Question

A bag contains 15 balls of which X are blue and remaining are red.If the number of red balls are increased by 5,the probability of drawing the red balls doubles.Find : i) P (red balls)
ii) P(blue balls)
iii) P(blue ball it of 5 extra red balls are actually added)

Answer 1

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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Answer 2

⇒  Number of blue balls =15

⇒  Number of blue balss =x

⇒  Number of red balls   =15−x

∴  Probability of taking a red ball =  15−x /15

 

Now, after 5 red balls are added,

⇒  Number of balls =15+5=20

⇒  Number of blue balls =x

⇒  Number of red balls =(15−x)+5=20−x

⇒  Probability of taking a red ball =  20−x /20

According to the question,

15−x /15 × 2 =  20−x /20

30−2x /15 =  20−x/20

600−40x=300−15x

300=25x

∴    x=  300 /25 =12

∴  We have 12 blue balls and 3 red balls .

(i)  P(red ball)=  3/15 = 1/5

(ii)  P(blue balls)=  12 /15  = 4/5

(iii)  P(blue ball if 5 balls are actually added) =  12 /15+5  = 12/20 =   3/5