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Math Help - probability question

  1. #1
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    probability question

    I am studying GCSE maths - the question I am stuck on is a probabilty question, I have answered most questions so far in the chapter when you are given the figures like 0.6, 0.5 etc however the question i am stuck on is worded differently. This is the exact wording: Each day Mr Smith runs home. He has a choice of three routes: the road, the fields or the canal path. The road route is 4 miles, the fields route is 6 miles and the canal route is 5 miles. In a three day period, what is the probability that Mr Smith runs a total distance of:
    Code:
    a) exactly 17 miles b) exactly 13 miles c) exactly 15 miles d) over 17 miles
    Code:
    the answers in the back are: a) 1/9 b) 1/9 c) 7/27 d) 1/27
    I cant see how you get these answers, i have tried many different ways including tree diagrams and just listing all the possible route combinations (doing this way i get a and b correct but not c and d). A simplistic view could be that every day is an event and he has a choice of three routes, so 1/3 x 1/3 x 1/3 would seem feasable but then that would be the same for each answer, doesnt make sense. Can you please help me understand how to solve this question, I hate skipping things because I dont know them. thanks in advance
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  2. #2
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    Quote Originally Posted by luigibmth View Post
    I am studying GCSE maths - the question I am stuck on is a probabilty question, I have answered most questions so far in the chapter when you are given the figures like 0.6, 0.5 etc however the question i am stuck on is worded differently. This is the exact wording: Each day Mr Smith runs home. He has a choice of three routes: the road, the fields or the canal path. The road route is 4 miles, the fields route is 6 miles and the canal route is 5 miles. In a three day period, what is the probability that Mr Smith runs a total distance of:
    Code:
    a) exactly 17 miles b) exactly 13 miles c) exactly 15 miles d) over 17 miles
    Code:
    the answers in the back are: a) 1/9 b) 1/9 c) 7/27 d) 1/27
    I cant see how you get these answers, i have tried many different ways including tree diagrams and just listing all the possible route combinations (doing this way i get a and b correct but not c and d). A simplistic view could be that every day is an event and he has a choice of three routes, so 1/3 x 1/3 x 1/3 would seem feasable but then that would be the same for each answer, doesnt make sense. Can you please help me understand how to solve this question, I hate skipping things because I dont know them. thanks in advance
    There are 3 possible routes and he has 3 days. First of all you need to find the number of possible combinations regardless of distance. This is your sample space. You then need to find combinations of these routes which add up the given distance, and express it as a probability from the sample space. You say that 1/3 x 1/3 x 1/3 is feasable, but in reality is is not! He doesn't have a free choice of any 3 routes during any three day, because he is constrained by the distance he has to travel. There are 27 possible routes, but for example, when he wants to go 17 miles, some of those 27 routes do not add up to 17 miles... for example, day 1: 4 miles, day 2: 5 miles, day 3: 5 miles! That doesn't not add to 17 miles. So clearly there are some routes which do not meet his demands!

    What you are trying to do is work out the proportion of possible routes that meet his demands, and the proportion of possible routes that don't meet his demands, and from there work out the probability of his demands being met!

    Part a) For example:

    In total there are 27 possible combinations. Which of these add up to 17? Well given 4,5 & 6, only combinations of 6, 6 and 5 add up to 17. Giving 6, 6, 5... 5, 6, 6... 6,5,6.... 3 of them (Not that this is also 2!). Total probability is therefore 3/27, which is indeed 1/9.

    Part c) For example:

    Still 27 total possible outcomes. Which add up to 15? Well you have any combination of 5, 5, 5, ...6, 4, 5... Which gives 5, 5, 5...4, 5, 6...4, 6, 5...5,4,6...5,6,4...6,4,5...6,5,4.... 7 of them(1+3!). So the answer is 7/27.

    Part d) there is only one possible combination of 4,5,6 that adds up to greater than 17 and that's 6, 6, 6. 1/27.
    Last edited by Mush; December 29th 2008 at 06:06 AM.
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  3. #3
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    thank you mush - twice, once for replying so promptly, secondly for solving the problem, it has been swirling around in my head for 48 hours
    .

    thanks
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