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    Eggs are sold in boxes of six and it is likely that 1 percent of the eggs will be broken when they are unpacked. Find
    1) the probability that a box contains no broken eggs, i got answer of 0.9415 for this.
    2)the probability that a box contains no more than one broken egg.
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    Let X be the random variable, the number of eggs that are broken when they are unpacked
    So X~B(6,0.01)
    So you are asking what is p(X<=1) less than or equal to 1 this means
    This is equal to p(X=0) + p(X=1)
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    (Original post by B_9710)
    Let X be the random variable, the number of eggs that are broken when they are unpacked
    So X~B(6,0.01)
    So you are asking what is p(X<=1) less than or equal to 1 this means
    This is equal to p(X=0) + p(X=1)
    exactly, that's why I did and yet book says 0.9995 or something like that.
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    What did you get ?
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    (Original post by ihatehannah)
    Eggs are sold in boxes of six and it is likely that 1 percent of the eggs will be broken when they are unpacked. Find
    1) the probability that a box contains no broken eggs, i got answer of 0.9415 for this.
    2)the probability that a box contains no more than one broken egg.
    You can use the C method of expansion to find it 6C0 (0.99)^6, but this doesnt get either of the answers...
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    (Original post by BinaryJava)
    You can use the C method of expansion to find it 6C0 (0.99)^6, but this doesnt get either of the answers...
    Using PDF we have Bin(6,0.01) = 0.9415 when evaluated at 0

    Using CDF we have Bin(6,0.01) = 0.9985 when evaluated at 1
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    For this question I imagined a giant tree diagram in my head.
    1) Try doing (0.99)^6 because the probability that each egg is not broken is 0.99 and of course whether an egg breaks or not is an independent variable
    2) As mentioned earlier that p(X<=1) = p(X=0) + p(X=1). The first part of the question is the answer to p(X=0). For p(X=1) you want five eggs not broken and only one broken. That means if we looked at a tree diagram we would do (0.99)^5 x 0.01 (probability that 5 eggs survive and one is broken in a box of six). It is possible that any one of the six eggs could be broken within a box of six eggs so there are six arrangements altogether. Therefore, p(X=1) = (0.99)^5 x 0.01 x 6.
    Hopefully that answers your question.
 
 
 
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