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    Hi guys

    Im a little bit stuck on a certain question I was given for homework yet if I figure it out, many of my problems will become resolved lol.

    I think I have the answer but I want to make sure I've done it correctly.
    The question instructs that we use the "decision tree method" for the following question:

    " A road safety study found that in 77% of all accidents the driver was wearing a seatbelt. Accident reports indicated that 92% of those drivers escaped serious injury (defined as hospitalisation or death), but only 63% of the non-belted drivers were so fortunate. What is the probability that a driver who was seriously injured was not wearing a seatbelt?"

    Using a tree diagram ( hopefully this is the same as the decision tree method), I came up with the decimal 0.0851 or 8.51% as the answer to the question. But I have doubts this is correct.
    The way I did it was multiplying 0.23 with 0.37.

    Am I missing a step? Have I done this correctly?
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    (Original post by Fyer1234)
    Hi guys

    Im a little bit stuck on a certain question I was given for homework yet if I figure it out, many of my problems will become resolved lol.

    I think I have the answer but I want to make sure I've done it correctly.
    The question instructs that we use the "decision tree method" for the following question:

    " A road safety study found that in 77% of all accidents the driver was wearing a seatbelt. Accident reports indicated that 92% of those drivers escaped serious injury (defined as hospitalisation or death), but only 63% of the non-belted drivers were so fortunate. What is the probability that a driver who was seriously injured was not wearing a seatbelt?"

    Using a tree diagram ( hopefully this is the same as the decision tree method), I came up with the decimal 0.0851 or 8.51% as the answer to the question. But I have doubts this is correct.
    The way I did it was multiplying 0.23 with 0.37.

    Am I missing a step? Have I done this correctly?
    You haven't done it correctly.

    You need to start by drawing a decision tree. You can't just cross your fingers and multiply numbers together. First set of branches can be wearing a seatbelt and not wearing a seatbelt. Second set of branches can be seriously injured and not seriously injured.
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    (Original post by Fyer1234)
    Hi guys

    Im a little bit stuck on a certain question I was given for homework yet if I figure it out, many of my problems will become resolved lol.

    I think I have the answer but I want to make sure I've done it correctly.
    The question instructs that we use the "decision tree method" for the following question:

    " A road safety study found that in 77% of all accidents the driver was wearing a seatbelt. Accident reports indicated that 92% of those drivers escaped serious injury (defined as hospitalisation or death), but only 63% of the non-belted drivers were so fortunate. What is the probability that a driver who was seriously injured was not wearing a seatbelt?"

    Using a tree diagram ( hopefully this is the same as the decision tree method), I came up with the decimal 0.0851 or 8.51% as the answer to the question. But I have doubts this is correct.
    The way I did it was multiplying 0.23 with 0.37.

    Am I missing a step? Have I done this correctly?
    No, not correct. You are being asked for a conditional probability, i.e. the probability that he was wearintg a seat belt, given that he was seriuosly injured. This is calculated
    as Prob(seriously injured and wearing a seat belt) divided by Prob(He was seriously injured)
    (It is a simple case of Bayes Theorem
     \text{P}(A|B)=\frac{\text{P}(B|A  )\text{P}(A)}{\text{P}(B|A)\text  {P}(A)+\text{P}B|A')\text{P}(A')  }
    The various probabilities arec easi;ly obtained from your tree diagram.
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    (Original post by brianeverit)
    No, not correct. You are being asked for a conditional probability, i.e. the probability that he was wearintg a seat belt, given that he was seriuosly injured. This is calculated
    as Prob(seriously injured and wearing a seat belt) divided by Prob(He was seriously injured)
    (It is a simple case of Bayes Theorem
     \text{P}(A|B)=\frac{\text{P}(B|A  )\text{P}(A)}{\text{P}(B|A)\text  {P}(A)+\text{P}B|A')\text{P}(A')  }
    Did you fail to notice that the question asked for a decision tree? You have also misread the question.
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    Thanks Mr M

    That is exactly how I did it though.

    0.77 wearing a seatbelt followed by 0.92 escape serious injury & 0.08 with serious injury
    Drivers
    0.23 not wearing a seatbelt followed 0.63 escaping serious injury & 0.37 with serious injury.

    Is that not correct? I think the part I am missing is adding the two serious injury decimals and diving by something, that part I'm not sure.

    Please any advice would be great
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    (Original post by brianeverit)
    No, not correct. You are being asked for a conditional probability, i.e. the probability that he was wearintg a seat belt, given that he was seriuosly injured. This is calculated
    as Prob(seriously injured and wearing a seat belt) divided by Prob(He was seriously injured)
    (It is a simple case of Bayes Theorem
     \text{P}(A|B)=\frac{\text{P}(B|A  )\text{P}(A)}{\text{P}(B|A)\text  {P}(A)+\text{P}B|A')\text{P}(A')  }
    The various probabilities arec easi;ly obtained from your tree diagram.

    Mmm ok

    So, as far as I can see it should be 0.08 divided by 0.45? That doesn't seem right...
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    (Original post by Fyer1234)
    Thanks Mr M

    That is exactly how I did it though.

    0.77 wearing a seatbelt followed by 0.92 escape serious injury & 0.08 with serious injury
    Drivers
    0.23 not wearing a seatbelt followed 0.63 escaping serious injury & 0.37 with serious injury.

    Is that not correct? I think the part I am missing is adding the two serious injury decimals and diving by something, that part I'm not sure.

    Please any advice would be great
    All correct so far.

    Now calculate these probabilities:

    a) P(wearing a seatbelt and injured)

    b) P(not wearing a seatbelt and injured)

    c) P(injured) [this is the sum of the first two probabilities]

    Your final step is to divide P(not wearing a seatbelt and injured) by P(injured).
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    (Original post by Mr M)
    All correct so far.

    Now calculate these probabilities:

    a) P(wearing a seatbelt and injured)

    b) P(not wearing a seatbelt and injured)

    c) P(injured) [this is the sum of the first two probabilities]

    Your final step is to divide P(not wearing a seatbelt and injured) by P(injured).
    Thanks

    ok this is what I've come up with:

    a) 0.0616
    b) 0.0851
    c) 0.1467

    0.0851
    0.1467 = 0.58 or 58%

    Hopefully I've got it.....
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    (Original post by Fyer1234)
    Thanks

    ok this is what I've come up with:

    a) 0.0616
    b) 0.0851
    c) 0.1467

    0.0851
    0.1467 = 0.58 or 58%

    Hopefully I've got it.....
    Yep!
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    (Original post by Mr M)
    Yep!
    wow thankyou.

    Just one more question regarding how to display probability like this. Is there a certain rule on how to show this probability? Is it best to show it as a decimal, fraction or percentage?
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    (Original post by Fyer1234)
    wow thankyou.

    Just one more question regarding how to display probability like this. Is there a certain rule on how to show this probability? Is it best to show it as a decimal, fraction or percentage?
    I wouldn't use percentages. Fractions or decimals are fine.
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    (Original post by Mr M)
    I use fractions if the answer is rational. My second preference is decimals.
    Ah ok thanks for your help today, has really relieved a lot of anxiousness towards these type of questions
 
 
 
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