Question

A bag contains 15 balls of which x are blue and the remaining are red. If the number of red balls is increased by 5, the probability of drawing the red balls doubles. Find
(I) P (red balls)
(II) P (blue balls)
(III) P (blue balls if 5 red balls are actually added).

Answer 1

Hint: For this question, we will first find the number of red balls in terms of x and then calculate the probabilities of getting red and blue balls in terms of x. At last, we will find required probabilities, $ \text{Probability}=\dfrac{\text{Number of favorable outcomes}}{\text{total outcomes}} $ .

Complete step by step answer:
Here we are given the total number of balls as 15.
The number of blue balls is x.
And thus the number of red balls will be the difference between the total balls and the number of blue balls. So the number of red balls is equal to 15-x.
Now let us find probabilities of picking red balls and blue balls.
We know that, $ \text{Probability}=\dfrac{\text{Number of favorable outcomes}}{\text{total outcomes}} $ .
So probability of getting red balls $ \Rightarrow \dfrac{\text{Number of red balls}}{\text{total balls}}=\dfrac{15-x}{15}\cdots \cdots \left( 1 \right) $ .
Probability of getting blue balls $ \Rightarrow \dfrac{\text{Number of blue balls}}{\text{total balls}}=\dfrac{x}{15}\cdots \cdots \left( 2 \right) $ .
Now we are given that if the number of red balls is increased by 5, the probability of drawing red balls doubles.
Number of red balls then become = 5+15-x = 20-x.
Total number of balls become = 15+5 = 20.
Probability of picking red balls become $ \Rightarrow \dfrac{\text{New number of red balls}}{\text{New total balls}}=\dfrac{20-x}{20}\cdots \cdots \left( 3 \right) $ .
According to statement we can say that $ \dfrac{20-x}{20} $ becomes double of $ \dfrac{15-x}{15} $ so we get,
 $ \dfrac{20-x}{20}=2\left( \dfrac{15-x}{15} \right)\Rightarrow \dfrac{20-x}{20}=\dfrac{30-2x}{15} $ .
Cross multiplying we get,
 $ 15\left( 20-x \right)=20\left( 30-2x \right)\Rightarrow 300-15x=600-40x $ .
Rearranging the terms, we get,
 $ 40x-15x=600-300\Rightarrow 25x=300 $ .
Dividing both sides by 25 we get,
 $ x=\dfrac{300}{25}=12 $ .
So the value of x is 12.
(I) From (1) we see that, P (red balls) = $ \dfrac{15-x}{15} $ .
Putting x = 12, we get, P (red balls) $ \Rightarrow \dfrac{15-12}{15}=\dfrac{3}{15} $ .
Dividing numerator and denominator by 3 we get, P (red balls) = $ \dfrac{1}{5} $ .
(II) From (2) we see that, P (blue balls) = $ \dfrac{x}{15} $ .
Putting x = 12, we get P (blue balls) $ \Rightarrow \dfrac{12}{15} $ .
Dividing numerator and denominator by 3 we get, P (blue balls) = $ \dfrac{4}{5} $ .
(III) If 5 red balls are actually added then a number of blue balls remain the same i.e. x which is equal to 12 but a total number of balls increases from 15 to 15+5 = 20.
So P (blue balls if 5 red balls are added) = $ \dfrac{12}{20} $ .
Dividing the numerator and denominator by 4 we get,
P (blue balls if 5 red balls are added) = $ \dfrac{3}{5} $ .

Note:
Students should take care of the signs while solving the equation. Find the probabilities carefully considering every factor. Make sure that probability always lies between 0 and 1.