Please help

in Statistics Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

Let the letters a, b, c, d, e, f represent the numbers 1-6, but not necessarily a=1, b=2, etc. The three dice are thrown. The first shows value a. There are 10 ways to get another a (and no more):

aab, aac, aad, aae, aaf, aba, aca, ada, aea, afa.

And there are another 5 ways where the other two dice duplicate their values:

abb, acc, add, aee, aff.

So that’s 15 duplications out of a possible 36 combinations of the second and third dice, where a is the value of the first die.

What applies to a symmetrically applies to the other values b-f. So the probability is 15/36=5/12. (Or we could simply state this as 90/216 for all possible outcomes. That is, out of the 216 possible outcomes for throwing three dice, 90 of them contain exactly one pair of duplicated values.)

by Top Rated User (680k points)

Related questions

Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
83,674 questions
88,560 answers
1,985 comments
5,772 users