Let two fair six-faced dice A and B be thrown simultaneously.

Question

Let two fair six-faced dice $A$ and $B$ be thrown simultaneously. If ${E_1}$ is the event that die $A$ shows up four, ${E_2}$ is the event that die $B$ shows up two and ${E_3}$ is the event that the sum of numbers on both dice is odd, then which of the following statements is $NOT$ true?

Accepted Answer

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.

Solutions · 1