Probability that second of $20$ watches is broken given $2$ are broken and after discarding first based on $80\%$-correct expert advice? The problem goes this way. There are $20$ watches, $2$ of which are broken. And there is an expert who can say if the watch is broken. But his verdict is only correct with probability $0.8$ - in both cases, if the watch is broken and if it is not. So, then I choose one watch, show it to the expert and he says: it's broken. I choose then another watch of these $20$. What is the probability that this second watch is broken? Thanks in advance.
PS. I dont really understand how this information can help, but let it be. My math background: PhD, but it is 20 years ago, so I'm not an active mathematician anymore. The source of the problem: a student who i know. What I tried: naturally, Bayes theorem. But I still cannot solve it.
PPS. I found the solution. It goes without Bayes but with the law of total probability. Its rather technical.

Let $B_1$ denote the event that the first watch is broken and let $B_2$ denote the event that the second watch is broken.
To be found is $P(B_2)$ and we have:$$P(B_2)=P(B_2|B_1)P(B_1)+P(B_2|B_1^c)P(B_1^c)=\frac1{19}P(B_1)+\frac2{19}(1-P(B_1))=$$$$\frac2{19}-\frac1{19}P(B_1)$$
So from here it is enough to find $P(B_1)$.
Event $B_1$ can also be recognized as  the event that the expert is correct in his judgement so that $P(B_1)=0.8=\frac45$.
So our final outcome is:$$P(B_2)=\frac2{19}-\frac1{19}\frac45=\frac6{95}$$

Edit:
There is a flaw in the approach above.
The statement that event $B_1$ coincides with the event of correctness of the expert is false. Also see the comments on this question.
An approach without this flaw is:
Let $T$ denote the event that the expert is correct in his judgment.
Then $T$ and $B_1$ are independent which follows from $P(T|B_1)=P(T|B_1^c)$. Moreover it is evident that $T$ and $B_2$ are independent.
The event that the expert tells us that the first watch is broke can now be recognized as: $TB_1\cup T^cB_1^c$
and to be found is:$$P(B_2|TB_1\cup T^cB_1^c)=\frac{P(TB_1B_2)+P(T^cB_1^cB_2)}{P(TB_1)+P(T^cB_1^c)}=$$$$\frac{P(T)P(B_1)P(B_2|B_1)+P(T^c)P(B_1^c)P(B_2|B_1^c)}{P(T)P(B_1)+P(T^c)P(B_1^c)}=$$$$\frac{\frac45\frac1{10}\frac1{19}+\frac15\frac9{10}\frac2{19}}{\frac45\frac1{10}+\frac15\frac9{10}}=\frac{22}{247}\sim0.089$$
