Seven balls are drawn simultaneously from a bag containing 5 white and 6 green balls. The probability of drawing 3 white and 4 green balls is:
1. $\dfrac{7}{{}^{11}{{C}_{7}}}$
2. $\dfrac{\left( {}^{5}{{C}_{3}}+{}^{6}{{C}_{4}} \right)}{{}^{11}{{C}_{7}}}$
3. $\dfrac{\left( {}^{5}{{C}_{2}}\times {}^{6}{{C}_{2}} \right)}{{}^{11}{{C}_{7}}}$
4. $\dfrac{\left( {}^{6}{{C}_{3}}+{}^{5}{{C}_{4}} \right)}{{}^{11}{{C}_{7}}}$

Hint: Let us name the two colours as colour 1 and colour 2. It is given that the number of balls from the bag which are drawn are 5 white and 6 green. Now we have to find the probability for drawing 3 white and 4 green balls. Let us assume that picking up a ball, the probability for this is equal to $P$. Now we will find the value of $P$. Now we will find the possibilities to pick the ball as per the question. We know that the probability to have exactly $r$ trials among $n$ trials is equal to $P\left( X=r \right)$. Then, $P\left( X=r \right)={}^{n}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}$, where $q=1-p$. Now by using this formula, we find the required probability.

Complete step-by-step solution:
According to the question, a total of seven balls are drawn simultaneously from a bag containing 5 white and 6 green balls and we are asked to find the probability of drawing 3 white and 4 green balls.
Since the total number of balls present in the bag are = 5 white + 6 green = 11 balls.
And we have to select 7 balls.
Therefore, total cases = ${}^{11}{{C}_{7}}$.
So, favourable cases for white balls = ${}^{5}{{C}_{3}}$.
And for green ball = ${}^{6}{{C}_{4}}$.
So, clear favourable cases = ${}^{5}{{C}_{3}}.{}^{6}{{C}_{4}}={}^{5}{{C}_{2}}.{}^{6}{{C}_{2}}$. $[{}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}]$
The required probability = $\dfrac{\left( {}^{5}{{C}_{2}}\times {}^{6}{{C}_{2}} \right)}{{}^{11}{{C}_{7}}}$.
Hence the correct answer is option 3.

Note: Students may have a misconception that if the probability to have exactly $r$ trials among $n$ trials is equal to $P\left( X=r \right)$. Then, $P\left( X=r \right)={}^{n}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}$, where $q=1-p$. But we know that if the probability to have exactly $r$ trials among $n$ trials is equal to $P\left( X=r \right)$. Then, $P\left( X=r \right)={}^{n}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}$, where $q=1-p$. OS, the misconception will lead to confusion. So, it should be avoided.