Step 1: Consider all \(2^{10}=1024\) equally likely sequences of 10 coin tosses.
Step 2: We want sequences where toss 10 is Heads and exactly 6 of the first 9 tosses are Heads, so that the 7th Head lands exactly on toss 10.
Step 3: The number of such sequences equals the number of ways to place 6 heads among the first 9 positions, with the 10th position fixed as Head: $$\binom{9}{6} = 84$$
Step 4: So the probability is $$\frac{84}{1024} = \frac{21}{256}$$
\[\boxed{\frac{21}{256}}\]