Answer

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**Hint:**We will firstly find total number of balls then we will find probability for each ball and as it is given balls are drawn at random without replacement from the urn there are different case which arise over here and we will find out for case where it is probably the three balls picked out have different colours.

**Formula used:**For a random event, the probability of an event, say P(E) is defined as ratio of a number of favourable outcomes or chances and the total number of outcomes.

Probability,

$P(E) = \dfrac{{N(E)}}{{N(S)}}$

where $N(E)$ is the number of favourable outcomes, and $N(S)$ is the total number of outcomes. The Event of choosing r things among n things is ${}^n{C_r}$

**Complete step-by-step answer:**

Given, there are 9 balls in an urn.

The urn contains 3 red balls, 4 blue balls, and 2 green balls.

Event of choosing r things among n things is given by combination formula ${}^n{C_r}$

Therefore, the event of choosing 3 balls from 9 balls is ${}^9{C_3}$

Now,

The event of choosing 3 different colour balls will be

${ \Rightarrow ^3}{C_1} \times {}^4{C_1} \times {}^2{C_1}$

Here, $N(E)$ is $^3{C_1} \times {}^4{C_1} \times {}^2{C_1}$ and $N(S)$ is ${}^9{C_3}$.

Therefore, the probability that the three balls have different colour will be,

$

\Rightarrow P\left( E \right) = \dfrac{{N(E)}}{{N(S)}} \\

\Rightarrow P\left( E \right) = \dfrac{{^3{C_1} \times {}^4{C_1} \times {}^2{C_1}}}{{{}^9{C_3}}} \\

\Rightarrow P(E) = \dfrac{{3 \times 4 \times 2}}{{\dfrac{{9!}}{{3!6!}}}} \\

\Rightarrow P(E) = \dfrac{{24 \times 3!6!}}{{9!}} \\

\Rightarrow P(E) = \dfrac{2}{7} \\

$

**Hence, option (A) is the correct answer.**

**Note:**When it’s given that the event is taking place as without replacement in this case after taking out each ball one ball is subtracted from the total number of balls in urn. That is simply written in the form of combinations and then we apply the concept of combinations in solving probability and in combinations we go for factorial which saves a lot of time.

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