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    Hi - i was wondering if someone could help. I am working through the edexcel book and am stuck on the question above. The question is:
    the probability of a bolt being faulty is 0.3. Find the probability that in a random sample of 20 bolts there are
    a) exactly 2 faulty bolts - answer is 0.0279
    b) more than 3 faulty bolts - answer is 0.8929

    these bolts are sold in bags if 20. John buys 10 bags.
    c) find the probability that exactly 6 of these bags contain more than 3 faulty bolts.

    so i have: X~B(10,0.8929) P(X=6) (10/6)(0.8929^6)(0.1071^4) but am getting the wrong answer. What am i doing wrong?

    thanks
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    (Original post by Catful1)
    Hi - i was wondering if someone could help. I am working through the edexcel book and am stuck on the question above. The question is:
    the probability of a bolt being faulty is 0.3. Find the probability that in a random sample of 20 bolts there are
    a) exactly 2 faulty bolts - answer is 0.0279
    b) more than 3 faulty bolts - answer is 0.8929

    these bolts are sold in bags if 20. John buys 10 bags.
    c) find the probability that exactly 6 of these bags contain more than 3 faulty bolts.

    so i have: X~B(10,0.8929) P(X=6) (10/6)(0.8929^6)(0.1071^4) but am getting the wrong answer. What am i doing wrong?

    thanks
    Hi there I did this question just a day ago and I'll be able to help First of all your binomial parameters are completely correct, remember you are missing one significant thing here, your factorials \frac{10!}{4!6!} should be in place of 10/6 remember for binomial distribution \frac{n!}{x!(n-x)!}
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    That first bit of the binomial formula is

     \begin{pmatrix}p \\ q \end{pmatrix}

    That means  p \mathrm{C} q = \frac{p!}{(p - q)! \cdot q!} , not  \frac{p}{q} .

    See your mistake?
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    Perfect! Thank you!
 
 
 
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