All you math geniuses

[categ]

Below is a probability problem, that I can't come up with the same answer as ETS on the sample college math exam. Can anyone help?

[I]The faces of a fair cube are numbered 1 through 6; the probability of rolling any number from 1 through 6 is equally likely. If the cube is rolled twice, what is the probability that an even number will appear on the top face in the first roll or that the number 1 will appear on the top face in the second roll[/I]

So as explained in the "Cracking the Clep" book, I figured the first roll would be 3/6 and the second roll would be 1/6. Add those together and you get 4/6, reduced to 2/3. That is one of the answers but they say the correct answer is 7/12. Where am I going wrong?

Urg, I am never gonna get through this exam.

[Bob]

This is from Yahoo answers. Looks like the same problem.
Good Luck,
Bob


Fair cube probability question? - Yahoo! Answers

[categ]

You are my hero! How did you find that, it was exactly the same problem?

Thank you!

[Bob]

You are my hero! How did you find that, it was exactly the same problem?

I cut and pasted the first half of the problem only into google and that brought up a different problem, so I pasted the second half only and up came your problem. Glad I could help.
Bob

[will1030]

Once again. InstantCert Forum is my hero. Thank you!!!

[Thorne]

The question is asking what the odds are that one or  the other would happen. 2/3 is on the right track, but you forgot a step.

In fact, 2/3 would be the correct answer if the question was changed from:

The faces of a fair cube are numbered 1 through 6; the probability of rolling any number from 1 through 6 is equally likely. If the cube is rolled twice, what is the probability that an even number will appear on the top face in the first roll or that the number 1 will appear on the top face in the second roll?

to

The faces of a fair cube are numbered 1 through 6; the probability of rolling any number from 1 through 6 is equally likely. If the cube is rolled twice, what is the probability that an even number will appear on the top face in the first roll, that the number 1 will appear on the top face in the second roll, or that both would occur simultaneously?

The word "or" means that we actually need to exclude the option of both happening.

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If the probability of rolling an even face on the first roll is R1 and the probability of rolling a 1 on the second roll is R2, then we can calculate the likelihood for each.

R1 = Number of even faces / number of total faces = 3/6
R2 = Number of faces with a value of 1 / number of total faces = 1/6

The probability of the independent occurrence of either OR both events, removing constraints, is simply Rn = R1+R2+R3...
Or: 2/3

The intersection of probabilities can be calculated using Rn = R1*R2*R3...
Thus: R1*R2 = R3, R1 * R2 = 3/36 = 1/12
R3 = 1/12

To find the union of probabilities, you take the probability of independent occurrence minus the intersection of probabilities, or 2/3 - 1/12 = 8/12 - 1/12 = 7/12