Question

A box contains 6 green balls, 4 blue balls, and 5 yellow balls. A ball is drawn at random. Find the probability of getting a yellow ball and not getting a green ball.
(a) $\dfrac{1}{5},\dfrac{1}{3}$
(b) $\dfrac{4}{15},\dfrac{3}{15}$
(c) $\dfrac{1}{3},\dfrac{3}{5}$
(d) $\dfrac{2}{3},\dfrac{1}{15}$

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.
Before moving to the question, let us talk about probability.
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:
Now, let’s move to the solution to the above question.
Let us try to find the number of favourable outcomes for picking yellow ball:
So, whenever we select one out of the 5 yellow balls, it is counted as a favourable event.
We can mathematically represent this as:
Ways of selecting one out of 5 yellow balls = $^{5}{{C}_{1}}$ .
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, no matter which colour the ball is. Now if we represent this mathematically, we get
Ways of selecting one out of ( 5+4+6) balls present in the bag = $^{15}{{C}_{1}}$ .
Now, using the above results let us try to find the probability of drawing a yellow ball:
$\text{Probability}=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$
\[\Rightarrow \text{Probability}=\dfrac{^{5}{{C}_{1}}}{^{15}{{C}_{1}}}\]
Now Using the formula: $^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$ .
\[\Rightarrow \text{Probability}=\dfrac{\dfrac{5!}{\left( 5-1 \right)!1!}}{\dfrac{\left( 15 \right)!}{\left( 15-1 \right)!1!}}\]
\[\Rightarrow \text{Probability}=\dfrac{\dfrac{5!}{4!}}{\dfrac{15!}{14!}}\]
We know n! can be written as n(n-1)! , so our equation becomes:
\[\text{Probability}=\dfrac{\dfrac{5\times 4!}{4!}}{\dfrac{15\times 14!}{14!}}\]
\[\therefore \text{Probability}=\dfrac{5}{15}=\dfrac{1}{3}\]
So, the probability of drawing a yellow ball from the box of 15 balls is $\dfrac{1}{3}$ .
Similarly, the probability of drawing a green ball is:
$\text{Probability}=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$
\[\Rightarrow \text{Probability}=\dfrac{^{6}{{C}_{1}}}{^{15}{{C}_{1}}}=\dfrac{6}{15}=\dfrac{2}{5}\]
Therefore, the probability of not drawing a green ball can be calculated as:
Probability of an event occurring + probability of an event not occurring = 1
$\Rightarrow \dfrac{\text{2}}{\text{5}}\text{+probability of not drawing green ball=1}$
 $\Rightarrow \text{probability of not drawing green ball=}\dfrac{3}{5}$
So, the probability of not drawing a green ball from the box of 15 balls is $\dfrac{3}{5}$ .
Hence, the answer to the above question is option (c).

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.