You and your friend are rolling number cubes. each cube has 6 sides with the numbers 1 to 6. if a sum of 7 is rolled, your friend gets a point. however, you get a point if either one of the sums listed below is rolled. which pair of numbers would make the game fair?

Hello mate!

Ok, the 6 sides cubes can be thinked as two normal dice.

A fair game is that game where both players have the same chance to win, first, let's calculate the situations where you friend win:

this are, if the cubes have numers between 1 and 6:

1 + 6

6 + 1

2 + 5

5 + 2

3 + 4

4 + 3

where 1 + 6 is different to 6 + 1 because in the first one you rolled a 1 and your friend a 6, and in the second one you rolled a 6 and your friend a 1, so are different events.

Then you can see that you friend has 6 combinations where he wins.

So we need to find two numbers that between them, have the same number of combinations.

I can think, for example in 4 and 10:

the combinations for 4 are:

2 + 2

3 + 1

1 + 3

a total of 3 combinations:

for 10 are:

5 + 5

4 + 6

6+ 4

also 3 combinations.

So if you win a point if the sum is 4 or 10, you also have 6 combinations where you win, then the game is fair

Now, suppose the first number is 5, let's find the second:

the combinatios for 5 are:

4 + 1

1+ 4

2 + 3

3 + 2

a total of 4 combinations, this means that we need to find another number that only has two combinations, wich can be, for example, 3

the combinations for 3 are:

1 + 2

2 + 1

2 combinations, then if your numbers are 3 and 5, you have 6 combinations where you win, and the game is fair.

answer: answer

step-by-step explanation:

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: