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Edexcel S2 - binomial is pissing me off... (question) Watch

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    wow apparently i can't do the first question on the s2 paper.. no surprise though, binomials are confusing as ****.

    question:


    A fair coin is tossed 4 times. Find the probability that

    (a) an equal number of head and tails occur


    ok, usually I would have done 4C2 (0.5)^2 (0.5)^2 (which is the correct answer). but my teacher introduced me to this little formula yesterday for binomials:

    n!/r!(n-r)!

    so the working ends up being 4!/2!2! (0.5)^2 (0.5)^2 :confused:

    they give different results.

    1) what's the difference between nCr and n!/r!
    2) under what circumstances to i use nCr and under what circumstances do i use n!/r!(n-r)! ?

    if somebody can clear this up for me it would be much appreciated. this is a subject that really get's on my man tits. thanks in advance!
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    (Original post by ucasisannoying)
    wow apparently i can't do the first question on the s2 paper.. no surprise though, binomials are confusing as ****.

    question:


    A fair coin is tossed 4 times. Find the probability that

    (a) an equal number of head and tails occur


    ok, usually I would have done 4C2 (0.5)^2 (0.5)^2 (which is the correct answer). but my teacher introduced me to this little formula yesterday for binomials:

    n!/r!(n-r)!

    so the working ends up being 4!/2!2! (0.5)^2 (0.5)^2 :confused:

    they give different results.

    1) what's the difference between nCr and n!/r!
    2) under what circumstances to i use nCr and under what circumstances do i use n!/r!(n-r)! ?

    if somebody can clear this up for me it would be much appreciated. this is a subject that really get's on my man tits. thanks in advance!
    _nC_r = \frac{n!}{r!(n-r)!}

    So

    _4C_2 = \frac{4!}{2!(2)!}

    Basically, they're identical.

    When you do nCr on your calculator, that's what it's doing. So it can't give different results.
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    (Original post by Noble.)
    _nC_r = \frac{n!}{r!(n-r)!}

    So

    _4C_2 = \frac{4!}{2!(2)!}

    Basically, they're identical.

    When you do nCr on your calculator, that's what it's doing. So it can't give different results.
    oh, turns out i do end up with the same answer lol, thanks...

    p.s. i hate stats
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    another thing...

    (b) all the outcomes are the same (answer is 0.125)
    (c) the first tail occurs on the third throw.

    any idea how to do part (c)?

    (thanks again!)
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    (Original post by ucasisannoying)
    another thing...

    (b) all the outcomes are the same (answer is 0.125)
    (c) the first tail occurs on the third throw.

    any idea how to do part (c)?

    (thanks again!)
    Just the same as before, except there now aren't combinations. The first two throws have to be heads.

    So you'd have HHT...
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    (Original post by Noble.)
    Just the same as before, except there now aren't combinations. The first two throws have to be heads.

    So you'd have HHT...
    iyt iyt...

    so it's just 0.5^3?
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    (Original post by ucasisannoying)
    iyt iyt...

    so it's just 0.5^3?
    No. If you toss the coin four times, and the first two have to be heads, and the third tails, what can the fourth result be?
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    (Original post by Noble.)
    No. If you toss the coin four times, and the first two have to be heads, and the third tails, what can the fourth result be?
    could be either heads or tails...?
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    (Original post by ucasisannoying)
    could be either heads or tails...?
    Yes, so the probability of the first tails appearing on the third throw will be the sum of the two probabilities HHTH and HHTT.
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    (Original post by Noble.)
    Yes, so the probability of the first tails appearing on the third throw will be the sum of the two probabilities HHTH and HHTT.
    (0.5)^4 + (0.5)^4 is the same as (0.5)^3 lol

    your method makes more sense though, i'll use that if I get another question like this. thanks for everything man
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    (Original post by ucasisannoying)
    (0.5)^4 + (0.5)^4 is the same as (0.5)^3 lol

    your method makes more sense though, i'll use that if I get another question like this. thanks for everything man
    Ah yes.

    2(0.5)^4 = 0.5^{-1} \times 0.5^4 = 0.5^3

    I thought you were just doing

    P(HHT) = 0.5 \times 0.5 \times 0.5
 
 
 
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