Two coins are tossed simultaneously. The probability of getting at most one head is:
$
  {\text{A}}{\text{. }}\dfrac{1}{4} \\
  {\text{B}}{\text{. }}\dfrac{1}{2} \\
  {\text{C}}{\text{. }}\dfrac{3}{4} \\
  {\text{D}}{\text{. 1}} \\
$

Hint – To calculate the probability of at most one head we calculate the number of favorable outcomes and the total number of outcomes by using the formula for probability.

Complete step by step answer:
Given Data: The probability of getting at most one head. (At most = maximum)
(This means the probability of getting 1 head or no heads at all.)
Two coins are tossed simultaneously.
The total numbers of possibilities are 4.
(H, H), (H, T), (T, H), (T, T) where H – heads and T – tails.
Required at most one head, i.e. there can be no heads or one head.
⟹For no heads at all: one possibility - (T, T)
⟹For one head: possibilities - (H, T), (T, H)
Probability of an event = $\dfrac{{{\text{number of favorable outcomes}}}}{{{\text{total number of outcomes}}}}$
Hence, probability of getting at most one head = $\dfrac{3}{4}$
Hence Option C is the right answer.

Note: In order to solve this type of question the key is to calculate the total possibilities when two coins are tossed and more importantly the number of times zero or one heads occur. Here understanding the question is critical. At most = maximum and at least = minimum.