
Probability and Statistics Basic Probability and Statistics Math Forum 
 LinkBack  Thread Tools  Display Modes 
March 5th, 2018, 04:15 PM  #1 
Member Joined: Oct 2017 From: Rumba Posts: 34 Thanks: 0  Probability question
Three fair dice are rolled: one red die, one blue die and one green die. Let R, B and G be the numbers respectively shown by the red, blue and green die. Compute the probability P {R < B < G} Last edited by sita; March 5th, 2018 at 04:31 PM. 
March 5th, 2018, 06:53 PM  #2 
Senior Member Joined: Jul 2012 From: DFW Area Posts: 628 Thanks: 92 Math Focus: Electrical Engineering Applications 
Hi Sita, I think that this problem can be done by brute force, especially if each die has six sides, as is typical. I worked the problem by starting with the green die. Obviously, for G=1,2 there are no possibilities. For G=3, then only G=3, B=2, R=1 (denoted as 321) gives a true outcome. For G=4, then 421, 431, and 432 give a true outcome. For G=5, then 521, 531, 532, 541, 542, and 543 give a true outcome. Can you do likewise for G=6, and can you calculate the total number of possibilities? Adding the number of true outcomes for G=3,4,5,6 and dividing by the total number of possibilities will give the answer. 
March 5th, 2018, 07:09 PM  #3  
Senior Member Joined: Sep 2015 From: USA Posts: 2,197 Thanks: 1152  Quote:
First let's determine how many groupings include equality. There are 6 ways all three dice are equal. There are $(3)(6)(5) = 90$ ways that only 2 of the dice are equal. $3 = \dbinom{3}{2}$ There are 6 numbers that the two dice that are equal could be. There are 5 numbers that the third die could be and not equal the other two. So there are at total of 96 rolls that involve equality. There are $6^3 = 216$ total possible rolls. Thus there are $21696 = 120$ rolls where strict inequality is observed. Now... the problem is symmetric in the colors and thus each ordering will get 1/6 of these rolls, as there are 6 ways to arrange RGB. Thus the particular ordering of $R < B < G$ will have $\dfrac{120}{6}=20$ associated rolls and thus $P[R< B< G] = \dfrac{20}{216}= \dfrac{5}{54}$  
March 15th, 2018, 05:49 PM  #4 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 622 Thanks: 85 
What's quicker than what romsek said is that the answer is 6C3 = 20. In combinations, the order matters, and an item can only be chosen once, which is what we want.


Tags 
probability, question 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Probability Question ?  jbergin  Probability and Statistics  1  January 2nd, 2015 07:42 PM 
probability question  shin777  Probability and Statistics  1  March 26th, 2014 07:44 PM 
probability question  hoyy1kolko  Probability and Statistics  2  July 27th, 2011 06:36 AM 
Probability Question help!  Azntopia  Advanced Statistics  1  July 25th, 2011 11:10 AM 
Probability Question 2  duke  Advanced Statistics  2  November 4th, 2007 09:18 AM 