# Probability that a two is thrown

1. Aug 18, 2009

### Chewy0087

1. The problem statement, all variables and given/known data
A die is based so that the numbers 5 & 6 appear 3 times more often than the numbers 2 , 3 & 4. Calculate;
i)The probability that a two is thrown.
ii)The probability that two consecutive throws are > or equal to 10.

3. The attempt at a solution

5 5 5 6 6 6 2 3 4 , satisfies the criteria meaning a 2 would be 1/9, however i have a feeling that that's nonsense...:S

Then for ii) Combo's are 46 64 65 56 55 66 so (4 * 1/9) + (2 * (3/81) = (36/81+6/81) = (42/81)?

I don't know how to go about these questions, sorry for bugging everyone >.<

2. Aug 18, 2009

### jgens

Re: Probability

Well, here's another way of looking at (i) . . .

Let x be the probability that you roll a 2, then the probability that you roll a 3 or a 4 is also x and the probability that you roll a 5 or 6 is then 3x. We know that the sum of the probabilities of each possible outcome must be 1 so x + x + x + 3x + 3x = 9x = 1. Therefore x = 1/9.

3. Aug 18, 2009

### Elucidus

Re: Probability

Did you mean to say "a die is biased so that..."?

Does the die really have five sides (such dice do exist) or is it that this is a normal six-sided die with repeated markings?

What is the probability of rolling a 1 if this is in fact a classically marked six-sided die?

Without this information, the question may not be answerable.

--Elucidus

4. Aug 19, 2009

### Chewy0087

Re: Probability

sorry, yeah, having a 1 is impossible, i think it just means it's a wierdly shaped dice or something :S

edit: and thanks jgens that's a much simpler and better way of looking at it >.<

5. Aug 19, 2009

### Elucidus

Re: Probability

jgens has explained the first part. In order to answer the second part, you need to determine how the two dice might have a sum >= 10 and find the probability that those results could occur.

--Elucidus