Correct Answer - D
Since , `P(A)=(2)/(5)`
For independent events , `P(A cap B )= P(A)P(B)`
`implies P(A cap B ) le (2)/(5)`
`implies P(A cap B ) =(1)/(10),( 2)/(10), (3)/(10),(4)/(10) ` [maximum 4 outcomes may be in `A cap B` ]
(i) Now , `P(A cap B )=(1)/(10) `
`implies P(A) *P(B) =(1)/(10) `
`implies P(B)=(1)/(10) xx (5)/(2)=(1)/(4)` , not possible
(ii) Now , `P(A cap B ) =(2)/(10) implies (2)/(5) xx P(B) =(2)/(10)`
`implies P(B) =(5)/(10) `, outcomes of B=5
(iii) , `P(A cap B ) =(3)/(10)`
`implies P(A)P(B) =(3)/(10) implies (2)/(5) xx P(B) =(3)/(10) `
`P(B) =(3)/(4)` , not possible
(iv) Now , `P(A cap B ) =(4)/(10) implies P(A) *P(B) =(4)/(10)`
`implies P(B) =1 `, outcomes of B=10