Question

Consider the two random experiments. Throwing a fair die once. In each case find the probability of each elementary event and add them. What conclusion can we draw?

Answer 1

Throwing a die
$S=\{1,2,3,4,5,6\}\therefore n\left(S\right)=6$
Elementary events are,
$E_1=\left\{1\right\},E_2=\left\{2\right\}=E_3=\left\{3\right\},E_4=\left\{4\right\}=E_5=\left\{5\right\},E_6=\left\{6\right\}$
$\therefore P\left(E_1\right)=P\left(E_2\right)=P\left(E_3\right)=P\left(E_4\right)=P\left(E_5\right)=P\left(E_6\right)=\frac{1}{6}$
By adding the probability of all these elementary events, we get
$P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)+P\left(E_4\right)+P\left(E_5\right)+P\left(E_6\right)=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{6}{6}=1$

We can observe that the sum of probabilities of all the elementary events of an event is $1$ and the result holds true in general.