What is the probability of picking at least two blue marbles?

Question

A bag contains five red marbles and six blue marbles. A sample of three marbles is selected.

I am looking more for how to do this than the actual answer. I am using combinations. I know how to do at least one blue marble, 1-5 choose 3/11 choose 3. But I can't figure out how to do it with at least two.

Just to let you people posting these rude and smart aleck answers, I am reporting each one of you.

Answer 1

there are 11 marbles total. Select 3 at time, the total possible outcome is (11 * 10 * 9) / (3)! = 165 ways.

find the favor combinations.
there are two ways you can pick.
Blue Blue Red
Blue Blue Blue

find the combination of 2 BLue and 1Red
(6*5)/2 * 5 = 75 out comes

find the combination of 3 Blue
(6*5*4)/3! = 20 outcomes

so the probabiltity is (20 + 75) /165

= 19/33 or 57.575%

Answer 2

The following choices meet the condition:
RBB
BRB
BBR
BBB

The first three each have the same probability:
5/11 * 6/10 * 5/9 = 5/33
So the probability of two blue is 5/33 * 3 ways to do it = 5/11

BBB has this probability (with only one way to do it):
6/11 * 5/10 * 4/9 = 4/33

Add the probability of getting two to the probability of getting three 5/11 + 4/33 = 19/33

That's more than half, which is intuitively right, because there are more blue marbles than red.

To check, run the other possibilities that do not meet the condition:
RRB
RBR
BRR
RRR

Condition with two red marbles has probability of:
5/11 * 4/10 * 6/9 = 4/33
Three ways to do it makes 4/11

RRR has probability of 5/11 * 4/10 * 3/9 = 2/33

Probability of not getting at least two blue is 4/11 + 2/33 = 14/33

19/33 (success) + 14/33 (not) = 33/33
Checks

Answer 3

available = 5 red 6 blue
you are sampling without replacing (so after you pick one there are less to choose from)

if you choose at least two blue marbles in a selection of three then you can look at this as either
- choose 2 blue 1 red or choose 3 blue
- dont (choose 1 blue, or no blue)
which way you go is a matter of taste and ease depending on what the number are.

0 blue = 5/11 * 4/10 * 3/9 = 5C3 6C0 / 11C3
1 blue = 5/11*4/10*6/9 * 3 = 5C2 5C0 / 11C3
(because the blue could come any time)
P(<2 blue) = add them = 21*5*4/(11*10*9)

Answer 4

I could explain it but it woud be easier for you to follow this link which provides how to do those problems in detail.Hope this helps.

http://mathforum.org/library/drmath/view/56491.html

Answer 5

C (n,r) = n!/((n-r)!*r!)