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A bag contains 15 balls of which x are blue ball and the remaining are red . If the number of red balls are increased by 5 , the probability of drawing the red balls doubles.Find :--- (1) P(red balls) (2) P ( blue ball) (3) P ( blue ball if 5 extra red balls are actually added ).?
Verified Answer
A bag contains 15 balls of which x are blue ball and the remaining are...
Answer:
No. of balls = 15
No. of blue balls = x
No. of red balls = 15 - x
So , P (red ball) = 15 - x / 15
Now
No. of red balls increased by 5
P (red ball) = 15 - x + 5 / 15 + 5
= 20 - x / 20
According to Question,
20 - x / 20 = 2 (15 - x / 15)
> 20 - x / 40 = 15 - x / 15
> 60 - 3x = 120 - 8x
> 5x = 60
> x = 12
Hence,
No. of red balls = 15 - 12 = 3
No. of blue balls = 12
i. P (red ball) = 3 / 15 = 1 / 5
ii. P (blue ball) = 12 / 15 = 4 / 5
iii. P (blue ball if actually 5 red balls added) = 12 / 20 = 3 / 5
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Most Upvoted Answer
A bag contains 15 balls of which x are blue ball and the remaining are...
Given information:
- The bag contains 15 balls.
- Let x be the number of blue balls.
- The remaining balls are red.
- When the number of red balls is increased by 5, the probability of drawing red balls doubles.
To find:
(1) P(red balls)
(2) P(blue ball)
(3) P(blue ball if 5 extra red balls are added)
Solution:
1. P(red balls):
Let's assume the number of red balls initially is r.
Since the total number of balls in the bag is 15, we can write the equation:
x + r = 15
The probability of drawing a red ball from the bag initially is given by:
P(red balls initially) = r/15
When the number of red balls is increased by 5, the new number of red balls becomes r + 5.
According to the given condition, the probability of drawing a red ball doubles. So, we can write the equation:
2 * P(red balls initially) = P(red balls after increasing by 5)
Substituting the values, we get:
2 * (r/15) = (r + 5)/15
Simplifying the equation:
2r = r + 5
r = 5
Therefore, there are initially 5 red balls in the bag.
So, the probability of drawing red balls is:
P(red balls) = 5/15 = 1/3
2. P(blue ball):
We know that the total number of balls in the bag is 15, and the number of red balls is 5. Therefore, the number of blue balls can be calculated as:
x + 5 = 15
x = 10
So, there are initially 10 blue balls in the bag.
The probability of drawing a blue ball is given by:
P(blue ball) = x/15
P(blue ball) = 10/15
P(blue ball) = 2/3
3. P(blue ball if 5 extra red balls are actually added):
If 5 extra red balls are added, the total number of red balls becomes 5 + 5 = 10.
The total number of balls in the bag becomes 15 + 5 = 20.
The probability of drawing a blue ball in this case can be calculated as:
P(blue ball if 5 extra red balls are added) = x/(x + 10)
P(blue ball if 5 extra red balls are added) = 10/20
P(blue ball if 5 extra red balls are added) = 1/2
Summary:
- The probability of drawing red balls is 1/3.
- The probability of drawing blue balls is 2/3.
- If 5 extra red balls are added, the probability of drawing a blue ball becomes 1/2.
- The bag contains 15 balls.
- Let x be the number of blue balls.
- The remaining balls are red.
- When the number of red balls is increased by 5, the probability of drawing red balls doubles.
To find:
(1) P(red balls)
(2) P(blue ball)
(3) P(blue ball if 5 extra red balls are added)
Solution:
1. P(red balls):
Let's assume the number of red balls initially is r.
Since the total number of balls in the bag is 15, we can write the equation:
x + r = 15
The probability of drawing a red ball from the bag initially is given by:
P(red balls initially) = r/15
When the number of red balls is increased by 5, the new number of red balls becomes r + 5.
According to the given condition, the probability of drawing a red ball doubles. So, we can write the equation:
2 * P(red balls initially) = P(red balls after increasing by 5)
Substituting the values, we get:
2 * (r/15) = (r + 5)/15
Simplifying the equation:
2r = r + 5
r = 5
Therefore, there are initially 5 red balls in the bag.
So, the probability of drawing red balls is:
P(red balls) = 5/15 = 1/3
2. P(blue ball):
We know that the total number of balls in the bag is 15, and the number of red balls is 5. Therefore, the number of blue balls can be calculated as:
x + 5 = 15
x = 10
So, there are initially 10 blue balls in the bag.
The probability of drawing a blue ball is given by:
P(blue ball) = x/15
P(blue ball) = 10/15
P(blue ball) = 2/3
3. P(blue ball if 5 extra red balls are actually added):
If 5 extra red balls are added, the total number of red balls becomes 5 + 5 = 10.
The total number of balls in the bag becomes 15 + 5 = 20.
The probability of drawing a blue ball in this case can be calculated as:
P(blue ball if 5 extra red balls are added) = x/(x + 10)
P(blue ball if 5 extra red balls are added) = 10/20
P(blue ball if 5 extra red balls are added) = 1/2
Summary:
- The probability of drawing red balls is 1/3.
- The probability of drawing blue balls is 2/3.
- If 5 extra red balls are added, the probability of drawing a blue ball becomes 1/2.
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Question Description
A bag contains 15 balls of which x are blue ball and the remaining are red . If the number of red balls are increased by 5 , the probability of drawing the red balls doubles.Find :--- (1) P(red balls) (2) P ( blue ball) (3) P ( blue ball if 5 extra red balls are actually added ).? for Class 10 2025 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about A bag contains 15 balls of which x are blue ball and the remaining are red . If the number of red balls are increased by 5 , the probability of drawing the red balls doubles.Find :--- (1) P(red balls) (2) P ( blue ball) (3) P ( blue ball if 5 extra red balls are actually added ).? covers all topics & solutions for Class 10 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A bag contains 15 balls of which x are blue ball and the remaining are red . If the number of red balls are increased by 5 , the probability of drawing the red balls doubles.Find :--- (1) P(red balls) (2) P ( blue ball) (3) P ( blue ball if 5 extra red balls are actually added ).?.
A bag contains 15 balls of which x are blue ball and the remaining are red . If the number of red balls are increased by 5 , the probability of drawing the red balls doubles.Find :--- (1) P(red balls) (2) P ( blue ball) (3) P ( blue ball if 5 extra red balls are actually added ).? for Class 10 2025 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about A bag contains 15 balls of which x are blue ball and the remaining are red . If the number of red balls are increased by 5 , the probability of drawing the red balls doubles.Find :--- (1) P(red balls) (2) P ( blue ball) (3) P ( blue ball if 5 extra red balls are actually added ).? covers all topics & solutions for Class 10 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A bag contains 15 balls of which x are blue ball and the remaining are red . If the number of red balls are increased by 5 , the probability of drawing the red balls doubles.Find :--- (1) P(red balls) (2) P ( blue ball) (3) P ( blue ball if 5 extra red balls are actually added ).?.
Solutions for A bag contains 15 balls of which x are blue ball and the remaining are red . If the number of red balls are increased by 5 , the probability of drawing the red balls doubles.Find :--- (1) P(red balls) (2) P ( blue ball) (3) P ( blue ball if 5 extra red balls are actually added ).? in English & in Hindi are available as part of our courses for Class 10.
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