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    1) For 1(b), Why isn't it P(X=1)? Also, I understand it is 2 days to the mean is doubled, but couldn't using X~Po(1.6) also include exactly 2 days? So why is this correct for less than 2 days?

    2) For 1(c), please could someone explain this step by step?

    3) For 7), where does Var(Y)=(4n/25) come from?

    Probability is my worst topic and this is really bugging me.

    Thanks for any help.
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    (Original post by PhyM23)
    1) For 1(b), Why isn't it P(X=1)?
    X=1 is the probability that only one car was caught in the first 48 hours. There could be any number >= 1, and it still satisfies the criterion of catching a car within 48 hours.

    Also, I understand it is 2 days to the mean is doubled, but couldn't using X~Po(1.6) also include exactly 2 days? So why is this correct for less than 2 days?
    The parameter lambda is a continuous function (multiple) of the number of days.
    What would lambda be if the distribution was for 2 days minus 1 hour?
    Or two days minus 1 minute? Or minus 1 second?, Etc.

    Also, although the Poisson is a discrete distribution, it's modelling a continous time period. What's the probability of a car arriving at exactly 2 days? Zero!

    Don't feel I've really dealt with that bit properly, but hope it helps.

    2) For 1(c), please could someone explain this step by step?
    Conditional probabilty.

    P(1 caught one day 1 AND 3 caught on day 2 | 4 caught over 2 days.)

    The first two parts (red) refer to a single day period - hence lambda = 0.8.
    Last past (green) refers to a 2 day period - hence lambda = 1.6.

    Then standard conditional probability.

    = P(1 caught one day 1 AND 3 caught on day 2 ) / P( 4 caught over 2 days.)

    The two days are independent, so


    = P(1 caught one day 1) P( 3 caught on day 2 ) / P( 4 caught over 2 days.)

    3) For 7), where does Var(Y)=(4n/25) come from?

    Probability is my worst topic and this is really bugging me.

    Thanks for any help.
    The binomial, which you are approximating, has a variance of npq, or np(1-p)
    Hence n x (1/5) x (4/5) = 4n/25.
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    (Original post by ghostwalker)
    X=1 is the probability that only one car was caught in the first 48 hours. There could be any number >= 1, and it still satisfies the criterion of catching a car within 48 hours.



    The parameter lambda is a continuous function (multiple) of the number of days.
    What would lambda be if the distribution was for 2 days minus 1 hour?
    Or two days minus 1 minute? Or minus 1 second?, Etc.

    Also, although the Poisson is a discrete distribution, it's modelling a continous time period. What's the probability of a car arriving at exactly 2 days? Zero!

    Don't feel I've really dealt with that bit properly, but hope it helps.



    Conditional probabilty.

    P(1 caught one day 1 AND 3 caught on day 2 | 4 caught over 2 days.)

    The first two parts (red) refer to a single day period - hence lambda = 0.8.
    Last past (green) refers to a 2 day period - hence lambda = 1.6.

    Then standard conditional probability.

    = P(1 caught one day 1 AND 3 caught on day 2 ) / P( 4 caught over 2 days.)

    The two days are independent, so


    = P(1 caught one day 1) P( 3 caught on day 2 ) / P( 4 caught over 2 days.)



    The binomial, which you are approximating, has a variance of npq, or np(1-p)
    Hence n x (1/5) x (4/5) = 4n/25.
    Thank you very much for such a detailed reply.

    1) This makes sense and your explanation of the second part is exactly what I was looking for.

    2) This makes more sense, but why isn't P(4 caught over 2 days) in the numerator too?

    3) Ah yes I should have spotted that :facepalm:
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    (Original post by PhyM23)
    2) This makes more sense, but why isn't P(4 caught over 2 days) in the numerator too?
    Look up conditional probability in the S1 section of your formula booklet:

    P(A | B) = \frac{P(A \cap B)}{P(B)}
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    (Original post by Zacken)
    Look up conditional probability in the S1 section of your formula booklet:

    P(A | B) = \frac{P(A \cap B)}{P(B)}
    That's what's confusing me.

    Let P(1 caught on day 1) = P(A)
    P(3 caught on day 2) = P(B)
    P(4 caught over 2 days) = P(C)

    P(A and B|C) = P(A and B and C) / P(C)

    Why isn't it this?
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    (Original post by PhyM23)
    That's what's confusing me.

    Let P(1 caught on day 1) = P(A)
    P(3 caught on day 2) = P(B)
    P(4 caught over 2 days) = P(C)

    P(A and B|C) = P(A and B and C) / P(C)

    Why isn't it this?
    Ah, I see what you mean.

    It's just that A and B and C is (1 caught on day 1 and 3 caught on day 2 and 4 caught over 2 days).

    But (1 caught on day 1 and 3 caught on day 2) is already (4 caught over 2 days). So A \cap B \cap C = A \cap B.

    Kind of like \{1, 2, 3\} \cap \{3, 4, 5\} \cap \{1, 2, 3, 4, 5, 6, 7,...\} = \{1, 2, 3\} \cap \{3, 4, 5\}

    Edit: see below, ghostwalker explained it better.
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    (Original post by PhyM23)
    That's what's confusing me.

    Let P(1 caught on day 1) = P(A)
    P(3 caught on day 2) = P(B)
    P(4 caught over 2 days) = P(C)

    P(A and B|C) = P(A and B and C) / P(C)

    Why isn't it this?
    You're quite right. I missed a step out.

    P(A and B|C) = P(A and B and C) / P(C)

    The event "A and B" (i.e. 1 on day 1, and 3 on day 2), implies 4 over 2 days, i.e. C.

    So, the event "A and B" is a subset of event "C".

    A\cap B \subseteq C


    Hence P(A and B and C) = P(A and B).

    Edit: ninja'd
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    (Original post by Zacken)
    Ah, I see what you mean.

    It's just that A and B and C is (1 caught on day 1 and 3 caught on day 2 and 4 caught over 2 days).

    But (1 caught on day 1 and 3 caught on day 2) is already (4 caught over 2 days). So A \cap B \cap C = A \cap B.

    Kind of like \{1, 2, 3\} \cap \{3, 4, 5\} \cap \{1, 2, 3, 4, 5, 6, 7,...\} = \{1, 2, 3\} \cap \{3, 4, 5\}

    Edit: see below, ghostwalker explained it better.
    (Original post by ghostwalker)
    You're quite right. I missed a step out.

    P(A and B|C) = P(A and B and C) / P(C)

    The event "A and B" (i.e. 1 on day 1, and 3 on day 2), implies 4 over 2 days, i.e. C.

    So, the event "A and B" is a subset of event "C".

    A\cap B \subseteq C


    Hence P(A and B and C) = P(A and B).

    Edit: ninja'd
    Haha

    This is exactly what I was looking for; thank you both for your help
 
 
 
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