Question 1034977: A shipment of 1313 microwave ovens contains 33 defective units. A restaurant buys four of these units. What is the probability of the restaurant buying at least three nondefective units?
Answer by mathmate(429)   (Show Source):
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Question:
A shipment of 1313 microwave ovens contains 33 defective units. A restaurant buys four of these units. What is the probability of the restaurant buying at least three nondefective units?
Solution:
At least three non-defective is the same as three non-defective or four non-defective.
Using the hypergeometric distribution:
A=33,B=1313-33=1280
C(n,r)=n!/(r!(n-r)!)
Case one: 3 non-defectives
a=1, b=3
P(a,b)=C(A,a)*C(B,b)/C(A+B,a+b)
=C(33,3)*C(1280,1)/C(1313,4)
=287682912/3081777751
Case two: 4 non-defectives
a=0,b=4
P(a,b)=C(A,a)*C(B,b)/C(A+B,a+b)
=C(33,0)*C(1280,4)/C(1313,4)
=2783114232/3081777751
Probability of at least three non-defectives
=287682912/3081777751+2783114232/3081777751
=3070797144/3081777751
=0.99644 approximately
Note:
since 33 is small compared to 1313, binomial distribution would give a close approximation.
p=33/1313=0.025133
P(3, or 4 non-defective)
=C(4,1)*p^1*(1-p)^4+C(4,0)*p^0*(1-p)^4
=0.093142+0.903193
=0.99634
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