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Baye's Theorem

  1. Mar 6, 2008 #1
    Hi! I have some problem about the use of Baye’s Theorem . So I have been given the problem below that we were asked to use Baye’s Theorem. But I sue the simple manner of conditional probability. So can somebody show me how I can use that Baye’s theorem or if I’m wrong correct me:
    95% of car drivers wear seatbelt, 64% of car drivers involved in serious accident die if not wearing seatbelt whereas 12% of those that do wear a seatbelt die. Calculate, correct to 3 significant figures, the percentage of drivers who died and were not wearing seatbelts.
    Let A= “wear seatbelt”; B= die; A’= don’t wear seatbelt.
    P(A)= 95%
    P(B and A’)=?
    By formula of conditional probability P(B/A’)= [P(B and A’)]/[P(A’)]
    Then P(B and A’)= [P(B/A’)]*[P(A’)]
    P(B and A’)= 0.64*0.05=0.032=3.20%
  2. jcsd
  3. Mar 13, 2008 #2
    We know that P(A) = .95, thus P(A') = .05.
    Now let B be the event that someone died, like you put. Then you have
    P(B | A') = P(B and A')/P(A') = .64 Since 64% of the people that don't wear seats belts die. Thus for P(B and A') = (.05)(.64).

    Now we want P(A' | B) = P(A' and B)/P(B). Well you have P(A' and B) already, so now you just need to find the probability that someone died, P(B). There are two things to take into consideration here. The people that died and were wearing their seatbelts and the other side of the coin, the people who died and were not wearing their seatbelts.

    Hope this helps.
  4. Mar 14, 2008 #3
    We know that the drivers who died and were wearing seatbelt are 12% and there 5% who died and were not wearing seatbelt.

    So we have
    P(B)= P(A and B) + P(A' and B)= 12% + 5% = 17%
    P(A'/B)=P(A' and B)/P(B)=(.05)/(.17)
    Is it like this?
    But here i don't understand why you have calculated again that P(A'/B)! if i try to express it in words you meant the probability of A' conditionned by B or again "the drivers who were not wearing seatbelt but who died before"? Just here can you explain me how we can express that P(A'/B)?
  5. Mar 14, 2008 #4
    Wait, I already told you that P(A' and B) = (.05)(.64) So why do you say that it's 5% above.
    Think of it this way....of the 5% of the people that don't wear seatbelts 64% of them die, that is P(A' and B).

    Now, we look at P(A and B); of the 95% that wear seatbelts, only 12% of them die.

    Is this more clear?
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