Question

A bag contains 4 red, 3 blue and 9 green balls. How many green balls should be removed such that the
difference between propability of drawing a green ball and probability of drawing a red ball is 1/12

Answer 1

Answer:

Red balls =4

Blue balls = 3

Green balls=9

so, if 4 green balls will be removed out then the difference between the probability of drawing a green ball and probability of drawing a red ball is 1/12

here is the verification:

after removing out 4 green balls we will have

4 red balls

3 blue balls

5 red balls

now,

P(G)=5/12 and P(R)=4/12

difference will be,

=>5/12 - 4/12

=>1/12

Answer 2

Given:

A bag contain 4 red , 3 blue and 9 green balls.

To Find :

No. of green balls should be removed so that difference between propability of drawing a green ball and probability of drawing a red ball is 1/12

Solution:

Let us remove x green balls from the bag

Now ,

• No. of green balls = (9-x)
• No. of blue balls = 3
• No. of red balls = 4

Total number of balls = 9-x + 4 + 3

= (16-x)

Probability of getting 1 green ball

$= \sf{ \frac{9 - x}{16 - x}}$

Probability of getting 1 red ball

$\sf{ \frac{4}{16 - x} }$

$\bold{ \frac{9 - x}{16 - x} - \frac{4}{16 - x} } \: = \frac{1}{12} \\ \\ \bold{ \: \implies \frac { (9 - x) - 4}{(16 - x)}} = \frac{1}{12} \\ \\ { \bold { \implies \frac{5 - x}{16 - x} = \frac{1}{12} }} \: \\ \\ \bold{ 60 - 12x = 16 - x} \: \sf(by \: cross \: multiplication) \\ \bold { \: - 12x + x = 60 - 16} \\ \bold {\implies{ \: 11x = 44}} \\ \bold {x = { \frac{44}{11} }} \\ \\ \bold {\boxed {\underline{\sf \red {\underline{ = \: 4 \: balls}}}}}$