Three counters are used for a board game. if the counters are tossed, how many ways can at least one counter with side a occur?

We use the equation for repeated trials written below:

Probability = n!/r!(n-r)! * p^(n-r) * q^r

The p is the probability of getting a side A in one toss. Since a counter has only two side, p = 0.5. The q is the probability of not getting side A in one toss, which is also q = 0.5. Now, r is the number of success per n trials. There are 3 tosses so, n=3. The question is getting "at least 1" counter. So, r=1, r=2 and r=3.

Probability for r=1: 3!/1!(3-1)! * (0.5)^(3-1) * (0.5)^1= 0.375
Probability for r=2: 3!/2!(3-2)! * (0.5)^(3-2) * (0.5)^2= 0.375
Probability for r=1: 3!/3!(3-3)! * (0.5)^(3-3) * (0.5)^3= 0.125

Total probability = 0.375 + 0.375 + 0.125 = 0.875

Agc and bgd are vertical angles