A bag contains 15 ball of which X are blue and the remaining are red . If yhe number of balld are increased by 5, the probability of drawing the red ball doubles, find (1) P(red ball) (2) P (Blue ball) (3) P( Blue ball it of 5 extra ted balls are actually added)
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In First Case,
No Of Balls = 15
No Of Blue Balls = x
No Of Red Balls = 15 - x
Probability of taking a red ball is; P(E) = 15 - x
15
After 5 red balls are added,
No of balls = 15 + 5 = 20
No of blue balls = x
No of Red Balls = (15 - x) + 5 = 20 - x
Probability of taking a red ball is; P(E) = 20 - x
20
According to your question, 15 - x x 2 = 20 - x
15 20
30 - 2x = 20 - x
15 20
600 - 40 x = 300 - 15 x
300 =25 x
x = 12
No Of Balls = 15
No Of Blue Balls = x
No Of Red Balls = 15 - x
Probability of taking a red ball is; P(E) = 15 - x
15
After 5 red balls are added,
No of balls = 15 + 5 = 20
No of blue balls = x
No of Red Balls = (15 - x) + 5 = 20 - x
Probability of taking a red ball is; P(E) = 20 - x
20
According to your question, 15 - x x 2 = 20 - x
15 20
30 - 2x = 20 - x
15 20
600 - 40 x = 300 - 15 x
300 =25 x
x = 12
Answers
Answered by
173
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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