
Probability and Statistics Basic Probability and Statistics Math Forum 
 LinkBack  Thread Tools  Display Modes 
June 4th, 2015, 08:21 PM  #1 
Newbie Joined: Jun 2015 From: Kiribati Posts: 5 Thanks: 0 Math Focus: Statistics  Probability problem
I am in high school and my Maths skills is worst. Two cubical dice are thrown and their scores added together. If X = "The sum of the scores on the two dice", what is P(X is divisible by 4)? I know that 6*6= 36 because there were two dice. My answer is this ?/36 I can't find the numerator for this one it is pretty hard for sure in my capability. 
June 4th, 2015, 08:46 PM  #2 
Senior Member Joined: Oct 2013 From: Far far away Posts: 422 Thanks: 18 
There are 6 * 6 = 36 possible permutations The permutations that are divisible BY 4 are: (1,3), (3, 1), (2, 2), (2, 6), (6, 2), (3, 5), (5, 3), (4, 4)...a total of 8. So P(sum divisible by 4) = 8/36 = 2/9 
June 4th, 2015, 09:05 PM  #3  
Newbie Joined: Jun 2015 From: Kiribati Posts: 5 Thanks: 0 Math Focus: Statistics  Quote:
and (1,3) and (3,1) aren't the same? I thought you should only take (1,3), (2,2), (2,6), (3,5), (4,4) anyway thanks again. Can I get a brief explanation why (1,3) and (3,1) are both counted or both pemutations.  
June 5th, 2015, 04:30 AM  #4 
Senior Member Joined: Feb 2010 Posts: 694 Thanks: 132  
June 5th, 2015, 04:38 AM  #5  
Senior Member Joined: Feb 2010 Posts: 694 Thanks: 132  Quote:
 
June 5th, 2015, 06:00 AM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,446 Thanks: 2499 Math Focus: Mainly analysis and algebra 
The main reason you have to count $\{1,3\}$ and $\{3,1\}$ is that your 36 possibilities which you identified in the denominator include both $\{1,3\}$ and $\{3,1\}$.

June 7th, 2015, 07:02 PM  #7 
Senior Member Joined: Oct 2013 From: Far far away Posts: 422 Thanks: 18  

Tags 
probability, problem 
Search tags for this page 
Click on a term to search for related topics.

Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Probability Problem / Insurance problem  bsjmath  Advanced Statistics  2  March 1st, 2014 02:50 PM 
Another Probability Problem  p99410  Probability and Statistics  9  January 28th, 2014 11:29 PM 
Probability Problem (Not sure)  Jakarta  Probability and Statistics  6  June 14th, 2012 08:47 PM 
Probability problem. Plz help me?  sysiphus  Probability and Statistics  1  February 28th, 2010 04:30 PM 
Help with a probability problem  al_teothc  Algebra  1  October 19th, 2009 02:48 PM 