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