A bag contains five blue marbles, five red marbles, five white marbles, and five black marbles. You draw a marble from the bag, and then draw another marble from the bag. Do you have a greater probability of drawing two blue marbles if you replace the first marble after you draw it or if you don't replace the marble?

Your answer should include both the mathematical proof and a written statement.

hey! i think i know the answer haha.

3x-x+2=4. hope that helps!

Step-by-step explanation:

20 marbles in all of which: 5 are blue, 5 are red, 5 are white and 5 are black.Drawing 2 marbles without replacin the 1st blue drawn.

P(2 blue) = 5/20*4/19 = 20/380 = 1/19.Drawing 2 marbles with replacement of the first blue one

P(2 blue) = 5/20*5/20 = 25/400 = 1//16. 1/16 > 1/19

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

typically if you want , you would actually put questions here for us to you. but hey, that's just me. maybe someone will magically answer the questions for you without knowing what they are.