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BMAD

A boy playing with his balls

Question

A boy has four red balls and eight blue balls. He arranges his twelve balls randomly, in a ring. What is the probability that no two red balls are adjacent?

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11 answers to this question

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p = 7/33 = 0.2121 ....

Place eight black balls in a circle.

There are 8 locations to place a red ball in the circle. None of them put two red balls together.

p1 (placing red ball 1 correctly) = 8/8

There are 9 locations to place another red ball in the circle. Two of them make adjacent red balls.

p2 (placing red ball 2 correctly) = 7/9

There are 10 locations to place another red ball in the circle. Four of them make adjacent red balls.

p3 (placing red ball 3 correctly) = 6/10

There are 11 locations to place another red ball in the circle. Six of them make adjacent red balls.

p4 (placing red ball 4 correctly) = 5/11

p (no adjacent red balls) = 8/8 x 7/9 x 6/10 x 5/11 = 7/33

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Are the balls identical? If not, i think this should work, though not sure.

8C4 / 12C4

The balls are identical in size.

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Clearly, we can generalize to r red balls and b ge.gif r − 1 blue balls, arranged in a line. For balls arranged in a ring, where we require b ge.gif r, replace b by b − 1 in the working below.

We have r − 1 fixed blue balls, and r red balls (dividing lines), around which the remaining b − r + 1 blue balls must be distributed.

Therefore, for balls arranged in a line, the probability that there are no two adjacent red balls is:

b+1Cr / b+rCr = [b! (b + 1)!] / [(b + r)! (b + 1 − r)!]

Hence, for example, in a lottery where 6 balls are drawn (without replacement) from balls numbered 1 through 49, the probability that no two winning numbers are consecutive is:

(43! 44!) / (49! 38!) approx.gif 0.5048.

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What if the question was, no. of ways you can form a ring with the balls?

Are we to take your question that each ball is unique or that there are identical balls with two distinct colors?

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I have it being 33 total combinations. Assuming there is a definite start and stop in the ring. I would possible remove some combinations if rotational symmetry is considered.

Edited by BMAD

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