The problem is as such below
I will try solve (i) with total probability:
$A -$ car fail test
$B_1-$ emit excessive pollutants that fail
$B_2-$ 17% not emit excessive pollutants that fail
Then $P(A)=P(B_1)P(A|B_1)+P(B_2)P(A|B_2)=(.25)(.99)+(.25)(.17)=0.29$
and (ii): $P(B_1|A)=\frac{P(B_1)P(A|B_1)}{P(A)}=\frac{(.25)(.99)}{0.29}=0.85$(2 decimal places)
My first answer of 0.29 seems about right as if 99% of some population makes up close to 25% then the remaining 17% of those that fail but shouldn't fail would make the probability go a bit higher. And if a car were to fail the test than it is likely that it is from $B_i$ so a probability of 0.85 makes sense. Did I make any mistakes?
Best Answer
In (i) it should have been $\mathbb{P}(A) = \dots = 0.25 \times 0.99 + 0.75 \times 0.17 = 0.375$. This results a minor mistake in (ii): $$ \mathbb{P}(B_1 ~|~ A) = \dots = \frac{0.25 \times 0.99}{0.375} = 0.66. $$