In three tosses of a fair coin, we have a total of \(2^3 = 8\) possible outcomes.
These outcomes are: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
Counting the number of heads in each outcome, we have:
HHH: 3 heads
HHT: 2 heads
HTH: 2 heads
HTT: 1 head
THH: 2 heads
THT: 1 head
TTH: 1 head
TTT: 0 heads
To find the mean, we sum up the number of heads in each outcome and divide by the total number of outcomes:
\(\frac{{3 + 2 + 2 + 1 + 2 + 1 + 1 + 0}}{8} = \frac{12}{8} = 1.5\)
Therefore, the mean number of heads in three tosses of a fair coin is 1.5. The correct option is (1) 1.5.