Question 934869: 74. Twelve items are independently sampled from a production line. If the probability any given item is defective is 0.1, the probability of at most two defectives in the sample is closest to …
o 0.3874
o 0.9872
o 0.7361
• 0.8891
Answer is 0.8891. But I didn't get why:) Please, help me, anyone)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the binomial probability formula is:
p(x) = c(n,x) * p^x * q^(n-x)
c(n,x) is the combination formula of n things taken x at a time.
p^x is the probability of occurrence raised to the x power.
q^(n-x) is the probability of not occurrence raised to the n-x power.
in your problem:
n = 12
x = 0 or 1 or 2
p = .1
q = 1 - .1 = .9
c(12,0) = 1
c(12,1) = 12
c(12,2) = 66
the probability of at most 2 defectives equals the probability of 0 defectives plus the probability of 1 defective plus the probability of 2 defective.
this becomes:
p(0) + p(1) + p(2)
p(0) = c(12,0) * .1^0 * .9^12 = .2824
p(1) = c(12,1) * .1^1 * .9^11 = .3766
p(2) = c(12,2) * .1^2 * .9^10 = .2301
add them up and you get .8891
all values were rounded to 4 decimal places.
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