# SOLUTION: two dice are thrown, the probability that the number on the red exceeds the number showing on the green by exactly two is a) 1/18 b) 1/4 c) 1/9 d)1/36 e) 1/24

Algebra ->  Algebra  -> Probability-and-statistics -> SOLUTION: two dice are thrown, the probability that the number on the red exceeds the number showing on the green by exactly two is a) 1/18 b) 1/4 c) 1/9 d)1/36 e) 1/24      Log On

 Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!

 Algebra: Probability and statistics Solvers Lessons Answers archive Quiz In Depth

 Question 208632: two dice are thrown, the probability that the number on the red exceeds the number showing on the green by exactly two is a) 1/18 b) 1/4 c) 1/9 d)1/36 e) 1/24Found 2 solutions by stanbon, Edwin McCravy:Answer by stanbon(57387)   (Show Source): You can put this solution on YOUR website! two dice are thrown, the probability that the number on the red exceeds the number showing on the green by exactly two is a) 1/18 b) 1/4 c) 1/9 d)1/36 e) 1/24 ----------------------------------------------- # of possible outcomes with two dice: 36 patterns with red exceeding green by 2:1;4,2;5,3;6, ========================================= Probability is 4/36 = 1/9 ==================================== Cheers, Stan H. ----------------- PS: Respond to stanbon@comcast.net as the algebra.com feedback program is down. --------------------------- Answer by Edwin McCravy(8909)   (Show Source): You can put this solution on YOUR website!Two dice are thrown, the probability that the number on the red exceeds the number showing on the green by exactly two is a) 1/18 b) 1/4 c) 1/9 d)1/36 e) 1/24 ``` Here are all the possible rolls with a red die and a green die: (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) There are 36 possible rolls. Now I will underline just those in which the number on the red exceeds the number showing on the green by exactly two: (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) There are 36 possible rolls. So there are 4 possible rolls out of the 36 which have that property. Therefore the probability is which reduces to Edwin```