If two coins are tossed simultaneously, the probability that at least one head will come up is:

In this problem, we must first determine the sample space when two coins are tossed simultaneously.

The sample space when two coins are tossed simultaneously $=\left\{\right(\mathrm{H}, \mathrm{H}),(\mathrm{H}, \mathrm{T}),(\mathrm{T}, \mathrm{H}),(\mathrm{T}, \mathrm{T}\left)\right\}$

Thus, total number of outcomes = 4

Now, for having atleast one head we need $=\left\{\right(\mathrm{H}, \mathrm{T}),(\mathrm{T}, \mathrm{H}),(\mathrm{H}, \mathrm{H}\left)\right\}$

Thus, number of favourable outcomes = 3

$\therefore \mathrm{P}(\mathrm{getting} \mathrm{atleast} \mathrm{one} \mathrm{head})=\frac{\mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}=\frac{3}{4}$

Hence, the correct answer is $\frac{3}{4}$.