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

$\left(i\right)$ If a die is thrown, then the sample space is
$S=\{1,2,3,4,5,6\}$

Let $A$ be the event that a prime number will appear
$A=\{2,3,5\}$

$n\left(S\right)=6\text{and}n\left(A\right)=3$

$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{3}{6}=\frac{1}{2}$

Hence the required probability that a prime number will occur is $\frac{1}{2}$.


$\left(ii\right)$ If a die is thrown, then the sample space is
$S=\{1,2,3,4,5,6\}$

Let $A$ be the event that a number greater than or equal to three will appear $A=\{3,4,5,6\}$

$n\left(S\right)=6\text{and}n\left(A\right)=4$

$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{4}{6}=\frac{2}{3}$

Hence the required probability that a number greater than or equal to three will appear is $\frac{2}{3}$.


$\left(iii\right)$ If a die is thrown, then the sample space is
$S=\{1,2,3,4,5,6\}$

Let $A$ be the event that a number less than or equal to one will appear $A=\left\{1\right\}$

$n\left(S\right)=6\text{and}n\left(A\right)=1$

$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{1}{6}$

Hence the required probability that a number less than or equal to one will appear is $\frac{1}{6}$.


$\left(iv\right)$ If a die is thrown, then the sample space is
$S=\{1,2,3,4,5,6\}$

Let $A$ be the event that a number more than $6$ will appear $A=\varphi$

$n\left(S\right)=6\text{and}n\left(A\right)=0$

$P\text{(a number more than 6 will appear)}=\frac{\text{favorable outcome}}{\text{total outcome}}$

$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{0}{6}=0$

Hence the required probability that a number more than $6$ will appear is $0$.


$\left(v\right)$ If a die is thrown, then the sample space is
$S=\{1,2,3,4,5,6\}$

Let $A$ be the event that a number less than $6$ will appear $A=\{1,2,3,4,5\}$

$n\left(S\right)=6\text{and}n\left(A\right)=5$

$P\text{(A)}=\frac{\text{favorable outcome}}{\text{total outcome}}$

$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{5}{6}$

Hence the required probability that a number less than 6 will appear is $\frac{5}{6}$.