Two coins are tossed simultaneously. Find the probability that either both heads or both tails occur.

Hint: In this problem, first we need to find the probabilities of occurrence of both heads and both tails. Next, add the two probabilities to obtain the probability of occurrence of both heads and both tails.

Complete step-by-step solution -
Two coins are tossed simultaneously; we can obtain the combination of sample space as shown below.
$(\left( {HH,HT,TH,TT} \right)$
The number of sample space $n\left( S \right)$ is 4.
Assume that $A$ denotes an event that both coins appear head. From sample space, it can be observed that there is only one condition appears that shows head on both coins. Therefore,
$n\left( A \right) = 1$
The probability of occurrence of head on both coins is as follows:
$ P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}} \\$
 $\Rightarrow P\left( A \right)\, = \dfrac{1}{4} \\ $
Assume that $B$ denotes an event that both coins appear tail. From sample space, it can be observed that there is only one condition appears that shows tail on both coins. Therefore,
$n\left( B \right) = 1$
The probability of occurrence of tail on both coins is as follows:
  $P\left( B \right) = \dfrac{{n\left( B \right)}}{{n\left( S \right)}} \\$
  $\Rightarrow P\left( B \right)\, = \dfrac{1}{4} \\ $
Add the above two probabilities to obtain the probability of both heads or both tails.
$P\left( A \right) + P\left( B \right) \\$
$\Rightarrow \dfrac{1}{4} + \dfrac{1}{4} \\$
$\Rightarrow \dfrac{1}{2} \\ $
Thus, the probability of occurrence of both heads or both tails is $\dfrac{1}{2}$.

Note: In this case, both the events are mutually exclusive because both cannot occur at the same time. Here we have used the addition of both probabilities as these are independent of each other, we already know that for independent events we use the additional rule of probability.