Total possible outcome with two six faced dice = 6
2 = 36
When dice A shows up 4, the possible cases are
E
1 = { (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) } = 6 cases
∴ P(E1)=366=61
When dice B shows up 2, the possible cases are
E
2 = { (1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2) } = 6 cases
P(E2)=366=61
E1∩E2 = Common in both in E
1 and E
2 = { (4, 2) }
P(E1∩E2)=361
And
P(E1).
P(E2) =
61.
61 =
361
∴P(E1∩E2) =
P(E1).
P(E2)
∴ E
1 and E
2 are independent.
E3 = [ (1, 2), (1, 4), (1, 6),
(2, 1), (2, 3), (2, 5),
(3, 2), (3, 4), (3, 6),
(4, 1), (4, 3), (4, 5),
(5, 2), (5, 4), (5, 6),
(6, 1), (6, 3), (6, 5) ] = 18 cases
E1∩E3 = { (4, 1) (4, 3) (4, 5) } = 3 cases
∴ P(E1∩E3)=363=121 =
61×21 =
P(E1) × P(E3)
∴ E
1 and E
3 are independent.
E2∩E3 = { (1, 2) (3, 2) (5, 2) } = 3 cases
∴ P(E2∩E3)=363=121 =
61×21 =
P(E2) × P(E3)
∴ E
2 and E
3 are independent.
E1∩E2∩E3 = 0
∴ P(E1∩E2∩E3) = 0
P(E1)× P(E2)×P(E3) =
61 × 61 × 21 =
721
∴ E1, E2 and
E3 are not independent.
∴ Option (D) is correct.