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    Hi, So I've been working through a past paper, but got stuck on question 3 part 2 and question 9 part 1 b on the following paper. http://www.ocr.org.uk/Images/79328-q...atistics-1.pdf

    For question 3 i worked out the probability of X=2 which is 0.27527...

    Then I though since we are given two values to find the probability of only 1 of them we divide by 2, but it's wrong.

    For question 9, I just don't know how to do it, applications of perms and combs always confuse me.
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    (Original post by Music99)
    Hi, So I've been working through a past paper, but got stuck on question 3 part 2 and question 9 part 1 b on the following paper. http://www.ocr.org.uk/Images/79328-q...atistics-1.pdf

    For question 3 i worked out the probability of X=2 which is 0.27527...

    Then I though since we are given two values to find the probability of only 1 of them we divide by 2, but it's wrong.

    For question 9, I just don't know how to do it, applications of perms and combs always confuse me.
    For q 3 part 2 find P(X=2) then set up a second distribution - with Y~B(2,P(X=2))
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    (Original post by Music99)
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    For 3 (i),

    P(X < 2) = P(X=0) + P(X=1)

    for the second part, follow the advice above.

    For 9,

    use

    \displaystyle \binom{n}{r}

    "n choose r"

    and note that odd numbers must end with an odd digit.

    The last part involves simple probabilities.
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    (Original post by Indeterminate)
    For 3 (i),

    P(X < 2) = P(X=0) + P(X=1)

    for the second part, follow the advice above.

    For 9,

    use

    \displaystyle \binom{n}{r}

    "n choose r"

    and note that odd numbers must end with an odd digit.

    The last part involves simple probabilities.
    It's the second part of 3. I tried the method above and i didn't get the correct answer. I'm retrying question 9 now.
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    (Original post by Music99)
    It's the second part of 3. I tried the method above and i didn't get the correct answer. I'm retrying question 9 now.
    Given

     B(2, P(X=2))

    let

    p=P(X=2), q=1-p

    P(Y=1) = p^1 q^{2-1}\binom{2}{1}
 
 
 
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