+0  
 
0
46
2
avatar

1- What is the probability of an odd number of sixes turning up in a random toss of 10 fair dice?
2-What is the probability of an even number of sixes turning up in a random toss of 10 fair dice?
3-Are the two numbers equal? Why, or why not?
Thank you for helping.

Guest Aug 18, 2018
 #1
avatar+88775 
+1

1- What is the probability of an odd number of sixes turning up in a random toss of 10 fair dice?

We  can either have 1 - 3 - 5 - 7  or  9 6s

We have a (1/6) probability of a six on any roll  and a  (5/6) probability of  "not 6" on any roll...so....

P( one 6)  = C(10,1)(5/6)^9 (1/6) = 9765625/30233088  ≈ 32.3%

P(three 6s)  =  C(10,3) (5/6)^7 ( 1/6)^3   =  390625/2519424 ≈ 15.5%

P ( five 6s)  = C(10,5)(5/6)^5 (1/6)^5   = 21875/1679616 ≈ 1.3%

P ( seven 6s)  = C(10,7) (5/6)^3 (1/6)^7  = 625/2519424 ≈ .02%

P ( nine 6s)  = C(10,9)(5/6) (1/6)^9   = 25/30233088 ≈  0%

 

Adding all the fractions, we get   58025/118098  ≈ 49.1%

 

2-What is the probability of an even number of sixes turning up in a random toss of 10 fair dice?

We can have 2 - 4 - 6 - 8  or 10  6s

P(two 6s) =   C(10,2) (5/6)^8 (1/6)^2  = 1953125/6718464 ≈ 29.1%

P(four 6s)  =  C(10,4) (5/6)^6(1/6)^4  = 546875/10077696 ≈ 5.42%

P( six 6s)  = C(10,6) (5/6)^4 (1/6)^6  = 21875/10077696 ≈ .21%

P(eight 6s) = C(10,8) (5/6)^2(1/6)^8  = 125/6718464 ≈ 0%

P (ten 6s)  =  (1/6)^10  = 1/60466176 ≈ 0%

 

Adding all these fractions, we get   20991751/60466176  ≈ 34.7%

 

They are not equal....notice that P( one 6 ) > P (two 6s)  and  P ( three 6s) > P (four 6s)  and P( five 6s) > P (six 6s)...etc.

Thus it appears that  P(2n - 1  6s)  > P(2n  6s)  for  n = 1 - 5  

 

cool cool cool

CPhill  Aug 18, 2018
 #2
avatar
0

This problem appears in a Mathematical Site on the Internet. It is classified under "Probability problem #113." Both the problem and the solution are posted there. Here is the main site:

http://www.cut-the-knot.org/content.shtml

Note: The small error in CPhill's counting of EVEN sixes neglected to take "0" into account as follows: C(10,0)*(5/6)^10*(1/6)^0 =0.1615056......., which would give a total of: 0.50867....etc., so that both odd sixes and even sixes must ADD up to 1.

Guest Aug 18, 2018
edited by Guest  Aug 18, 2018

22 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.