What is the probability of getting heads OR a 2 on a single, fair six-sided die ONLY ONCE in 3 attempts?

What is the probability of getting heads OR a 2 on a single, fair six-sided die ONLY ONCE in 3 attempts?

OK, so to clarify, getting heads AND a 2 WOULD NOT be the desired outcome. Getting heads or a 2 more than once WOULD NOT a desired outcome either. It doesn’t matter which one of the attempts you get a heads or a 2 (it could be the 1st, 2nd, or 3rd).

Here is how I approached the problem:

The coin toss and die roll are independent events. There are 2 possibilities for the coin toss and 6 possibilities for the die roll. Therefore, if a single trial consists of a coin toss and a die roll then there are $12$ possible outcomes ($6 \cdot 2 = 12$) per trial. If $3$ trials are conducted there are $1728$ ($12^3$) possible outcomes.

If a coin toss is done $3$ times, there are $8$ possible outcomes:

1. HHH No

2. TTT No

3. HTT Yes

4. THT Yes

5. TTH Yes

6. HHT No

7. HTH No

8. THH No

Therefore, the probability of getting heads only once in $3$ trials is $3/8$. Btw, I know there is an easier way to do this than to write out all the possibilities, I just don’t know how to apply it here. Also writing out all the possibilities gets annoying when there are many trials. So if I would appreciate it if someone could remind me of the formula.

I know that if I had to find the probability of getting heads in $3$ trials at least once that would be easy. I would have to find the probability of getting tails $3$ times and subtract that value from $1$. But there are cases that are more difficult.

The die roll is a little tricky and I’m not sure if I did this right. There are $216$ ($6^3$) possible outcomes for rolling a die $3$ times. The probability of getting a 2 is $1/6$ and the probability of not getting a 2 is $5/6$ (this is obviously in a single trial). So the probability of getting a 2 in the first trial when $3$ trials are conducted is:

$$\frac{1}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} = \frac{25}{216}$$

But since it is a favorable event to get a 2 in either the 2nd or 3rd trial you have to account for that.

$$25 \cdot \frac{3}{216} = \frac{75}{216}$$

So now we have to put these together.

$$\frac{3}{8} \cdot \frac{75}{216} = \frac{225}{1728} = 0.1302 = 13.02\%$$

I feel like I am missing something here. If I am not mistaken I have found the probability of getting heads only once and a 2 only once in $3$ trials. I don’t wanna get both heads and a 2, so would I have to subtract from the numerator to be completely correct? If so would I subtract 3? Btw, I picked 3 because that’s the number of trials so that means I’d remove the outcome of a heads and a 2 in the 1st, 2nd, and 3rd trials.

Also, if anyone has any idea how to approach this formulaically, let me know because trying to figure this out was exhausting. Also, ideally I’d like to know how to do this for any given number of trials, not just $3$ trials.