A bucket contains blue balls and yellow balls. the probability of removing two blue balls without replacement is 2/5 and he probability of removing three blue balls without replacement is 1/5. Why must there be two more blue balls then yellow balls? How many red balls were in the bucket. show that b=y+2

b/(b+y) * (b-1)/(b+y-1) = 2/5

b/(b+y) * (b-1)/(b+y-1) * (b-2)(b+y-2) = 1/5 = 1/2 * 2/5

so,

(b-2)/(b+y-2) = 1/2
2(b-2) = b+y-2
2b-4 = b+y-2
b = y+2

no idea about any red balls, but

(y+2)/(2y+2) * (y+1)/(2y+1) = 2/5

(y+2)/2(2y+1) = 2/5
5y+10 = 8y+4
3y = 6
y=2
so, b=4

4/6*3/5 = 2/5
2/5*2/4 = 1/5

To understand why there must be two more blue balls than yellow balls, let's analyze the problem step by step.

Let's assume that there are 'b' blue balls, 'y' yellow balls, and 'r' red balls in the bucket.

1. Probability of removing two blue balls without replacement is 2/5:
When we remove the first blue ball, there are (b-1) blue balls left in the bucket. After removing the second blue ball, there are (b-2) blue balls left. The total number of balls removed is 2.
So, the probability of removing the first blue ball is (b/(b+y+r)), and the probability of removing the second blue ball is ((b-1)/(b-1+y+r)).
Thus, according to the problem, the probability is (b/(b+y+r)) * ((b-1)/(b-1+y+r)) = 2/5.

2. Probability of removing three blue balls without replacement is 1/5:
When we remove the first blue ball, there are (b-1) blue balls left in the bucket. After removing the second blue ball, there are (b-2) blue balls left. Removing the third blue ball leaves (b-3) blue balls. The total number of balls removed is 3.
So, the probability of removing the first blue ball is (b/(b+y+r)), the probability of removing the second blue ball is ((b-1)/(b-1+y+r)), and the probability of removing the third blue ball is ((b-2)/(b-2+y+r)).
According to the problem, the probability is (b/(b+y+r)) * ((b-1)/(b-1+y+r)) * ((b-2)/(b-2+y+r)) = 1/5.

Now, let's analyze why there must be two more blue balls than yellow balls:

To solve these equations, let's create a system of equations based on the probabilities:

Equation 1: (b/(b+y+r)) * ((b-1)/(b-1+y+r)) = 2/5

Equation 2: (b/(b+y+r)) * ((b-1)/(b-1+y+r)) * ((b-2)/(b-2+y+r)) = 1/5

By simplifying equation 2, we get:

(b-2)/(b+y+r) = 1/2 --equation 3

Now, let's solve for the number of red balls:

Since the probability of removing two blue balls is 2/5, we can write equation 1 as:

b(b-1) = 2(b+y+r)(b-1+y+r)

Expanding and simplifying, we get:

b^2 - b = 2(b^2 + b(y+r) + y+r)^2

b^2 - b = 2b^2 + 2b(y+r) + 2(y+r)^2

Canceling out common terms, we get:

0 = b^2 + 2b(y+r) + 2(y+r)^2

Simplifying further:

0 = b^2 + 2by + 2br + 2y^2 + 2yr + 2r^2

Now, solving for the number of red balls (r):

2br = 0

Therefore, there must be 0 red balls in the bucket.

Simplifying equation 3, we get:

b-2 = b+y+r/2

b-2 = b + (y+r)/2

-(y+r) = 2

y + r = -2

Since the number of balls cannot be negative, we assume that (y+r) = 0, which means y = -r.

Substituting this into equation 3:

-(y+r) = 2

-(-r+r) = 2

0 = 2

This is not possible, so our assumption that (y+r) = 0 was incorrect.

Therefore, we conclude that there must be two more blue balls than yellow balls: b = y + 2.