Question

A bag contains 15 balls of which some are white and others are red. If the probability of drawing a red ball is twice that of a white ball, find the number of white balls in the bag?

Answer 1

Hint: Here the question is related to the probability. They have given the total balls in the bag. So first we consider any either color ball as x and the other one will be represented as \[15 - x\] . We have the condition that the probability of drawing a red ball is equal to twice of drawing the white ball. So by writing the terms in the probability and writing in the terms of the given condition, we can obtain the required solution.

Complete step by step solution:
Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e., how likely they are to happen, using it. Probability can range in from 0 to 1, where 0 means the event to be an impossible one and 1 indicates a certain event.
The probability formula is defined as the probability of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes.
\[Probability of event to happen P\left( E \right) = \dfrac{{Number\;of \;favourable \;outcomes}}{{Total\; Number \;of \;outcomes}}\]
Now we will consider the given question. A bag contains white and red balls.
The total number of balls present in the bag = 15
We don’t know how many are red and white.
Let us consider, there are x white balls in the bag.
Therefore, the number of white balls = x
If there are x white balls there means, then remaining will be the red balls.
Therefore the number of red balls = \[15 - x\]
From the question we have that the probability of drawing a red ball is twice that of a white ball.
The probability of drawing a white ball = \[\dfrac{x}{{15}}\]
The probability of drawing a red ball = \[\dfrac{{\left( {15 - x} \right)}}{{15}}\]
From the given condition the probability of drawing a red ball is twice that of a white ball.
This can be written as
 \[ \Rightarrow \dfrac{{\left( {15 - x} \right)}}{{15}} = 2 \times \dfrac{x}{{15}}\]
In the RHS 2 is multiplied to the numerator.
 \[ \Rightarrow \dfrac{{\left( {15 - x} \right)}}{{15}} = \dfrac{{2x}}{{15}}\]
The number 15 which is present in the denominator on either side, we can cancel it.
 \[ \Rightarrow 15 - x = 2x\]
Add x on the both sides
 \[ \Rightarrow 15 - x + x = 2x + x\]
On simplifying we have
 \[ \Rightarrow 15 = 3x\]
On dividing by 3 we have
 \[ \Rightarrow x = 5\]
 \[ \Rightarrow 15 - x = 15 - 5 = 10\]
The x and \[15 - x\] represents the number of white balls therefore there are 5 white balls and 10 red balls present in the bag.
So, the correct answer is “Option B”.

Note: Suppose if we think we have x red balls, then we have \[15 - x\] white balls. The given condition for this can be written as
 \[ \Rightarrow \dfrac{x}{{15}} = 2 \times \dfrac{{(15 - x)}}{{15}}\]
In the RHS 2 is multiplied to the numerator.
 \[ \Rightarrow \dfrac{x}{{15}} = \dfrac{{2(15 - x)}}{{15}}\]
 \[ \Rightarrow \dfrac{x}{{15}} = \dfrac{{30 - 2x}}{{15}}\]
The number 15 which is present in the denominator on either side, we can cancel it.
 \[ \Rightarrow x = 30 - 2x\]
Add 2x on the both sides
 \[ \Rightarrow x + 2x = 30 - 2x + 2x\]
On simplifying we have
 \[ \Rightarrow 3x = 30\]
On dividing by 3 we have
 \[ \Rightarrow x = 10\]
Here x will represents as a red balls
Therefore there are 10 red and 5 white balls in a bag.
So either way we can solve this problem.