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?

Danika Mckay

Danika Mckay

Answered question

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 ) = 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 ( A n C ) / P ( C ) = P ( A ) P ( C | A ) / P ( C ) = ( 0.04 ) ( 0.82 ) / 0.0616 = 0.0328 / 0.0616
P ( A | C ) = 0.532467532

Answer & Explanation

blogpolisft

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
P ( F ) = 0.04 P ( A | F ) = 0.82 P ( A | F ) = 0.03
Step 2
The things you're asked to calculate are:
(a) A one car accident will be attributed to faulty brakes
P ( A ) = P ( A | F ) P ( F ) + P ( A | F ) P ( F ) = 0.82 × 0.04 + 0.03 × 0.96 = 0.0616
and
(b) a one-car accident attributed to faulty brakes was actually due to faulty brakes
P ( F | A ) = P ( A , F ) P ( A ) = P ( A | F ) × P ( F ) P ( A ) = 0.04 × 0.82 0.0616 = 0.532...

Do you have a similar question?

Recalculate according to your conditions!

New Questions in High school probability

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?