Nose picking probability 
A questionnaire is carried out to find the percentage of men that pick
  their nose. Since most men are embarrassed to answer this questions,
  they are told to do the following. Secretly roll a dice. If the number
  is 1 always say I do not pick my nose regardless of what the truth is,
  if the number is 2 always say I pick my nose regardless of what the
  truth is, if the number is 3, 4, 5 or 6 then tell the actual truth.
  This way, it is not possible to ever be sure which men pick their nose
  or not. If we assume that 2/3 of men answered I pick my nose, what is
  the probability that a man picks their nose? If instead we assume that
  fraction of men that pick their nose is 3/4, what is the probability
  that a man picks his nose given that he answered I do pick my nose.

So far I have that (for part a). $P(t|y) = \frac{P(y|t)\cdot P(t)}{P(y)}$ where $y=$ yes they pick nose. $t = $ telling the truth... However, I can't find $P(y|t)$ Thanks.
 A: Hint:
$$
P(\text{yes})=\overbrace{\frac23P(\text{pick})}^{\text{rolled $3$-$6$ and pick}}+\overbrace{\ \ \ \ \ \frac16\ \ \ \ \ }^{\text{rolled $2$}}
$$

There are two disjoint cases where people say yes and are telling the truth:


*

*roll $3$-$6$ and pick their nose $=\frac23P(\text{pick})$

*roll $2$ and pick their nose $=\frac16P(\text{pick})$


Therefore,
$$
\begin{align}
P(\text{yes }\land\text{ truth})
&=\frac23P(\text{pick})+\frac16P(\text{pick})\\
&=\frac56P(\text{pick})
\end{align}
$$
There are three cases where people tell the truth:


*

*roll $3$-$6$ $=\frac23$

*roll $2$ and pick their nose $=\frac16P(\text{pick})$

*roll $1$ and don't pick their nose $=\frac16(1-P(\text{pick}))$


Therefore,
$$
P(\text{truth})=\frac56
$$
This makes sense since the only time they wouldn't tell the truth is when they rolled $1$ or $2$ depending on whether they did or did not pick their nose.
Thus,
$$
P(\text{yes}\mid\text{truth})=P(\text{pick})
$$
A: 
If we assume that 2/3 of men answered I pick my nose, what is the probability that a man picks their nose?
If instead we assume that fraction of men that pick their nose is 3/4, what is the probability that a man picks his nose given that he answered I do pick my nose.

First, be clear on the events you are measuring.   Labelling one event "telling the truth" is a misdirection - it is not something you are directly given.   You are comparing what they say and what they do.
You want your $y$ to be the event of saying "I do", and $t$ to be the event of nose picking.
The probability of someone saying yes given that they do pick their nose is $5/6$, by reason that they answer yes on a roll of $\{2, 3,4,5,6\}$.
$$\mathsf P(y\mid t) =\frac 5 6$$
So for (a)
$$\mathsf P(y) = \mathsf P(y\mid t)~\mathsf P(t)+\mathsf P(y\mid \neg t)(1-\mathsf P(t)) \\ \tfrac 2 3 = \tfrac 5 6 \mathsf P(t)+ \tfrac 2 6 (1-\mathsf P(t)) \\ \vdots$$
