A bag contains 10 blue marbles, 20 black marbles and 30 red marbles. A marble is drawn from the bag, its color recorded and it is put back in the bag. This process is repeated 3 times. Find the probability that no 2 of the marbles drawn have the same color.

A bag contains 10 blue marbles, 20 black marbles and 30 red marbles. A marble is drawn from the bag, its color recorded and it is put back in the bag. This process is repeated 3 times. The probability that no 2 of the marbles drawn have the same color is

(A) $\dfrac{1}{36}$ 

(B) $\dfrac{1}{6}$ 

(C) $\dfrac{1}{4}$ 

(D) $\dfrac{1}{3}$ 

(GATE 2005) 

Answer: (B) $\dfrac{1}{6}$ 

Explanation: 

Let $E_1$, $E_2$ and $E_3$ be the events of drawing a blue marble, a black marble and a red marble, respectively. 

Here, three marbles are drawn from the bag one-by-one with replacements. 

Therefore, $P(E_1) = \dfrac{10}{60}$, $P(E_2) = \dfrac{20}{60}$, $P(E_3) = \dfrac{30}{60}$. 

Therefore, the required probability is 

$P(E_1 \cap E_2 \cap E_3) + P(E_1 \cap E_3 \cap E_2) + P(E_2 \cap E_1 \cap E_3) \\+ P(E_2 \cap E_3 \cap E_1) + P(E_3 \cap E_1 \cap E_2) + P(E_3 \cap E_2 \cap E_1)$ 

$= 6 \times \left[P(E_1) \times P(E_2) \times P(E_3)\right]$, since $E_1$, $E_2$ and $E_3$ are independent events

$= 6 \times \left[\dfrac{10}{60} \times \dfrac{20}{60} \times \dfrac{30}{60}\right]$ 

$= \dfrac{1}{6}$