die rolled, 'success' is 1,2; prob. exactly 1 succ. in 9 rol

Assume that the die is rolled 9 times,
and the roll is called a success if the result is in {1,2}

What is the probability that there are exactly 1 successes or
exactly 1 failures in the 9 rolls?

I tried doing ((1/6)^9)+((1/6)^9)=.000000198, but that was not correct

Re: Probability

Does this help?
There are two ways to get a 1 and a 2 on one roll of the dice. Say one die is red and the other is white. On one roll there are 36 possible outcomes. There are two outcome for a 1 and a 2, a red 1 and a white 2 or a red 2 and a white 1.

Re: Probability

I'm not sure what to do next. Would I multiply that answer I got before by 2, since there are 2 ways to get a 1 and 2?

Re: Probability

The probability of "exactly one success" will be when one of the rolls is a 1 or a 2, and the rest are 3,4,5, or 6.

(1/6 + 1/6)^1 * (1/6 + 1/6 + 1/6 + 1/6)^8 = (2/6)(4/6)^8

The probability of "exactly one failure" will be when one of the rolls is a 3,4,5 or 6, and the rest are 1 or 2.

(1/6 + 1/6 + 1/6 + 1/6)^1 * (1/6 + 1/6)^8 = (4/6)(2/6)^8

Since the events are mutually exclusive, the probability of one or the other happening is their sum.

Re: Probability

Thanks, that helped a lot.