Danika Mckay

2022-10-24

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

$P(A)=0.04$

$P(B)=0.82$

$P(D)=0.03$

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

(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;

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

$P(A|C)=0.532467532$

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;

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

$P(A|C)=0.532467532$

blogpolisft

Beginner2022-10-25Added 10 answers

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

$$\begin{array}{rl}P(F)& =0.04\\ P(A|F)& =0.82\\ P(A|{F}^{\prime})& =0.03\end{array}$$

Step 2

The things you're asked to calculate are:

(a) A one car accident will be attributed to faulty brakes

$$\begin{array}{rl}P(A)& =P(A|F)P(F)+P(A|{F}^{\prime})P({F}^{\prime})\\ & =0.82\times 0.04+0.03\times 0.96\\ & =0.0616\end{array}$$

and

(b) a one-car accident attributed to faulty brakes was actually due to faulty brakes

$$\begin{array}{rl}P(F|A)& =\frac{P(A,F)}{P(A)}\\ & =\frac{P(A|F)\times P(F)}{P(A)}\\ & =\frac{0.04\times 0.82}{0.0616}\\ & =\mathrm{0.532...}\end{array}$$

There are two circumstances, "attribution to faulty brakes" and "has faulty brakes". Let's call these events A and F. Then you should write

$$\begin{array}{rl}P(F)& =0.04\\ P(A|F)& =0.82\\ P(A|{F}^{\prime})& =0.03\end{array}$$

Step 2

The things you're asked to calculate are:

(a) A one car accident will be attributed to faulty brakes

$$\begin{array}{rl}P(A)& =P(A|F)P(F)+P(A|{F}^{\prime})P({F}^{\prime})\\ & =0.82\times 0.04+0.03\times 0.96\\ & =0.0616\end{array}$$

and

(b) a one-car accident attributed to faulty brakes was actually due to faulty brakes

$$\begin{array}{rl}P(F|A)& =\frac{P(A,F)}{P(A)}\\ & =\frac{P(A|F)\times P(F)}{P(A)}\\ & =\frac{0.04\times 0.82}{0.0616}\\ & =\mathrm{0.532...}\end{array}$$

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