QJEEM16Maba9a: JEE Main 2016 Mathematics - Class 12 JEEM16Maba9a Solution: Let two fair six-faced dice and be thrown simultaneously. If is …

Total possible outcome with two six faced dice = 6² = 36

When dice A shows up 4, the possible cases are
E₁ = { (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) } = 6 cases
$\therefore$ $P\left( {{E_1}} \right) = {6 \over {36}} = {1 \over 6}$

When dice B shows up 2, the possible cases are
E₂ = { (1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2) } = 6 cases

$P\left( {{E_2}} \right) = {6 \over {36}} = {1 \over 6}$

${E_1} \cap {E_2}$ = Common in both in E₁ and E₂ = { (4, 2) }

$P\left( {{E_1} \cap {E_2}} \right) = {1 \over {36}}$

And $P\left( {{E_1}} \right)$ . $P\left( {{E_2}} \right)$ = ${1 \over 6}$ . ${1 \over 6}$ = ${1 \over 36}$

$\therefore$ $P\left( {{E_1} \cap {E_2}} \right)$ = $P\left( {{E_1}} \right)$ . $P\left( {{E_2}} \right)$

$\therefore$ E₁ and E₂ are independent.

${E_3}$ = [ (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

${E_1} \cap {E_3}$ = { (4, 1) (4, 3) (4, 5) } = 3 cases

$\therefore$ $P\left( {{E_1} \cap {E_3}} \right) = {3 \over {36}} = {1 \over {12}}$ = ${1 \over 6} \times {1 \over 2}$ = $P\left( {{E_1}} \right)$ $\times$ $P\left( {{E_3}} \right)$

$\therefore$ E₁ and E₃ are independent.

${E_2} \cap {E_3}$ = { (1, 2) (3, 2) (5, 2) } = 3 cases

$\therefore$ $P\left( {{E_2} \cap {E_3}} \right) = {3 \over {36}} = {1 \over {12}}$ = ${1 \over 6} \times {1 \over 2}$ = $P\left( {{E_2}} \right)$ $\times$ $P\left( {{E_3}} \right)$

$\therefore$ E₂ and E₃ are independent.

${E_1} \cap {E_2} \cap {E_3}$ = 0

$\therefore$ $P\left( {{E_1} \cap {E_2} \cap {E_3}} \right)$ = 0

$P\left( {{E_1}} \right) \times$ $P\left( {{E_2}} \right) \times$ $P\left( {{E_3}} \right)$ = ${1 \over 6}$ $\times$ ${1 \over 6}$ $\times$ ${1 \over 2}$ = ${1 \over {72}}$

$\therefore$ ${E_1},$ ${E_2}$ and ${E_3}$ are not independent.

$\therefore$ Option (D) is correct.