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There are 5 blue balls and 4 red balls arranged in a circle. What is the probability that no two red balls are together?

 Dec 18, 2020

Best Answer 

 #3
avatar+112863 
+1

I managed to do this question the wrong way around.

I have 5 red and 4 blue beads. 

oh dear, they are meant to be balls, never mind, mine are balls with holes to turn them into beads.

 

I have thought of it as a bangle so rotations are the same.

Plus the bangle can put on either way so that cuts down on possibilities as well.

 

At first I just thought of it as a standard permutation question but it became clear to me that standard 'simple' formula methods were not going to work.

So I resorted to drawing all the possibilities.

This is what I came up with.

If anyone says I have missed any, or that I have double-counted I am all ears.

 

Only one of these has all the blue separate.

So if you know anything at all about probability you can finish it yourself.

 Dec 20, 2020
 #1
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1/9

Becuase if you add all the balls you get nine,

then i subtracted the red balls from the blue balls to get one

so all the balls together are the deniminator and the red balls subtracted from the blue balls are the numerator

Answer=1/9

 Dec 19, 2020
 #3
avatar+112863 
+1
Best Answer

I managed to do this question the wrong way around.

I have 5 red and 4 blue beads. 

oh dear, they are meant to be balls, never mind, mine are balls with holes to turn them into beads.

 

I have thought of it as a bangle so rotations are the same.

Plus the bangle can put on either way so that cuts down on possibilities as well.

 

At first I just thought of it as a standard permutation question but it became clear to me that standard 'simple' formula methods were not going to work.

So I resorted to drawing all the possibilities.

This is what I came up with.

If anyone says I have missed any, or that I have double-counted I am all ears.

 

Only one of these has all the blue separate.

So if you know anything at all about probability you can finish it yourself.

Melody Dec 20, 2020

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