new problem......lots of trouble with probability!!!!!!!!!
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pka
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A regular tetrahedron is a solid with four faces each of which is an equilateral triangle. That is your four-sided die. Tossed with a regular die there are 24 ordered pair outcomes.
(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).
Use these pairs to count the required events.
(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).
Use these pairs to count the required events.
Re: new problem......lots of trouble with probability!!!!!!!
Hello, Molly!
There must be more to the problem, but I'll take a guess.
\(\displaystyle \;\;\)(From your answer of "7", I assume you made the same guess.)
The six-sided die has faces marked {1,2,3,4,5,6}.
\(\displaystyle \;\;\)Both dice are rolled and their numbers are noted.
What is the probability that the two numbers differ by one?
There are \(\displaystyle 4\,\times\,6\:=\:24\) possible outcomes.
There are <u>seven</u> outcomes with a difference of one.
\(\displaystyle \;\;(\text{four, six})\;=\; (1,2),\,(2,3),\,(3,4),\,(4,5)\,(2,1),\,(3,2)\,(4,3)\)
Therefore: \(\displaystyle \,P(\text{diff. of one}) \;= \;\frac{7}{24}\)
Hello, Molly!
There must be more to the problem, but I'll take a guess.
\(\displaystyle \;\;\)(From your answer of "7", I assume you made the same guess.)
The four-sided die has faces marked {1,2,3,4}.
The six-sided die has faces marked {1,2,3,4,5,6}.
\(\displaystyle \;\;\)Both dice are rolled and their numbers are noted.
What is the probability that the two numbers differ by one?
There are \(\displaystyle 4\,\times\,6\:=\:24\) possible outcomes.
There are <u>seven</u> outcomes with a difference of one.
\(\displaystyle \;\;(\text{four, six})\;=\; (1,2),\,(2,3),\,(3,4),\,(4,5)\,(2,1),\,(3,2)\,(4,3)\)
Therefore: \(\displaystyle \,P(\text{diff. of one}) \;= \;\frac{7}{24}\)