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  • Five had oranges are accidently mixed with 20 good ones. if four oranges are drawn one by one successively with replacement, then find the probability distribution of number of bad oranges drawn. Hence find the mean and variance of the distribution.

Five had oranges are accidently mixed with 20 good ones. if four oranges are drawn one by one successively with replacement, then find the probability distribution of number of bad oranges drawn. Hence find the mean and variance of the distribution.

 

 

 

 
 
 
 
 

Answers (1)

Let x denotes the number of bad oramges in a draw of 4 ornages from a group of 20 good oranges and 5 bad oranges. Since there are 5 bad orange in the group, therefore x can take value 0,1,2,3,4 Now p(x = 0) = Probability of getting 4 good oranges \Rightarrow \left ( \frac{20}{25} \right )^{4}\, ^{4}C_{0}
p(x = 1) = Probability of getting 1 bad oranges
\Rightarrow \frac{5}{25}\times \left ( \frac{20}{25} \right )^{3}\, ^{4}C_{1}
p(x = 2) = Probability of getting 2 bad oranges
=\left ( \frac{5}{25} \right )^{2}\times \left ( \frac{20}{25} \right )^{2}\, ^{4}C_{2}
 p(x = 3) = Probability of getting three bad oranges
=\left ( \frac{5}{25} \right )^{3}\times \left ( \frac{20}{25} \right )\, ^{4}C_{3}
p(x = 4) = Probability of getting four bad oranges
=\left ( \frac{5}{25} \right )^{4}\, ^{4}C_{4}
computation of Mean and Variance
 

Xi Pi =p(X=xi) pixi pixi2
0 256/625 0 0
1 256/625 256/625 256/625
2 96/625 192/625 384/625
3 16/625 48/625 144/625
4 1/625 4/625 16/625
    \sum pixi= \frac{500}{625} \sum pixi^{2}= \frac{800}{625}

we have, \sum pixi= \frac{500}{625}= \frac{4}{5} and \sum pixi^{2}= \frac{800}{625}= \frac{32}{25}
\therefore \tilde{X}= Mean= \sum pixi= \frac{4}{5}
and var\left ( X \right )= \sum pixi^{2}= \left ( \sum pixi \right )^{2}= \frac{32}{25}-\frac{16}{25}= \frac{16}{25}
Hence, mean= \frac{4}{5}\; and\;\: variance= \frac{16}{25}
 

Posted by

Ravindra Pindel

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