JEE Main 2016 — Probability Question with Solution

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
$$\therefore$$ $$P\left(E_1\right)=\frac{6}{36}=\frac{1}{6}$$

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

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

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

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

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

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

$$\therefore$$ 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

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

$$\therefore$$ E₁ and E₃ are independent.

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

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

$$\therefore$$ 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

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

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

$$\therefore$$ Option (D) is correct.