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probability math
a fair dice has three faces marked 1, one face marked 2, and two faces marked 4.
another fair dice has four faces marked 1, and two faces marked 2. these two dice are thrown and the total score recorded. Set up a sample space for the possible totals. Hence find the probability that the score is even. |
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You have two things that are going to happen: 1) you're going to throw the first die, 2) you're going to throw the second die. You want to set up the outcomes of each of those. That is, the outcomes of the first die are 1, 1, 1, 2, 4, 4. Don't forget that even though there are duplicates, they are six different sides of the die. So the 3 1's are actually 3 different things. To get the sample spaces, start with the first outcome of the first die: 1. Now if you throw that, then you have 6 possible outcomes of the second die. So that's 6 things that could happen right there. I'll start you off on that one. You could get (1, 1)(1, 1)(1, 1)(1, 1)(1, 2) or (1, 2). (Again, remember those duplicates are still different sides of the die.) Then keep going in this fashion. Go to the next possible outcome on the first die and match that up with the six possible outcomes on the second die. Etc. Since you have 6 possible outcomes on the 1st die, and 6 possible outcomes on the 2nd die, how many total possibilities do you have? (Multiplication rule.) That's how many sample spaces you should have. When you get done, check how many of those sample spaces give you an even total. A probability is the total "favorable" outcomes divided by the total possible outcomes. (Favorable meaning the ones you're looking for -- even totals in this case.) You should already have figured the total possible at this point. Then when you look through them and see how many give you an even total, you'll also have the favorable outcomes. Then just divide. |
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There's another way of looking at it. It's by having a probability table. I'll post it later on. |
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Er, um, thanks. :-) (You have a probability table? Never heard of such a thing.) |
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Ok, here it is (I don't really know how to call it :o) There are all the possible outcomes here, six from each die, making a total of 36 outcomes, some of which are even. Count the number of even outcomes, and put it over the whole number of outcomes. (The shaded circles indicate that the sum are even) |
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