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    This question sounds really simple, but I can't understand how you'd do it. Please help:

    "over a long period of time i have worked out the probability that my train is late on a friday is 0.1.
    (I) what is the probability that in my next 4 journeys at least one of the trains is late.
    (II) What is the probability that in my next 10 journeys at least one of the trains is late"
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    Do I draw a probability tree?
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    (Original post by Alex-Torres)
    Do I draw a probability tree?
    Have you done the Binomial Distribution
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    (Original post by Alex-Torres)
    This question sounds really simple, but I can't understand how you'd do it. Please help:

    "over a long period of time i have worked out the probability that my train is late on a friday is 0.1.
    (I) what is the probability that in my next 4 journeys at least one of the trains is late.
    (II) What is the probability that in my next 10 journeys at least one of the trains is late"
    P(At least one of the trains is late) = 1 - P(None of the trains are late)
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    You could draw a probability tree if you need to, but it'd be very repetitive. Seperate it into probability of "train late" and "train not late" Since it has to be one or the other, P(train not late) = 1 - P(train late) = 1 - 0.1 = 0.9

    Go from there
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    (Original post by TenOfThem)
    Have you done the Binomial Distribution
    No not yet
    (Original post by notnek)
    P(At least one of the trains is late) = 1 - P(None of the trains are late)
    I have already done this, it equals 0.9.
    (Original post by savethegoldfish)
    You could draw a probability tree if you need to, but it'd be very repetitive. Seperate it into probability of "train late" and "train not late" Since it has to be one or the other, P(train not late) = 1 - P(train late) = 1 - 0.1 = 0.9

    Go from there
    As above, I've already found not late equals 0.9, but what now?
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    (Original post by Alex-Torres)
    No not yet

    I have already done this, it equals 0.9.

    As above, I've already found not late equals 0.9, but what now?
    No 0.9 is the probability that on a certain Friday, the train is not late. I'm saying that over the 4 week period:

    P(At least one of the trains is late) = 1 - P(None of the trains are late)

    None of the trains are late means that on each of the four Friday's, the train is not late. What's the probability of this?

    Then subtract your answer from 1.
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    (Original post by Alex-Torres)
    No not yet

    Ok

    If none of the 4 trains are late then the probability is

    P(Not) * P(Not) * P(Not) * P(Not)

    i.e. that branch on an imagined tree diagram

    Can you take it from there?
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    I assume that you are talking about the next journey's on fridays only!
    for the next four journeys the probability of the rain being late on atleast one is = ( 0.1 x 0.1 x 0.1 x 0.1)+(0.9 x 0.1 x 0.1 x 0.1)x4 + (0.9 x 0.9 x 0.1 x 0.1)x (4!/(2!x2!)) + (0.9 x 0.9 x 0.9 x 0.1) x 4 = 0.3439

    OR you could calculate the probability of the train being on time for all the 4 days and subtract it from 1. i.e
    1-(0.9 x 0.9 x 0.9 x 0.9) = 0.3439
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    (Original post by notnek)
    No 0.9 is the probability that on a certain Friday, the train is not late. I'm saying that over the 4 week period:

    P(At least one of the trains is late) = 1 - P(None of the trains are late)

    None of the trains are late means that on each of the four Friday's, the train is not late. What's the probability of this?

    Then subtract your answer from 1.
    How do I find the probability of "none trains are late", is it 0.1?
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    (Original post by Alex-Torres)
    How do I find the probability of "none trains are late", is it 0.1?
    Read post 8
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    (Original post by Alex-Torres)
    How do I find the probability of "none trains are late", is it 0.1?
    Imagine the tree diagram in your head and look at TenOfThem's post #8.
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    (Original post by TenOfThem)
    Read post 8
    (Original post by notnek)
    Imagine the tree diagram in your head and look at TenOfThem's post #8.
    Oh ok, so the answer to Question (I) is 0.9*0.9*0.9*0.9 ?
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    (Original post by Alex-Torres)
    Oh ok, so the answer to Question (I) is 0.9*0.9*0.9*0.9 ?
    NO, read post 9
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    (Original post by Alex-Torres)
    Oh ok, so the answer to Question (I) is 0.9*0.9*0.9*0.9 ?
    No

    That is the probability that none are late

    Read your question
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    (Original post by Alex-Torres)
    Oh ok, so the answer to Question (I) is 0.9*0.9*0.9*0.9 ?
    No that's P(None of the trains are late) but the question asks for P(At least one of the trains is late).

    Look back to my post #7
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    (Original post by sghoury)
    NO, read post 9
    Oh yes, sorry, I forgot you subtract it from one!
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    (Original post by Alex-Torres)
    Oh yes, sorry, I forgot you subtract it from one!
    Have a go at (II) to see if you fully understand the method.
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    (Original post by TenOfThem)
    No

    That is the probability that none are late

    Read your question
    (Original post by notnek)
    No that's P(None of the trains are late) but the question asks for P(At least one of the trains is late).

    Look back to my post #7
    Oh ok, so it asks for one is late. Therefore, I do 0.1*0.9*0.9*0.9 ?
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    (Original post by Alex-Torres)
    Oh ok, so it asks for one is late. Therefore, I do 0.1*0.9*0.9*0.9 ?
    NO, you'll have to multiply it by 4 because it does not specify on which of the days its late.
 
 
 
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