Here is a hint:
P(A | B) = P(A and B)/P(B)
You want to know the probability of an accident (A) given that they took defensive driving last year (B).
I have this question that I believe I have the correct answer for, however, I am unsure as to exactly what the proper formula is (and obviously unsure if my answer is correct as well) and my prof is super sticky on the proper format to answer the problem. If anyone could help me learn this stuff it would be much appreciated!!!
Problem: Suppose the proporation of drivers in a city who took defensive driving lessons last year was 0.101, the proportion who experienced an accident in the current year was 0.051, and the proportion who took defensive driving and experienced an accident this year is 0.001. Find the probability that a randomly selected person had an accident in the current year given that he/she took defensive driving lessons last year.
I divided 0.51 by .101 and got .505 but that looks high to me and I don't have a clue on the formula to use to even start this mess....
Here is a hint:
P(A | B) = P(A and B)/P(B)
You want to know the probability of an accident (A) given that they took defensive driving last year (B).
Ok so should I be dividing .051 by .101? Im still lost Dawson....
Dason gave you the correct formula in the right format. You got the denominator right but not the numerator.
I'll try in words:
You want the probability of one thing happening given that the other thing happens: For this, divide the probability of both things happening by the probability of the other thing happening. Look again at the numbers you have in the text and have another try.
Ok...so I multipy .051 by .101 and divide by .101 correct????
Oh ok...now I understand!!! P(A given B) = P(A & B)/P(B)
I apologize if I seem totally clueless...I am a 33 year old going back to university and I really am lost!! I appreciate the help more than I can say!
I have another question that I believe requires Bayes Theorem that I will post....thank you!
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