Question

A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of red balls. Find the number of blue balls in the bag?
(a) 10
(b) 5
(c) 15
(d) 20

Answer 1

For finding the probability of drawing a red ball, the number of favourable cases is equal to the number of ways of selecting 1 ball out of the 5 red balls, while for finding the probability of drawing a blue ball is equal to the number of ways of selecting one out of the x blue balls. The total number of possible outcomes is equal to the number of ways of selecting one out of (5+x) balls. So, get the probabilities and equate probability of drawing blue balls with twice the probability of drawing black balls, to get x and hence, the required result.

Complete step-by-step answer :
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.
Given:
Number of red balls in the bag = 5
We let the number of blue balls be x.
Let us try to find the number of favourable outcomes for drawing red ball:
So, whenever we select one out of the 5 red balls, it is counted as a favourable event.
We can mathematically represent this as:
Ways of selecting one out of 5 red balls = $^{5}{{C}_{1}}$ .
Similarly, we can say that the favourable outcomes for drawing blue balls is equal to the number of ways of selecting one out of x blue balls, i.e., $^{x}{{C}_{1}}$ .
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 red or a blue ball.
Now let us try to represent it mathematically.
We get;
Ways of selecting one out of ( 5 + x) balls present in the bag = $^{5+x}{{C}_{1}}$ .
Now, using the above results let us try to find the probability of drawing red balls:

$\text{Probability}=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$
$\Rightarrow \text{Probability}=\dfrac{^{5}{{C}_{1}}}{^{5+x}{{C}_{1}}}$
Similarly we can say probability of drawing blue ball is $\dfrac{^{x}{{C}_{1}}}{^{5+x}{{C}_{1}}}$
Now as it is given that probability of drawing blue balls is double that of red balls, we can say:
$\dfrac{^{x}{{C}_{1}}}{^{5+x}{{C}_{1}}}=\dfrac{2{{\times }^{5}}{{C}_{1}}}{^{5+x}{{C}_{1}}}$
${{\Rightarrow }^{x}}{{C}_{1}}=2{{\times }^{5}}{{C}_{1}}$
$\Rightarrow x=2\times 5$
$\Rightarrow x=10$
Hence, the answer to the above question is option (a).

Note : Be very careful about whose probability is double of the other and don’t misinterpret it as $\dfrac{2{{\times }^{x}}{{C}_{1}}}{^{5+x}{{C}_{1}}}=\dfrac{^{5}{{C}_{1}}}{^{5+x}{{C}_{1}}}$ . If you want you can also think about the question as all balls are similar and have equal chances, so for the probability of some colour to be double of the other must have twice the number of balls of that colour, so the answer is $2\times 5=10$ .