SOLUTION: A bag initially had blue, red, and purple gumballs in the ration 2 : 3 : 5. Let N be the number of red gumballs added to the bag, and 2N be the number of purple gumballs added to

Question 1167077: A bag initially had blue, red, and purple gumballs in the ration 2 : 3 : 5. Let N be the
number of red gumballs added to the bag, and 2N be the number of purple gumballs
added to the bag. How many blue gumballs we need to add , in the bag so that the
probability of drawing a blue gumball, a red gumball and a purple gumball is now 1/6,
2/6, and 3/6, respectively?
A) is equal to N
B) is equal to 2N
C) is less than N
D) is greater than N but less than 2N
E) cannot be determined

Answer by greenestamps(13206) (Show Source): You can put this solution on YOUR website!

A curious problem -- I have to wonder if the information given is not correct.

Defining N as the number of red gumballs added to the bag suggests that N should be a positive integer; but it turns out to be negative....

Given the original ratio 2:3:5...

let 2x = number of blue originally
let 3x = number of red originally
let 5x = number of purple originally

let A be the number of blue to be added. Then after adding more gumballs,

2x+A = number of blue
3x+N = number of red
5x+2N = number of purple

After more gumballs are added, the probability of drawing a blue is 1/6, the probability of drawing a red is 2/6, and the probability of drawing a purple is 3/6. That means the number of red is 2 times the number of blue, and the number of purple is 3 times the number of blue.

So now

So the answer to the problem as stated is answer A: the number of blue gumballs to be added is equal to N.

But now let's put some actual numbers in the problem.

The number of blue gumballs is now 2x+N, and the number of red gumballs is 3x+N.

And since the probability of drawing a red is 2/6 while the probability of drawing a blue is 1/6, the number of red gumballs is twice the number of blue gumballs:

Now what we have is this:

original number of blue gumballs: 2x
original number of red gumballs: 3x
original number of purple gumballs: 5x

number of blue gumballs "added": -x
number of red gumballs "added": -x
number of purple gumballs "added": -2x

ending number of blue gumballs: 2x-x = x
ending number of red gumballs: 3x-x = 2x
ending number of purple gumballs: 5x-2x = 3x

And we see from this that the final condition of the problem is satisfied:

P(blue) = x/6x = 1/6
P(red) = 2x/6x = 2/6
P(purple) = 3x/6x = 3/6

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