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5 bad 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 drwan.hence find the mean and variance of the distribution

Dear Student,

Let X be the number of bad oranges drawn of 4 oranges from group of  20 good oranges and 5 bad ones∴ X can be 0,1,2,3,4P(X=0)=probability of getting no bad orange=probabilty of getting 4 good oranges            =20254.C04=256625P(X=1)=probability of getting 1 bad orange            =525.20253.C14=256625P(X=2)=probability of getting 2 bad oranges            =5252.20252.C24=96625P(X=3)=probability of getting 3 bad oranges            =5253.2025.C34=16625P(X=4)=probability of getting 4 bad oranges            =5254.C44=1625

Now we have,∑i=14 pixi=500625=45∑i=14pixi2=800625=3225Mean=∑i=14xipi=45Variance=∑i=14pixi2-∑i=14pixi2=3225-1625=1625

Regards

  • 10
View Full Answer
take the probability n=4 ,p=4/5 and q=1/5
now the distribution table would be like 
              0                         1                      2                        3                           4                  
  4C0(4/5)0(1/5)4  4C1(4/5)1(1/5)3  ​4C2(4/5)2(1/5)2 ​  4C3(4/5)3(1/5)1  ​4C4(4/5)4(1/5)0 
Mean = np=16/5    Variance=npq=16/25
  • 4
Please see answer below
  • -7
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