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Question

There are 5 defective items in a large bulk of 100 items. What is the probability that a sample of 10 items will include not more than one defective item?

A

(1920)10
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B

(920)91920
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C

(1920)92920
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D
(920)101920
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Solution

The correct option is C
(1920)92920
Let there be X number of defectve items in a sample of ten items drawn successively.
Now, as we can see that the drawing of the items is done with replacement. Thus, the trials are Bernoulli trials.
Now, pobability of getting a defective item, p=5100=120
Thus, q=1120=1920
We can say that x has binomial distribution, where n=10 and p=120
Thus, P(X=x)=nCxqnxpx, where x=0,1,2,...n
=10Cx(1920)10x(120)x
Probability of getting not more than one defective item =P(X1)
=P(X=0)+P(X=1)=10C0(1920)10(120)0+10C1(1920)9(120)1=(1920)9[1920+1020]=(1920)9×(2920)

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