# Probability of randomly selecting three big blue marbles?

## Homework Statement

A bag contains 40 blue marbles, 40 red marbles, 20 green marbles. 60 of the marbles are big. Three marbles are selected without replacement.

What is the probability of randomly selecting three big blue marbles?

## Homework Equations

p(x,y/theta=p(x)p(y/x)=p(y)p(x/y)

## The Attempt at a Solution

(40/60)(39/59)(38/58)/(40/100)(39/99)(38/98)

first being big blue: p1 = (40/60)(60/100)
second: p2 = (39/60)(59/99)
third: p3 = ...

all 3 = p1*p2*p3
... right?

wouldn't p2 = (39/59)(59/99) since there is 1 less big marble left?

def (thats what i intended), sorry.
p2 = (39/59)(59/99)

after multiplying p1*p2*p3 do I need to divide the sum by another number?

Why would you think that you need to divide by another number?

Because the answer when p1*p2*p3=.061. yet my worksheet has choices of a. .0125, b. .0240, c. .2323, d. .2400

well on second thought you are write, it can't be the above.

B=blue, b=big.

P(B)*P(b) is the probability of the marble being big and blue and it initially = 40/100*60/100 since being big and being blue are independent of each other.

So I get 40/100*60/100*39/99*59/99*38/98*58/98 for all 3 to be B&b, since they are drawn in succession and without replacement. That = 0.0129, which is close to A?

Thanks I got it and I really appreciate your help!

so was A the answer? I am just curious.

yes thanks!