Question

A bag contains 15 balls of which x are black and the remaining are red. If the number of red balls is increased by 5, the probability of drawing red ball doubles, then the probability of drawing a red ball is
(A) $\dfrac{1}{5}$
(B) $\dfrac{4}{5}$
(C) $\dfrac{3}{5}$
(D) $\dfrac{2}{5}$

Answer 1

The total number of red balls in the bag is 15. In the bag, there are red balls and black balls. We have two cases. In the ${{1}^{st}}$ case, we have 15 balls in a bag. There are x black balls and the remaining balls are red. Since the number of black balls is x so the number of red balls is $\left( 15-x \right)$ . Now, use the formula, $\text{Probability=}\dfrac{\text{Number}\,\text{of}\,\text{possible}\,\text{outcomes}}{\text{Sample}\,\text{space}}$ and calculate the probability in the ${{1}^{st}}$ case. In the ${{2}^{nd}}$ case, we have added 5 more red balls in a bag. So, the total number of balls is 20.
Since 5 more red balls are added to the bag, the number of red balls is 5 more than the number of red balls in the ${{1}^{st}}$ case. Now, get the number of red balls in the ${{2}^{nd}}$ case. Then, use the formula, $\text{Probability=}\dfrac{\text{Number}\,\text{of}\,\text{possible}\,\text{outcomes}}{\text{Sample}\,\text{space}}$ and calculate the probability in the ${{2}^{nd}}$ case. It is given that after adding 5 red balls, the probability of drawing red ball doubles. Now, form an equation using this information and solve it further to get the value of x. Put the value of x in the probability of drawing the red balls in the ${{1}^{st}}$ case.

Complete step-by-step answer :
According to the question, we have two cases.
In the ${{1}^{st}}$ case, we have 15 balls in a bag. There are x black balls and the remaining balls are red.
The total number of balls in the bag = 15 …………………………….(1)
The number of black balls = x ……………………………………..(2)
The number of red balls = $\left( 15-x \right)$ ……………………………………(3)
We know the formula of the probability, $\text{Probability=}\dfrac{\text{Number}\,\text{of}\,\text{possible}\,\text{outcomes}}{\text{Sample}\,\text{space}}$ …………………………..(4)
The total number of balls in the bag is the sample space.
For calculating the probability of red balls, the number of possible outcomes is equal to the number of red balls in the bag.
From equation (1) and equation (3), we have the total number of balls in the bag and the number of red balls.
$\text{Probability=}\dfrac{\left( 15-x \right)}{15}$ ……………………………………….(5)
In the ${{2}^{nd}}$ case, we have added 5 more red balls in a bag. So, the total number of balls is 20.
The total number of balls in the bag = 20 …………………………….(6)
Since 5 more red balls are added to the bag, the number of red balls is 5 more than the number of red balls in the ${{1}^{st}}$ case.
From equation (3), we have the number of red balls in the ${{1}^{st}}$ case.
The number of red balls = $\left( 15-x \right)+5=\left( 20-x \right)$ ……………………………………(7)
We know the formula of the probability, $\text{Probability=}\dfrac{\text{Number}\,\text{of}\,\text{possible}\,\text{outcomes}}{\text{Sample}\,\text{space}}$ …………………………..(8)
The total number of balls in the bag is the sample space.
For calculating the probability of red balls, the number of possible outcomes is equal to the number of red balls in the bag.
From equation (6) and equation (7), we have the total number of balls in the bag and the number of red balls.
$\text{Probability=}\dfrac{\left( 20-x \right)}{20}$ ……………………………………….(9)
From equation (5) and equation (9), we have the probability of drawing red balls.
It is given that after adding 5 red balls, the probability of drawing red ball doubles. It means that the probability in the ${{2}^{nd}}$ case is double of the probability in ${{1}^{st}}$ case.

& \Rightarrow \dfrac{\left( 20-x \right)}{20}\text{=2 }\times \dfrac{\left( 15-x \right)}{15} \\\ & \Rightarrow \dfrac{\left( 20-x \right)}{4}=\dfrac{2\left( 15-x \right)}{3} \\\ & \Rightarrow 3\left( 20-x \right)=4\times 2\left( 15-x \right) \\\ & \Rightarrow 60-3x=120-8x \\\ & \Rightarrow 8x-3x=60 \\\ & \Rightarrow 5x=60 \\\ & \Rightarrow x=\dfrac{60}{5} \\\ & \Rightarrow x=12 \\\ \end{aligned}$$ Now, putting the value of x in equation (5), we get $$\text{Probability=}\dfrac{\left( 15-12 \right)}{15}=\dfrac{3}{15}=\dfrac{1}{5}$$ . Therefore, the probability of drawing a red ball is $$\dfrac{1}{5}$$ . Hence, the correct option is (A). **Note** :In this question, one might do a silly mistake while forming the equation using the information that after adding 5 red balls the probability of drawing red balls doubles. One might form the equation as $$\text{2 }\times \dfrac{\left( 20-x \right)}{20}\text{=}\dfrac{\left( 15-x \right)}{15}$$ . This is wrong because the probability of drawing red balls after adding 5 red balls is already twice the initial probability of drawing red balls. So, we have to multiply by 2 in the initial probability of drawing red balls to make it equal to the probability of drawing red balls after adding 5 red balls.