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# A statistics/probability problem watch

1. 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?
2. (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.
3. (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

The various probabilities arec easi;ly obtained from your tree diagram.
4. (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
Did you fail to notice that the question asked for a decision tree? You have also misread the question.
5. 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.

6. (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

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...
7. (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.

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).
8. (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.....
9. (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!
10. (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?
11. (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.
12. (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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