A bag contains 4 white, 5 red, and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is
A. \[\dfrac{1}{{22}}\]
B. \[\dfrac{3}{{22}}\]
C. \[\dfrac{2}{{91}}\]
D. \[\dfrac{2}{{77}}\]

Hint: Here we need to find the probability that all the balls drawn are red. For that, we will find the number of ways to draw the required number of balls out of the total number of balls. Then we will find the number of ways to draw the red balls. To obtain the required probability, we will calculate the ratio of both of them.

Complete step by step solution:
Here we need to find the probability that all the balls drawn are red.
It is given that the number of red balls is equal to 5, number of white balls is 4 and number of blue balls is equal to 6.
Therefore, the total number of balls \[ = 5 + 4 + 6 = 15\]
Let \[S\] be the sample space and this is equal to the number of ways to draw 3 balls out of the 15 balls.
\[n\left( S \right) = {}^{15}{C_3}\]
Using the formula of combination \[{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)! \times r!}}\], we get
\[ \Rightarrow n\left( S \right) = \dfrac{{15!}}{{\left( {15 - 3} \right)! \times 3!}}\]
Subtracting the terms in the denominator, we get
\[ \Rightarrow n\left( S \right) = \dfrac{{15!}}{{12! \times 3!}}\]
Now, we will expand the terms by finding their factorials. So, we get
\[ \Rightarrow n\left( S \right) = \dfrac{{15 \times 14 \times 13 \times 12!}}{{12! \times 3 \times 2 \times 1}}\]
On further simplification, we get
\[ \Rightarrow n\left( S \right) = 455\]
Let \[E\] be the event to get all of the 3 red balls.
Therefore,
\[n\left( E \right) = {}^5{C_3}\]
Using the formula of combination \[{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)! \times r!}}\], we get
\[ \Rightarrow n\left( E \right) = \dfrac{{5!}}{{\left( {5 - 3} \right)! \times 3!}}\]
Subtracting the terms in the denominator, we get
\[ \Rightarrow n\left( E \right) = \dfrac{{5!}}{{2! \times 3!}}\]
Now, we will expand the terms by finding their factorials.
\[ \Rightarrow n\left( E \right) = \dfrac{{5 \times 4 \times 3!}}{{2 \times 1 \times 3!}}\]
On further simplification, we get
\[ \Rightarrow n\left( E \right) = 10\]
Now, we will find the required probability that all the balls drawn are red.
\[P\left( E \right) = \dfrac{{n\left( E \right)}}{{n\left( S \right)}}\]
Now, we will substitute the obtained values in the above equation. Therefore, we get
\[ \Rightarrow P\left( E \right) = \dfrac{{10}}{{455}}\]
On further simplifying the terms, we get
\[ \Rightarrow P\left( E \right) = \dfrac{2}{{91}}\]
Therefore, the required probability is equal to \[\dfrac{2}{{91}}\]

Hence, the correct option is option C.

Note:
We know that the probability is defined as the ratio of the favorable outcomes or the desired outcomes to the total number of outcomes. The value of probability is always greater than 0 and less than 1. That means the probability of an event cannot be negative. If the probability of an event is 1 then it is called a sure event, however if the probability is 0 then the event is called an impossible event. In real life, probability is used to calculate the occurrence of rainfall in an area, winning a match etc.