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    A motorcycle company has identified that there is a 10% chance of their new AX125 mopeds having a faulty clutch.

    This week 48 recalled mopeds are to be tested for faulty clutches.

    Calculate the probability that exactly 3 of these mopeds will have faulty clutches.

    Calculate the probability that 41 or more of these mopeds will not have a faulty clutches.

    7 replacement clutches are available from stock. Calculate the probability that this will be insufficient.
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    I see the binomial distribution in your future.
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    This is the binomial distribution with 48 trials and probability 0.1.
    Thus P(3 faulty) = 48C3 * 0.1^3 * 0.9*45, etc.
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    I have managed to complete parts a) and partb)

    Part C is the part I am having trouble getting a possible number
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    (Original post by constellarknight)
    This is the binomial distribution with 48 trials and probability 0.1.
    Thus P(3 faulty) = 48C3 * 0.1^3 * 0.9*45, etc.
    I have calculated parts a and b. Part C i cannot achieve a reasonable answer
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    Answer to Part C follows immediately from the answer to Part B, doesn't it? If 41+ mopeds have working clutches, then 7 replacement clutches are sufficient (I suppose assuming that all of the replacement clutches work, rather than themselves having a 10% chance of failure).
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    P(7 insufficient) = P(Number of failures > 7). Now can you proceed?

    (Original post by derekweir2980)
    I have calculated parts a and b. Part C i cannot achieve a reasonable answer
 
 
 
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