A boy playing with his balls

[BMAD]

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?

[bonanova]

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

[vistaptb]

Dat title

[dark_magician_92]

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

8C4 / 12C4

Edited May 11, 2013 by dark_magician_92

[BMAD]

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

8C4 / 12C4

The balls are identical in size.

[dark_magician_92]

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

8C4 / 12C4

The balls are identical in size.

My answer was wrong, anyhow. WIll give a shot soon.

[Aaryan]

I'll be honest - I didn't come to this page to answer a riddle.

[BMAD]

Clearly, we can generalize to r red balls and b r − 1 blue balls, arranged in a line. For balls arranged in a ring, where we require b 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+1C_r / _b+rC_r = [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!) 0.5048.

[dark_magician_92]

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

[BMAD]

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?

[dark_magician_92]

all identical except colour. 8 blue, 4 red

[BMAD]

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 May 15, 2013 by BMAD