An urn contains 4 white and 3 red balls. Three balls are drawn with replacement from this urn. Then, the standard deviation of the number of red balls/drawn is

Options

A. `(6)/(7)`
B. `(36)/(49)`
C. `(5)/(7)`
D. `(25)/(49)`

Answer

A

Explanation

Let X denotes the number of red balls.
Here, probability of getting red balls, `p=(3)/(7)` and probability of getting no red balls, `q=(4)/(7)`
(i) `P_(1)(X=0)=""^(3)C_(0)((3)/(7))^(0)((4)/(7))^(3)=(64)/((7)^(3))`
(ii) `P_(2)(X=1)=""^(3)C_(1)((3)/(7))^(1)((4)/(7))^(2)=(144)/((7)^(3))`
(iii) `P_(3)(X=2)=""^(3)C_(2)((3)/(7))^(2)((4)/(7))^(1)=(108)/((7)^(3))`
(iv) `P_(4)(X=3)=""^(3)C_(3)((3)/(7))^(3)=(27)/((7)^(3))`
`:." Variance"=underset(i=0)overset(3)sumP_(i)x_(i)^(2)-(underset(i=0)overset(3)sumP_(i)x_(i))^(2)`
`=[(64)/((7)^(3))xx0+(144)/((7)^(3))xx(1)^(2)+(180)/((7)^(3))xx(2)^(2)+(27)/((7)^(3))xx(3)^(2)]-[(64)/((7)^(3))xx0+(144)/((7)^(3))xx1+(108)/((7)^(3))xx2+(27)/((7)^(3))xx3]^(2)`
`=[0+(144)/(343)+(432)/(343)+(243)/(343)]-[0+(144)/(343)+(216)/(343)+(81)/(343)]^(2)`
`=(819)/(343)-((441)/(343))^(2)`
`=(280971-194481)/((343)^(2))=(36)/(49)`
Now, standard deviation `=sqrt("variance")`
`=sqrt((36)/(49))=(6)/(7)`