Question 893837: 2% of the items of a factory are defective.The items are packed in
boxes.What is the probability that there will be
i.2 defective items
ii.At least three defective items?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! since 2% of the items are defective, then the probability that any one item will be defective is .02.
the probability of any one item not being defective is equal to 1 minus .02 which is equal to .98
you need to know how many items are packed in each box.
that information is missing.
i'll assume 100 packages in the box so i can show you how to do the calculations.
you would need to replace 100 with the actual number that you have in order to find the answer you need.
the probability that 0 out of the 100 will be defective is equal to:
p(0) = C(100,0) * .02^0 * .98^100
the probability that 1 out of the 100 will be defective is equal to:
p(1) = C(100,1) * .02^1 * .98^99
the probability that 2 out of the 100 will be defective is equal to:
p(2) = C(100,2) * .02^2 * .98^98
to find the probability that exactly 2 are defective, you would use the formula for exactly 2 defective which is:
p(2) = C(100,2) * .02^2 * .98^98 which is equal to .2734...
to find the probability that at least 3 are defective you would find the probability of 1 minus the probability that 0 or 1 or 2 are defective.
1 minus the probability that 0 or 1 or 2 are defective gives you the probability that 3 or more are defective.
3 or more are defective means the same thing as at least 3 are defective.
the formulas you would use are:
p(0) = C(100,0) * .02^0 * .98^100 which is equal to .1326...
p(1) = C(100,1) * .02^1 * .98^99 which is equal to .2706...
p(2) = C(100,2) * .02^2 * .98^98 which is equal to .2734...
add these up and you get p(0 or 1 or 2) = .6766...
take 1 minus .6766... and you get .3233...
that would be your answer if you packed 100 items in each box.
C(n,x) is the combination formula of n! / (x! * (n-x)!)
for example:
C(100,2) is equal to 100! / (2! * 98!)
this is equivalent to (100 * 99 * 98!) / (2! * 98!)
the 98! in the numerator and denominator cancel out and you are left with:
C(100,2) = (100 * 99) / 2 which is equal to 50 * 99 which is equal to 4950
the formula of C(100,2) * .02^2 * .98^98 becomes:
4950 * .0004 * .1380878... which is equal to .2734.....
this number should agree with what i got before and it does.
a number shown as .2734.... means that i'm using the stored number in the calculator rather than the rounded number.
rounding is done only at the end.
your solution for p(greater than or equal to 3) is .3323 rounded to 4 decimal places.
that's equivalent to 33.23 percent rounded to 2 decimal places.
that is only your solution if you have packed 100 items per box.
you solution will be different if you packed a different amount of items per box.
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