Re: A die is rolled thrice

Quote:

Originally Posted by

**usre123** three dice are thrown. what is the probability that the first two show the same number, and the last one a different number?

So I did 1/6 * 1/6 * 5/6 (first two show the same number, the last one another number) = 5/216. then i multiply by 3 and get 15/216. the answer is 15/36. why?

Why do you think that the answer is 15/36?

As you noted there are 216 possible triples.

How many of those look like $\displaystyle (X,X,Y)$ where $\displaystyle X\ne Y~?$.

Divide that number by 216.

Re: A die is rolled thrice

The "probability the first two show the same number" is NOT "1/6*1/6". That is the probability that the first two numbers are a **specific** number from 1 to 6. Here, the first two numbers can be any number 1 to 6 as long as they are the same. The first number can be **any** number. The probability of that is, of course, 1. Then the second number must be the same as the first- the probability of that is 1/6 to the probability that the first two numbers are the same is 1(1/6)= 1/6.

Re: A die is rolled thrice

the official answer is 15/36. I'm at a loss. I get 5/216

Re: A die is rolled thrice

Quote:

Originally Posted by

**usre123** the official answer is 15/36. I'm at a loss. I get 5/216

Of the 216 triples, there are 30 with the form $\displaystyle (X,X,Y)$ where $\displaystyle X\ne Y~.$

That gives the probability $\displaystyle \frac{30}{216}=\frac{5}{36}$.

It seems that the given answer is incorrect.

That a suggested answer is for the question: *The probability that two of the dice are equal and the third different.*

Did you translate the question correctly?