Bayes Theorem

can anyone help me see where i went wrong?

If he fails overall, there are three options, (pass, fail), (fail, pass), (fail, fail), this eliminates (pass, pass)
the one we want is (pass, fail)
The probability of this with no other evidence is 0.42, as you calculated
With the evidence, it becomes $\frac{0.42}{0.42 + 0.3(0.2) + 0.3(0.8)} = \frac{0.42}{0.42 + 0.3} = 0.583333$

ok so how do you do this one?

are you familiar with Bayesian probability?
$P(S|W') = \frac{P(W'|S)P(S)}{P(W')}$
$\therefore P(S|W') = \frac{(0.45)(0.9)}{1 - 0.53}$
$\therefore P(S|W') = 0.86(2 dp)$
sorry about the wait, read loses as wins the first time