Question

The probability of drawing a green colored ball from a bag containing $6$red and $5$black balls is:

$$A.0 \\\ B.1 \\\ C.\dfrac{5}{{11}} \\\ D.\dfrac{6}{{11}} \\\$$

Answer 1

Hint: We will use the step by step probability method to find the solution. At first we are going to find the total number of outcomes and the favourable outcomes individually to avoid any kind of mistake and then We will use the formula where probability of an event is $\dfrac{{Number{\text{ }}of{\text{ }}favourable{\text{ }}outcomes}}{{Number{\text{ }}of{\text{ }}all{\text{ }}possible{\text{ }}outcomes}}$

Complete step by step Answer :

We are given the experiment that we have to draw a ball from a bag containing numerous balls.
Let us define an event $A$ where we draw a green-colored ball from the bag.
The probability of an event denotes the chance or the likelihood of its happening.
We see that the bag contains $6$ red and $5$ black balls but not any green colored balls, i.e., $0$green balls.
There is no likelihood of the event that a green colored ball might be drawn from the bag as there are no green colored balls in the bag.
Clearly the total number of possible outcomes $= 5 + 6 = 11$
The total number of favourable outcomes is $0$ as there are no green colored balls to be drawn.
Therefore, the probability of drawing a green colored ball from a bag containing $6$red and $5$black balls$= \dfrac{{Number{\text{ }}of{\text{ }}favourable{\text{ }}outcomes}}{{Number{\text{ }}of{\text{ }}all{\text{ }}possible{\text{ }}outcomes}} = \dfrac{0}{{11}} = 0$.

Thus, the answer is option A.

Note: This question is framed as such that we get confused but it is a very simple and clear cut problem on probability. Another way to solve this problem might be to use the combination formula, so the probability of drawing a green-colored ball from a bag containing $6$red and $5$black balls

= \dfrac{{Number{\text{ }}of{\text{ }}favourable{\text{ }}outcomes}}{{Number{\text{ }}of{\text{ }}all{\text{ }}possible{\text{ }}outcomes}} \\\ = \dfrac{{{}^{11}{C_0}}}{{{}^{11}{C_1}}} = \dfrac{0}{{\dfrac{{11!}}{{1!\left( {11 - 1} \right)!}}}} \\\ = \dfrac{0}{{\dfrac{{11!}}{{10!}}}} = \dfrac{0}{{11}} = 0 \\\ $$.