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    There's a question on a wroksheet I was struggling with,

    In an experiment, a biased coin is thrown 10 times.
    The probability of obtaining a heads is 0.4.
    - If the experiment is performed 7 times, what's the probability that exactly 5 heads are obtained on exactly 2 occaisions?

    taken from Q2 part iv of section 1 exercise level 2 from MEI introducing the binomial distribution.

    so X~B(10, 0.4) and P(X=5) but as far as I know that only works for one experiment, so would you then have to times the number once by itself, and then five times by its complement to mimic obtaining the result twice, and then not obtaining it five times out of seven?

    sorry if this is lengthy or if it's more obvious then I realise, I'd appreciate any and all help.
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    (Original post by Usii101)
    There's a question on a wroksheet I was struggling with,

    In an experiment, a biased coin is thrown 10 times.
    The probability of obtaining a heads is 0.4.
    - If the experiment is performed 7 times, what's the probability that exactly 5 heads are obtained on exactly 2 occaisions?

    taken from Q2 part iv of section 1 exercise level 2 from MEI introducing the binomial distribution.

    so X~B(10, 0.4) and P(X=5) but as far as I know that only works for one experiment, so would you then have to times the number once by itself, and then five times by its complement to mimic obtaining the result twice, and then not obtaining it five times out of seven?

    sorry if this is lengthy or if it's more obvious then I realise, I'd appreciate any and all help.
    Almost there, you are right. It's a binomial distribution made from a binomial distribution.

    You get your 'p' from the first part of the question (n=10) then it forms its own distribution with n=7
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    So, let's say that Y is the number of times 5 heads are obtained on 2 occaisons.
    The experiment is performed 7 times, so n=7
    Probability of success is the probabilty of 5 heads in one experiment (P(X=5) = 0.201)
    And we want to achieve this twice i.e. 2 success, r=2

    And so to answer this question, you would use Y~B(7,0.201), P(Y=2) Which would be 0.276
 
 
 

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