Question

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

Answer 1

When two coins are tossed, sample space $S$ is given by {$HH,HT,TH,TT$} and therefore, $n\left(S\right)=4$.

Let $A$ denote the event that both head appear that is {$HH$} and $n\left(A\right)=1$, therefore, probability of both head appear is:

$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{1}{4}$

Let $B$ denote the event that both tail appear that is {$TT$} and $n\left(B\right)=1$, therefore, probability of both tail appear is:

$P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{1}{4}$

Intersection of $A$ and $B$ is the common elements between $A$ and $B$ which is none, thus, $n(A\cap B)=0$ and

$P(A\cap B)=\frac{n(A\cap B)}{n\left(S\right)}=\frac{0}{4}=0$

Therefore, the events are mutually exclusive.

The probability of either both head or both tail occur is $P(A\cup B)$ and we know that for mutually exclusive event, $P(A\cup B)=P\left(A\right)+P\left(B\right)$ that is:

$P(A\cup B)=P\left(A\right)+P\left(B\right)=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$

Hence, probability that either both heads or both tails occur is $\frac{1}{2}$.