Question

A box contains 4 green balls, 5 blue balls, and 3 red balls. A ball is drawn at random. Find the probability of getting a red ball and not getting a green ball.
(A). $\dfrac{1}{4},\dfrac{2}{3}$
(B). $\dfrac{4}{1},\dfrac{2}{11}$
(C). $\dfrac{5}{6},\dfrac{5}{6}$
(D). $\dfrac{13}{6},\dfrac{2}{9}$

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.

Complete step-by-step solution -
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 $=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$ .
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 3 red balls = $^{3}{{C}_{1}}$ .
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 ( 4+5+3 ) balls present in the bag = $^{12}{{C}_{1}}$ .
Now, using the above results let us try to find the probability of drawing a red ball:
$\text{Probability}=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$
\[\Rightarrow \text{Probability}=\dfrac{^{3}{{C}_{1}}}{^{12}{{C}_{1}}}\]
Now using the formula: $^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$ .
\[\Rightarrow \text{Probability}=\dfrac{\dfrac{3!}{\left( 3-1 \right)!1!}}{\dfrac{\left( 12 \right)!}{\left( 12-1 \right)!1!}}\]
\[\Rightarrow \text{Probability}=\dfrac{\dfrac{3!}{2!}}{\dfrac{12!}{11!}}\]
We know n! can be written as n(n-1)! , so our equation becomes:
\[\text{Probability}=\dfrac{\dfrac{3\times 2!}{2!}}{\dfrac{12\times 11!}{11!}}\]
\[\therefore \text{Probability}=\dfrac{3}{12}=\dfrac{1}{4}\]
So, the probability of drawing a red ball from the box of 12 balls is $\dfrac{1}{4}$ .
Similarly, the probability of drawing a green ball is:
$\text{Probability}=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$
\[\Rightarrow \text{Probability}=\dfrac{^{4}{{C}_{1}}}{^{12}{{C}_{1}}}=\dfrac{4}{12}=\dfrac{1}{3}\]
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
$\Rightarrow \dfrac{1}{3}\text{+probability of not drawing green ball=1}$
 $\Rightarrow \text{probability of not drawing green ball=}\dfrac{2}{3}$
So, the probability of not drawing a green ball from the box of 12 balls is $\dfrac{2}{3}$ .
Hence, the answer to the above question is option (a).

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.