The probability that a one-car accident is due to faulty brakes is 0.04, the probability that a one-car accident is correctly attributed to faulty brakes is 0.82, and the probability that a one-car accident is incorrectly attributed to faulty brakes is 0.03. What is the probability that (a) a one-car accident will be attributed to faulty brakes; (b) a one-car accident attributed to faulty brakes was actually due to faulty brakes?
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')=0.96
(a) I use total probability rule i.e.
$P(C)=P(A)P(B|A)+P(A')P(D|A')$
$P(c)=(0.04)(0.82)+(0.96)(0.03)$
$P(C)=0.0328+0.0288$
$P(C)=0.0616$
(b) For this I used Bayes theorem; $$P(A|C)=P(AnC)/P(C)=P(A)P(C|A)/P(C)\\ =(0.04)(0.82)/0.0616=0.0328/0.0616$$
$P(A|C)=0.532467532$
You've arrived at the correct answers, but the way you've phrased the problem is a muddle.
There are two circumstances, "attribution to faulty brakes" and "has faulty brakes". Let's call these events $A$ and $F$. Then you should write $$ \begin{align} P(F)&=0.04\\ P(A|F)&=0.82\\ P(A|F')&=0.03\\ \end{align} $$ The things you're asked to calculate are:
(a) A one car accident will be attributed to faulty brakes
$$ \begin{align} P(A)&=P(A|F)P(F)+P(A|F')P(F')\\ &=0.82\times0.04+0.03\times0.96\\ &=0.0616 \end{align} $$ and
(b) a one-car accident attributed to faulty brakes was actually due to faulty brakes
$$ \begin{align} P(F|A)&={P(A, F)\over P(A)} \\ &= {P(A|F)\times P(F)\over P(A)}\\ &= {0.04\times0.82\over0.0616}\\ &=0.532...\\ \end{align} $$ You shouldn't have introduced all those other symbols, but somehow you made it all work.