Question

A bag contains 8 green balls and 6 red balls. If one ball is drawn from the bag, what is the probability that it is green?

Answer 1

Hint: Use permutations and combinations to get the number of favourable outcomes and the total number of outcomes. Think of the basic interpretation of combination.
Probability in simple words is the possibility of an event to occur.
Probability can be mathematically defined as $=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$

Complete step-by-step answer:
Given:
Number of green balls in the bag = 8
Number of red balls in the bag = 6
Let us try to find the number of favourable outcomes:
So, whenever we select one out of the eight green balls, it is counted as a favourable event.
We can mathematically represent this as:
Ways of selecting one out of eight green balls = $^{8}{{C}_{1}}$ .
Similarly,
Now let us try to calculate the total number of possible outcomes.
So, it is counted as one of the possible outcomes whenever we draw a ball, whether it is green or a red ball.
Now let us try to represent it mathematically.
We get;
Ways of selecting one out of ( 6+8) balls present in the bag = $^{6+8}{{C}_{1}}{{=}^{14}}{{C}_{1}}$ .
Now, using the above results let us try to find the probability of drawing green balls:
$\text{Probability}=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$
$\Rightarrow \text{Probability}=\dfrac{^{8}{{C}_{1}}}{^{14}{{C}_{1}}}$
Now Using the formula: $^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$ .
$\Rightarrow \text{Probability}=\dfrac{\dfrac{8!}{\left( 8-1 \right)!1!}}{\dfrac{\left( 14 \right)!}{\left( 14-1 \right)!1!}}$
$\Rightarrow \text{Probability}=\dfrac{\dfrac{8!}{7!}}{\dfrac{14!}{13!}}$
We know n! can be written as n(n-1)! , so our equation becomes:
$\text{Probability}=\dfrac{\dfrac{8\times 7!}{7!}}{\dfrac{14\times 13!}{13!}}$
$\therefore \text{Probability}=\dfrac{8}{14}=\dfrac{4}{7}$
So, the probability of drawing a green ball from a bag of 8 green balls and 6 red balls is $\dfrac{4}{7}$.

Note: It is preferred that while solving a question related to probability, always cross-check the possibilities, as there is a high chance you might miss some or have included some extra or repeated outcomes. Also, when a large number of outcomes are to be analysed then permutations and combinations play a very important role as we see in the above solution.