I want to know the answer of this question URGENTLY please:

Q:
State the number of possible outcomes if a fair dice is drown two times. Write down all the possible outcomes that give the following events and calculate then the probability such that:
a)The sum of the two tries is above 5

b) The draws have the same outcome.

Thanks a lot

There are 36 outcomes. List them out. 1,1; 1,2; 1,3; 1,4; 1,5; 1,6;... and so on.

If there were 3 dice, then there would be 6^3 outcomes. If there were 10 dice, there would be 6^10 outcomes. See the pattern?

There is a quick way to find how many outcomes sum to more than 5.

That means they would have to sum to 6,7,8,9,10,11,12.

It is called a generating function.



If we expand this we get

Look at the exponents and their corresponding coefficients(numbers in front of the x's).

Look at the exponents larger than 5 and add up their coefficients. Each number it fron of the x is how many times that sum occurs. For instance, how many ways sum to 7? The coefficient corresponding to x^7 is 6. There are 6 ways the two dice sum to 7. So, the probability we get a sum of 7 is 6/36=1/6.
See? Do this with your case and divide by 36 to fnd the probability. Or, you can list them out and count them up. But I do not like brute force. Especially with counting problems. It defeats the purpose.

So the pr(A), where A is the sm of the two tries above 5 = 27/36

and the probability of the second Q is 6/36 (the probabilti of the two of tries are equal)

Thank you very much for your patient and your helpful answers.

I got 26/36!

I used a probability table.

Yes, Unknown has it. Just add up the coefficients.

They sum to 26

So, we ave 26/36=13/18

what is the probability of getting one six if a die is thrown four times?

if a die is thrown four times what is the probability that the nunber is six?

Start your own thread sakidbzu. And post the answer you think is good.

Oh OK thanks,
And the draws have the same outcome will it be 6/36?

Yes, exactly! :) Well done!

Thank you very much :)

You're welcome! :)