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Mathematics 88 Online
OpenStudy (anonymous):

You roll 1 red and 1 white dice. What is the probability that the number on the red die is larger than the number on the white die? can anyone help me?:(

OpenStudy (ragingsquirrel):

This one is easy. Try your guess.

hero (hero):

There's a possibility they could be equal so don't forget that while you're calculating your probability

OpenStudy (ragingsquirrel):

I don't think that would matter much, would it?

hero (hero):

Of course it matters. You have to take into account all possibilities when calculating probability

OpenStudy (anonymous):

how would i start solving it?:( im really not smart in math:/

OpenStudy (ragingsquirrel):

try eliminating the colors, as color of the dice dont matter.

OpenStudy (ragingsquirrel):

theres 3 possibilitys of what will happen so...

OpenStudy (anonymous):

The ability to count the number of outcomes is what makes this problem hard/easy. So first, do you know how many outcomes there are total? For example, red - 3, white - 2, thats one outcome, so on and so forth.

OpenStudy (anonymous):

i still dont get it:/ i dnt get these kind of math problems:[

OpenStudy (ragingsquirrel):

im not completely sure, but what i'm getting is %100 divided by 3 (or %33.33)

OpenStudy (anonymous):

i think you might be oversimplifying it Squirrel. First we need to know how many possible outcomes there are for rolling the die. Since the red one can be a 1 through 6, and the white can also be a 1 through 6, there are 6*6=36 possible outcomes for the die after we roll them. Now we need to count how many of those outcomes have a number of the red die greater than the number on the white die. thats a little trickier.

OpenStudy (anonymous):

i still dnt get it ://

OpenStudy (anonymous):

Think back to other probability questions you've done. It comes down to being able to count outcomes. If you want to probability of getting a 3 on one die, you say: There is only 1 three. There are 6 sides total. The probability is: \[\frac{1}{6}\] the number of successes divided by the total number of outcomes.

OpenStudy (anonymous):

So for the problem you posted, it comes down to: How many ways are there to roll the two die? In how many of those rolls will the red die be greater than the white die? If you can count the possibilities, you can answer this question.

OpenStudy (anonymous):

is it 5/6?

OpenStudy (anonymous):

its a little harder than that. There are 36 possible outcomes for the die after we roll them. It could be: R W 1 1 1 2 1 3 1 4 so on and so forth. Do you see why there will be 36 different outcomes?

OpenStudy (anonymous):

Now you have to count how many of those rolls have a red number bigger than a white number. For example, the ones i posted above all fail, since 1 is not bigger than 1, 2, 3, or 4. R W 3 1 3 2 These two would be good, since 3 is bigger than 1 and 2. You need to count all the ones that have red number bigger than the white one.

OpenStudy (anonymous):

i dont know how how but i got 12/36

OpenStudy (anonymous):

thats not good if you dont know how lol =/ I know it might be lengthy, but i suggest writing out all 36 possible outcomes, and circling the ones with the red number larger than the white number: R w 1 1 1 2 1 3 1 4 1 5 1 6 2 1 (this one is good) 2 2 2 3 2 4 2 5 2 6 3 1 (this one too) 3 2 (and this one) etc etc. I think you might still be looking for a short/fast way to do this problem, when the reality is we just have to get our hands dirty and count.

OpenStudy (anonymous):

im doing the problem right now lets see what i get..LOL

OpenStudy (anonymous):

15/36!?!?!^_^

OpenStudy (anonymous):

that is correct, great job :)

OpenStudy (anonymous):

really?!?! it is ?! omgg your a lifesaver xd i have more problems to ask u too ..LOL thank you for helping meeeexD

OpenStudy (anonymous):

\[\frac{1}{6}\frac{1}{6}+\frac{1}{6}\frac{2}{6}+\frac{1}{6}\frac{3}{6}+\frac{1}{6}\frac{4}{6}+\frac{1}{6}\frac{5}{6}\text{=}\frac{5}{12}\]The odds of winning is 5/12. Used the idea that one die can be used to solve the problem. Roll the die and record the number showing. Roll again and compare the current number with the first one. It turns out that 15/36 =5/12

OpenStudy (anonymous):

Roll a 2, 1 is smaller than 2 Roll a 3, 1 and 2 are smaller Roll a 4, 1,2 and3 are smaller and so on.

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