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Five balls (2 red , 1 blue , 1 green , 1 yellow )
are arranged in a circle, Find the probability that the red balls are not together
please help me with the answer, my exam is about 11 hours
Wisdom is a tree which grows in the heart and fruits on the tongue
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In circular type questions the first red may be anywhere as all spaces are equal. So now consider how to place the rest.
The space to the left of the red may be any of 3. We'll assume we use one of these. The space to the right of the red may be any from the remaining two. Now to place the rest. Two ways to do that.So that's three x two x two = 12 ways.
Now to consider the total number of possibilities.
We've placed one red. Left may be one from four and right one from three and once again two ways to place the rest. That's 24 ways.
So P = 12/24
Bob
Good luck with the exam.
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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