# 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?

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
$P\left(A\right)=0.04$
$P\left(B\right)=0.82$
$P\left(D\right)=0.03$
$P\left({A}^{\prime }\right)=0.96$
(a) I use total probability rule i.e.
$P\left(C\right)=P\left(A\right)P\left(B|A\right)+P\left({A}^{\prime }\right)P\left(D|{A}^{\prime }\right)$
$P\left(c\right)=\left(0.04\right)\left(0.82\right)+\left(0.96\right)\left(0.03\right)$
$P\left(C\right)=0.0328+0.0288$
$P\left(C\right)=0.0616$
(b) For this I used Bayes theorem;
$P\left(A|C\right)=P\left(AnC\right)/P\left(C\right)=P\left(A\right)P\left(C|A\right)/P\left(C\right)\phantom{\rule{0ex}{0ex}}=\left(0.04\right)\left(0.82\right)/0.0616=0.0328/0.0616$
$P\left(A|C\right)=0.532467532$
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Step 1
There are two circumstances, "attribution to faulty brakes" and "has faulty brakes". Let's call these events A and F. Then you should write
$\begin{array}{rl}P\left(F\right)& =0.04\\ P\left(A|F\right)& =0.82\\ P\left(A|{F}^{\prime }\right)& =0.03\end{array}$
Step 2
The things you're asked to calculate are:
(a) A one car accident will be attributed to faulty brakes
$\begin{array}{rl}P\left(A\right)& =P\left(A|F\right)P\left(F\right)+P\left(A|{F}^{\prime }\right)P\left({F}^{\prime }\right)\\ & =0.82×0.04+0.03×0.96\\ & =0.0616\end{array}$
and
(b) a one-car accident attributed to faulty brakes was actually due to faulty brakes
$\begin{array}{rl}P\left(F|A\right)& =\frac{P\left(A,F\right)}{P\left(A\right)}\\ & =\frac{P\left(A|F\right)×P\left(F\right)}{P\left(A\right)}\\ & =\frac{0.04×0.82}{0.0616}\\ & =0.532...\end{array}$