I'm posting one puzzle, riddle, math, or statistical problem a day. Try to answer each one and post your answers in the comments section. I'll post the answer the next day. Even if you have the same answer as someone else, feel free to put up your answer, too!
Tuesday, March 16, 2010
I am Feeling a Little Bit Probabilistic Today
A boy has four red marbles and eight blue marbles. He arranges his twelve marbles randomly, in a ring. What is the probability that no two red marbles are adjacent?
The total number of possible ball combinations where 4 are red and 8 are blue is: 12!/(8! * 4!) = 495
The number of combinations where red doesn't repeat is 105.
I'm sure there's a formula for this, but I just counted out each scenario. For example, if marble 1 is red, there are 35 ways to have 3 more red marbles that aren't consecutive. The same is true for marble 2 being red. Then for marble 3 being red, there are 20 ways without having consecutive red marbles or a red marble 1. Etc...you eventually end up with:
35+35+20+10+4+1=105
So that makes the probability that no two red marbles are adjacent:
Nice work Andy. I came at it a little bit differently, but came up with the same answer.
Select any blue marble, then arrange the 11 remaining marbles in a line. That's the same as putting them into a ring.
The number of ways to choose k out of n is (N choose k) n!/(k!(n-k)!)
Choosing 4 red marbles out of 11 is 330.
I basically did the same as you, Andy, and counted up the number of ways to divide up the marbles, and came up with 70 distinct combinations (8 choose 4).
Therefore the probability that no two red marbles are adjacent is 70/330 = 7/33.
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I came up with 7/33 ~ 21.2%
ReplyDeleteThe total number of possible ball combinations where 4 are red and 8 are blue is:
12!/(8! * 4!) = 495
The number of combinations where red doesn't repeat is 105.
I'm sure there's a formula for this, but I just counted out each scenario. For example, if marble 1 is red, there are 35 ways to have 3 more red marbles that aren't consecutive. The same is true for marble 2 being red. Then for marble 3 being red, there are 20 ways without having consecutive red marbles or a red marble 1. Etc...you eventually end up with:
35+35+20+10+4+1=105
So that makes the probability that no two red marbles are adjacent:
105/495 = 7/33
Nice work Andy. I came at it a little bit differently, but came up with the same answer.
ReplyDeleteSelect any blue marble, then arrange the 11 remaining marbles in a line. That's the same as putting them into a ring.
The number of ways to choose k out of n is (N choose k) n!/(k!(n-k)!)
Choosing 4 red marbles out of 11 is 330.
I basically did the same as you, Andy, and counted up the number of ways to divide up the marbles, and came up with 70 distinct combinations (8 choose 4).
Therefore the probability that no two red marbles are adjacent is 70/330 = 7/33.