Conditional Probability- hard Hi guys can someone please help me with the following question:

I have come up with the following:
a. P(\$20) = (2/3)
b. Since Tom states that Jenny used a $10 note we have to assume he didnt correctly identify it and by assuming independence between these two events; P(Jenny used 20) = 0.1*(2/3) = 1/15.
c. Jenny must have correctly identified the 20 note thus P(Jenny used 20) = 0.8*(1/15)=4/75.
 A: For (2), essentially you need to deal with both cases which make it a USD 20 note.
Either it was a 20 (2/3 probability like you stated), and he guessed incorrectly that it was a 10. Like you wrote, that hs a 1/15 probability of occurring. The alternative, is that it was a 10, and he guessed correctly that this was the case. This has a 0.9*(1/3) probability of occurring. Now you just need to likelyhood of the foremer out of the sample P(20|tom says 10) = (0.1*(2/3))/(0.1*(2/3)+0.9*(1/3))
A: I agree 100% with @Moshe (+1) on b.
Regarding C.  I am assuming that Jenny’s and Tom’s percentages are conditionally independent of one another.
$$P(20|TomSaid10 and JennySaid20) = \frac{P(TomSaid10 and JennySaid20|20)*P(20)}{P(TomSaid10 and JennySaid20)}$$
$$P(20|TomSaid10 and JennySaid20) = \frac{0.1*0.8*\frac{2}{3}}{P(TomSaid10 and JennySaid20|20)*P(20)+ P(TomSaid10 and JennySaid20|10)*P(10)}$$
$$P(20|TomSaid20 and JennySaid10) = \frac{0.08*\frac{2}{3}}{0.1*0.8*\frac{2}{3}+0.9*0.3*\frac{1}{3}}$$
$P(20|TomSaid20 and JennySaid10) = \frac{16}{16+27}$$=\frac{16}{43}\approx0.37$
