Question

A die is thrown, find the probability of follwoing events:
(i) A prime number will appear.
(ii) A number greater than or equal to 3 will appear
(iii) A numbher less than or equal to one will appar.
(iv) A number more than 6 will appear.
(v) A number less tahn 6 will appear.

Answer 1

Here the sample space $S=\{1,2,3,4,5,6\}$
$\therefore n\left(S\right)=6$
(i) Let A be the event of gettinga prime number
$A=\{2,3,5\}\Rightarrow n\left(A\right)=3$
Thus $P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{3}{6}=\frac{1}{2}$
(ii) Let B be th event of getting a nmber greater than or equal to 3
$B=\{3,4,5,6\}\Rightarrow n\left(B\right)=4\text{thus}P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{4}{6}=\frac{2}{3}$
(iii) Let C be the event of getting a number les than or equal to 1
$C=\left\{1\right\}$ $\Rightarrow n\left(C\right)=1$
Thus $P\left(C\right)=\frac{n\left(C\right)}{n\left(S\right)}=\frac{1}{6}$
(iv) Let D be the event of getting a number more than 6
$D=\varphi \Rightarrow n\left(D\right)=0\text{Thus}P\left(D\right)=\frac{n\left(D\right)}{n\left(S\right)}=\frac{0}{6}=0$
(v) Let E be the event of getting a number less than 6
$E\{1,2,3,4,5\}\Rightarrow n\left(E\right)=5\text{Thus}P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{5}{6}$