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Q. 6.     From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.

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Total bulbs = 30

defective bulbs = 6

Non defective bulbs =30-6=24

P(defective \, \, \, \, \, bulbs)=\frac{6}{30}=\frac{1}{5}

P(Non \, \, \, \, defective \, \, \, \, \, bulbs)=\frac{24}{30}=\frac{4}{5}

4 bulbs is drawn at random with replacement.

Let X : number of defective bulbs 

4 Non defective bulbs and 0  defective bulbs : P(X=0)={4}\textrm{C}_0\frac{4}{5}.\frac{4}{5}.\frac{4}{5}.\frac{4}{5}=\frac{256}{625}

3 Non defective bulbs and 1  defective bulbs : P(X=1)={4}\textrm{C}_1\frac{1}{5}.\frac{4}{5}.\frac{4}{5}.\frac{4}{5}=\frac{256}{625}

2 Non defective bulbs and 2  defective bulbs : P(X=2)={4}\textrm{C}_2\frac{1}{5}.\frac{1}{5}.\frac{4}{5}.\frac{4}{5}=\frac{96}{625}

1 Non defective bulbs and 3  defective bulbs : P(X=3)={4}\textrm{C}_3\frac{1}{5}.\frac{1}{5}.\frac{1}{5}.\frac{4}{5}=\frac{16}{625}

0 Non defective bulbs and 4  defective bulbs : P(X=4)={4}\textrm{C}_4\frac{1}{5}.\frac{1}{5}.\frac{1}{5}.\frac{1}{5}=\frac{1}{625}

the probability distribution of the number of defective bulbs is as :

X 0 1 2 3 4
P(X) \frac{256}{625} \frac{256}{625} \frac{96}{625} \frac{16}{625} \frac{1}{625}

 

Posted by

seema garhwal

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