0 like 0 dislike
0 like 0 dislike
Pairs probability

2 Answers

0 like 0 dislike
0 like 0 dislike
I get C(9,4)×8!×10!/18! ≈ 1/347. Simulation suggests that this figure is correct.

There are C(9,4) distinct orderings of RRRRGGGGG. For each ordering, there are 8! ways to order the red balls and 10! ways to order the green balls. So C(9,4)×8!×10! ways to get 9 pairs. Out of 18! possible orderings in total.
0 like 0 dislike
0 like 0 dislike
Two methods come to mind:

1. Consider all 18! orders in which you might draw the balls. We want to count the number of orders which give the correct result. To decide on a good order you first decide which of the 9 pairs are the red pairs, then choose the relative order of the 8 red balls and then the relative order of the 10 green balls. Then divide the number of ways by 18!.

2. Let P(r, g) be the probability of succeeding if you start with r red balls and g green balls. You can set up a recurrence relating P(r, g) to P(r-2, g) and P(r, g-2), and you have base cases P(0, g) = P(r, 0) = 1, with that you can work your way up to P(8, 10). (Excel is helpful here)

I did both ways and they give the same answer.

No related questions found

33.4k questions

135k answers


33.7k users

OhhAskMe is a math solving hub where high school and university students ask and answer loads of math questions, discuss the latest in math, and share their knowledge. It’s 100% free!