A fair coin is tossed 3 times in succession. The set of equally likely outcomes is $\{\mathrm{HHH}, \mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}, \mathrm{HTT}, \mathrm{THT}, \mathrm{TTH}, \mathrm{TTT}\}$. Find the probability of getting a head on the first toss. The probability of getting a head on the first toss is (Type an integer or a simplified fraction.)

Step 1 ：The set of equally likely outcomes when a fair coin is tossed 3 times in succession is \(\{\mathrm{HHH}, \mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}, \mathrm{HTT}, \mathrm{THT}, \mathrm{TTH}, \mathrm{TTT}\}\).

Step 2 ：We want to find the probability of getting a head on the first toss.

Step 3 ：The probability of an event is calculated by dividing the number of favorable outcomes by the total number of outcomes.

Step 4 ：In this case, the favorable outcomes are the ones where the first toss results in a head. These are HHH, HHT, HTH, and HTT.

Step 5 ：The total number of outcomes is 8, as given in the question.

Step 6 ：So, the probability of getting a head on the first toss is \(\frac{4}{8} = \frac{1}{2}\).

Step 7 ：Final Answer: The probability of getting a head on the first toss is \(\boxed{\frac{1}{2}}\).