Is my work below correct?

My Approach:

Event A- Car accident is due to faulty break

Event B- It gets correctly attributed to faulty break

Event D- It gets incorrectly attributed to faulty break

Event C- It gets attributed to faulty breaks ; then

$P(A)=0.04$

$P(B)=0.82$

$P(D)=0.03$

$P({A}^{\prime})=0.96$

(a) I use total probability rule i.e.

$P(C)=P(A)P(B|A)+P({A}^{\prime})P(D|{A}^{\prime})$

$P(c)=(0.04)(0.82)+(0.96)(0.03)$

$P(C)=0.0328+0.0288$

$P(C)=0.0616$

(b) For this I used Bayes theorem;

$$P(A|C)=P(AnC)/P(C)=P(A)P(C|A)/P(C)\phantom{\rule{0ex}{0ex}}=(0.04)(0.82)/0.0616=0.0328/0.0616$$

$P(A|C)=0.532467532$