Probability Two tetrahedral dice are thrown

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Homework Statement


Two tetrahedral dice (four-sided dice) are thrown. What is the probability that the sum of the scores is:
a) even
b) prime
c) even or prime?

Homework Equations





The Attempt at a Solution


a) P(even) = 1/2
b) P(prime) = 9/16

c) c for confused :confused:

Can someone please explain the theory behind answering Question c?

Cheers.
 
on Phys.org
Have you written out the set of "outcomes"? A tetrahedron has four sides so it has the numbers 1, 2, 3, and 4. The possible outcomes when two are thrown are (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), and (4, 4). There are, of course, [itex]4^2= 16[/itex] outcomes and each is equally likely. How many have sums that are either even or prime?
 
HallsofIvy said:
Have you written out the set of "outcomes"? A tetrahedron has four sides so it has the numbers 1, 2, 3, and 4. The possible outcomes when two are thrown are (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), and (4, 4). There are, of course, [itex]4^2= 16[/itex] outcomes and each is equally likely. How many have sums that are either even or prime?

Ah... 2 is a prime number too.. I see now.
My mind has gone mad from all these probability questions..
 
It wasn't until now that I worked out the answer to (c) myself and I was very surprised at the answer!
 
HallsofIvy said:
It wasn't until now that I worked out the answer to (c) myself and I was very surprised at the answer!
There is an easy way to find this probability:
How many odd, non-prime numbers are possible as sum?
All numbers which are not [odd and not prime] are even or prime.
 

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