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A bag contains 6 red marbles, 6 blue marbles, 10 white marbles and 7 yellow marbles. You are asked to draw 4 marbles from the bag without replacement. In how many ways can you draw two blue marbles? what formula do you use for questions like these

Guest Nov 8, 2015

Best Answer 

 #1
avatar+90988 
+5

A bag contains 6 red marbles, 6 blue marbles, 10 white marbles and 7 yellow marbles. You are asked to draw 4 marbles from the bag without replacement. In how many ways can you draw two blue marbles? what formula do you use for questions like these

 

I am going to play with this question a little

If ALL the b***s are different. I mean, you have blue 1 blue 2 etc then

 

Number of ways of choosing 2 blue from 6 is 6C2 = 15

Number of ways of choosing 2 from the others 23C2 = 253

Total = 15*253 = 3795 ways

The number of ways that 4 marbles can be drawn with no restrictions is 29C4 = 23751

 

So the probablility of drawing 2 blue is    3795 / 23751 = 0.1598    (4 dp)

This is the way it must be done if you are looking at probablilities

 

BUT you are not concerned with probablilities and I think below is what the question may really be asking.

 

 

Lets see,

6 red marbles, 6 blue marbles, 10 white marbles and 7 yellow marbles

you could have 

 

 

BlueWhiteRedYellowTHE 4 B***S
BBWW  BBWW
BBWR BBWR
BBW YBBWY
BB RR BBRR
BB RYBBRY
BB  YYBBYY
    6 WAYS

 

So there are 6 different combinations that could include 2 blue b***s BUT they are NOT all equally likely to be drawn.  laugh

Melody  Nov 8, 2015
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2+0 Answers

 #1
avatar+90988 
+5
Best Answer

A bag contains 6 red marbles, 6 blue marbles, 10 white marbles and 7 yellow marbles. You are asked to draw 4 marbles from the bag without replacement. In how many ways can you draw two blue marbles? what formula do you use for questions like these

 

I am going to play with this question a little

If ALL the b***s are different. I mean, you have blue 1 blue 2 etc then

 

Number of ways of choosing 2 blue from 6 is 6C2 = 15

Number of ways of choosing 2 from the others 23C2 = 253

Total = 15*253 = 3795 ways

The number of ways that 4 marbles can be drawn with no restrictions is 29C4 = 23751

 

So the probablility of drawing 2 blue is    3795 / 23751 = 0.1598    (4 dp)

This is the way it must be done if you are looking at probablilities

 

BUT you are not concerned with probablilities and I think below is what the question may really be asking.

 

 

Lets see,

6 red marbles, 6 blue marbles, 10 white marbles and 7 yellow marbles

you could have 

 

 

BlueWhiteRedYellowTHE 4 B***S
BBWW  BBWW
BBWR BBWR
BBW YBBWY
BB RR BBRR
BB RYBBRY
BB  YYBBYY
    6 WAYS

 

So there are 6 different combinations that could include 2 blue b***s BUT they are NOT all equally likely to be drawn.  laugh

Melody  Nov 8, 2015
 #2
avatar+78577 
+5

We want to draw 2 blue ones out of 6 and out of the other 23, draw any 2 of them.

 

So.....the total number of ways to do this is :

 

C(6,2) * C(23,2)  = 3795 ways

 

 

cool cool cool

CPhill  Nov 8, 2015

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