Question

A die is thrown find the probability of following events:
(i) A prime number will appear
(ii) A number greater than or equal to $3$ will appear
(iii) A number less than or equal to one will appear
(iv) A number more than $6$ will appear
(v) A number less than $6$ will appear

Answer 1

The sample space of the given experiment is given by,
$S=\{1,2,3,4,5,6\}$
(i) Let $A$ be the event of the occurrence of a prime number $\Rightarrow A=\{2,3,5\}$
$\therefore P\left(A\right)=\frac{NumberofoutcomesfavourabletoA}{Totalnumberofpossibleoutcomes}=\frac{n\left(A\right)}{n\left(S\right)}=\frac{3}{6}=\frac{1}{2}$
(ii) Let $B$ be the event of the occurrence of a number greater than or equal to $3$ $\Rightarrow B=\{3,4,5,6\}$
$\therefore P\left(B\right)=\frac{NumberofoutcomesfavourabletoB}{Totalnumberofpossibleoutcomes}=\frac{n\left(B\right)}{n\left(S\right)}=\frac{4}{6}=\frac{2}{3}$
(iii) Let $C$ be the event of the occurrence of a number less than or equal to one $\Rightarrow C=\left\{1\right\}$
$\therefore P\left(C\right)=\frac{NumberofoutcomesfavourabletoC}{Totalnumberofpossibleoutcomes}=\frac{n\left(C\right)}{n\left(S\right)}=\frac{1}{6}$
(iv) Let $D$ be the event of the occurrence of a number greater than $6$ $\Rightarrow D=\varphi$
$\therefore P\left(D\right)=\frac{NumberofoutcomesfavourabletoD}{Totalnumberofpossibleoutcomes}=\frac{n\left(D\right)}{n\left(S\right)}=\frac{0}{6}=0$
(v) Let $E$ be the event of the occurrence of a number less than $6$ $\Rightarrow E=\{1,2,3,4,5\}$
$\therefore P\left(E\right)=\frac{NumberofoutcomesfavourabletoE}{Totalnumberofpossibleoutcomes}=\frac{n\left(E\right)}{n\left(S\right)}=\frac{5}{6}$