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I am using permutation knowledge to determine number of successful outcomes in following situation, but I am finding that my calculations are wrong according to solution at back of book.

QUESTION: What is wrong about my approach?

PROBLEM: 7 white balls and 3 black ball are placed randomly in a row. What is the probability that no 2 black balls are next to each other?

My approach is as below.

Total number of ways of placing in a row= 10!/(7! X 3!) = 120

We find number of ways in which 2 or 3 black balls are next to each other. We put 2 black balls together and consider it as one object. We now arrange 7 white balls, 1 black ball and 1 grouped black ball. 1 X 9!/(7! ) = 72 (Note: the grouped black balls can only be arranged in 1 way)

So, number of ways in which 2 or 3 black balls are not together = 120 - 72 = 48

Therefore, probability that 2 or 3 balls are not together is 48 / 120 = 0.4

BUT the answer in back of book is given as 7/15 which is .47.

1 Answers 1

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As you calculate, there are $120$ ways to place them if you don't care.

To place them such that no black balls are adjacent, place the white balls down first. Then in the eight gaps between or to the side of the white balls, choose three of them to be occupied by black balls.

There are then $\binom{8}{3}=56$ ways to arrange black and white so that blacks don't touch and therefore the probability is $\binom{8}{3}/120=56/120=7/15$

As for why your answer is incorrect, you have overcounted how many violate the condition. Note that $B(B\!\!_B)WWWWWWW$ is the same arrangement as $(B\!\!_B)BWWWWWWW$ but you have counted this twice. There are exactly eight times that you overcounted in this fashion (where all three blacks are adjacent). Correcting this and counting each only once, you have $120-(72-8)=56$ ways to arrange black and white so that blacks don't touch, matching the answer I already gave.

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    Thanks for your very nice explanation. If I used my approach then I think I would need to find number of ways when 3 black balls are together and then number of ways in which WBBW is always together do we get for the of only 2 black balls being together. Does that sound correct to you? – 2017-01-23
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    @Sunil but then you are missing where the two adjacent blacks are at one of the edges... – 2017-01-23
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    Wow. You are so good at this. I need to get my mind around what you said and understand this more. – 2017-01-23
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    So now I get it. If I used my approach I would need to determine for BBB and for WBBW and for BBW on left end and WBB on right end. This gives me the same answer as your approach. So now I fully understand where my original logic was wrong. I appreciate your excellent help. – 2017-01-23