
E[AY] &= E[A] \cdot E[1-1_{\{D = 0\}}] = E[A](1-P(D=0)) = \frac{7}{8} \frac{5}
6 = \frac{35}{48}
\end{align}
This gives us
$$
\frac{35}{48} = (\alpha + \beta) \frac{3}{4}. \quad (**)
$$
Putting (*) and (**), we can solve for $\alpha = 7/18$ and $\beta=7/12$. That is, we have
$$
P(A = 1 \mid Y) = E[A \mid Y] = \frac{7}{18} Y + \frac{7}{12},
$$ 
assuming that there was no numerical error!
A: Your approach and result for (a) are correct.
For (b),
\begin{eqnarray}
P(\text{accesses}\mid\text{yes})=P(\text{yes}\mid\text{accesses})\cdot\frac{P(\text{accesses})}{P(\text{yes})}\;,
\end{eqnarray}
and you can readily determine everything on the right-hand side.
