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.
 A: 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:$$P(B_2|TB_1\cup T^cB_1^c)=\frac{P(TB_1B_2)+P(T^cB_1^cB_2)}{P(TB_1)+P(T^cB_1^c)}=$$$$\frac{P(T)P(B_1)P(B_2|B_1)+P(T^c)P(B_1^c)P(B_2|B_1^c)}{P(T)P(B_1)+P(T^c)P(B_1^c)}=$$$$\frac{\frac45\frac1{10}\frac1{19}+\frac15\frac9{10}\frac2{19}}{\frac45\frac1{10}+\frac15\frac9{10}}=\frac{22}{247}\sim0.089$$
A: Here is a version of the solution that does not uses the set fromalism (which is not at all denied!) but uses an intuitive approach.
(I must say it is a corrected version : the first one  was erroneous)
This 2-steps problem can be considered as equivalent to the choice, in the second step, between

*

*an urn $U_1$ containing 1 Broken (Black) and 18 Working (White) watches [for which the probability of drawing a B watch is $p_1=1/19$]


*an urn $U_2$ containing 2 B and 17 W balls [for which the probability of drawing a B watch is $p_2=2/19$]
Let us first consider the case where we have no information about what has happened in the first step ; in this context, the probability of having a B watch in the secound step would be :
$$\underbrace{\dfrac{2}{20}}_{w_1} \times \underbrace{\dfrac{1}{19}}_{p_1} \ \  + \ \ \ \underbrace{\dfrac{18}{20}}_{w_2} \times \underbrace{\dfrac{2}{19}}_{p_2}\tag{1}$$
The "weights" $w_k$ sum up to $1$.
Now, how are modified these weights with the verdict of the expert ? They become resp.
$$w_1=\dfrac{2}{20} \times \dfrac{4}{5} \ \text{and} \ w_2=\dfrac{18}{20} \times \dfrac{1}{5}$$
We must take care of the fact that the sum $w_1$ and $w_2$ is not $1$ ; we have to normalize these weights by dividing them by their sum $\dfrac{26}{100}$:
$$w'_1=\dfrac{8}{26} \ \text{and} \ w'_2=\dfrac{18}{26}$$
Instead of (1), we have now:
$$\underbrace{\dfrac{8}{26}}_{w'_1} \times \underbrace{\dfrac{1}{19}}_{p_1} \ \  + \ \ \ \underbrace{\dfrac{18}{26}}_{w'_2} \times \underbrace{\dfrac{2}{19}}_{p_2}=\dfrac{44}{494}=\dfrac{22}{247} \approx 0.0891$$
A: The correct solution is as this.
Let's $A_1$, $A_2$ be the events "first (second) watch is NOT broken".
Let's $B_1$ be the event "expert says the watch is broken".
Then we know $P(B_1|A_1^c)=P(B_1^c|A_1)=0.8$.
What we are looking for is $P(A_2|B_1)$. So, we need $P(B_1)$ and $P(A_2\cap B_1)$.
It is: $P(B_1)=P(B_1|A_1^c)P(A_1^c)+P(B_1^c|A_1)P(A_1)=13/50$.
$P(A_2\cap B_1)$ is not so easy:
$P(B_1∩A_2)=P(B_1∩A_1∩A_2)+P(B1∩A_1^c ∩ A_2) =P(B_1∣A_1\cap A_2)P(A_1∩A_2)+P(B_1∣A_1^c\cap A_2)P(A_1^c ∩ A_2)=
P ( B_1  ∣ A_1  ) P(A_1∩A_2)   + P ( B_1  ∣ A_1^c  ) P(A_1^c ∩ A_2)$ and now we know all the parts.
In the end we have $P(A_2\cap B_1)=9/38$ and $P(A_2|B_1)=\frac{9}{38}\frac{50}{13}=\frac{225}{247}$.
But as the question was about probability that the second watch is broken, the answer for this question is $P(A_2^c|B_1)=1-P(A_2|B_1)=\frac{22}{247}$.
A: I did use Bayes' Theorem for this (along with the Law of Total Probability.)
Let $B_i$ be the event that the $i^{\text{th}}$ watch is Broken, and let $R_i$ be the event that the expert Reports that the $i^{\text{th}}$ watch is broken.  We want to find
$$
P(B_2 \vert R_1) = \frac{P(R_1\vert B_2)P(B_2)}{P(R_1)}
$$
$$
= \frac{P(R_1\vert B_2)P(B_2)}{P(R_1\vert B_2)P(B_2) + {P(R_1\vert B_2^c)P(B_2^c)}}.
$$
Certainly
$$
P(B_2) = \left(\frac{2}{20}\right)
$$
and
$$
P(B_2^c) = \left(\frac{18}{20}\right).
$$
It remains to calculate $P(R_1\vert B_2)$ and $P(R_1\vert B_2^c)$.
Part I: To find $P(R_1\vert B_2)$, let us simply assume that $B_2$ is true.  So for now, we assume $B_2$, and calculate $P(R_1)$ under this assumption.  Since the second watch is broken, we know that of the $19$ other watches, exactly $1$ is broken.  Now
$$
P(R_1) = P(R_1 \vert B_1) P(B_1) + P(R_1\vert B_1^c) P(B_1^c)
$$
$$
= (0.8) \left(\frac{1}{19}\right) + (0.2) \left(\frac{18}{19}\right)
$$
$$
= \frac{4.4}{19}.
$$
But this was calculated under the condition that $B_2$ is true.  Thus we have really shown that
$$
P(R_1\vert B_2) = \frac{4.4}{19}.
$$
Part II: To calculate $P(R_1\vert B_2^c)$, we simply assume that $B_2$ is false.
Under this assumption,
$$
P(R_1) = P(R_1 \vert B_1) P(B_1) + P(R_1\vert B_1^c) P(B_1^c)
$$
$$
= (0.8) \left(\frac{2}{19}\right) + (0.2) \left(\frac{17}{19}\right)
$$
$$
= \frac{5}{19}.
$$
Because this was calculated under the assumption that $B_2$ is false, we have really shown that
$$
P(R_1\vert B_2^c) = \frac{5}{19}.
$$
Collecting our results together, we find that
$$
P(B_2\vert R_1) = \frac{\frac{8.8}{380}}{\frac{8.8}{380} + \frac{90}{380}} = \frac{22}{247}.
$$
A: Let $B_1$ be the event that the first watch is broken,
$B_2$ be the event that the second watch is broken, and
$C_k$ be the event that the expert is correct about the $k$th watch.
Since the expert has an $80\%$ chance of being correct regardless of the condition of the watch, for each $k\in\mathbb N,\;P(C_k)=0.8.$ Thus, event $C_1$ is independent of event $B_1,$ and vice versa.

The required probability is \begin{align}&P(B_2|C_1B_1\cup\overline C_1\overline B_1)\\=&\frac{P(C_1B_1B_2,\overline C_1\overline B_1B_2)}{P(C_1B_1B_2,C_1B_1\overline B_2,\overline C_1\overline B_1B_2,\overline C_1\overline B_1\overline B_2)}\\=&\frac{0.8\times\frac2{20}\times\frac1{19}+0.2\times\frac{18}{20}\times\frac2{19}}{0.8\times\frac2{20}\times\frac1{19}+0.8\times\frac2{20}\times\frac{18}{19}+0.2\times\frac{18}{20}\times\frac2{19}+0.2\times\frac{18}{20}\times\frac{17}{19}}\\=&\frac{22}{247}.\end{align}
