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?
Thanks!
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.
Of the 216 triples, there are 30 with the form where
That gives the probability .
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?