Question

One number card is chosen randomly from the number cards $1$ to $25$. Find the probability that it is divisible by $3$ or $11$

Answer 1

Sample space $S$ of cards is given by {$1,2,3,\mathrm{4........},25$} and therefore, $n\left(S\right)=25$.

Let $A$ denote the event of getting a number divisible by $3$ that is {$3,6,9,12,15,18,21,24$} and $n\left(A\right)=8$, therefore, probability of getting a number divisible by $3$ is:

$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{8}{25}$

Let $B$ denote the event of getting a number divisible by $11$ that is {$11,22$} and $n\left(B\right)=2$, therefore, probability of getting a number divisible by $11$ is:

$P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{2}{25}$

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

$P(A\cap B)=\frac{n(A\cap B)}{n\left(S\right)}=\frac{0}{25}=0$

Therefore, the events are mutually exclusive.

The probability of getting a number divisible by $3$ or $11$ 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:

$P(A\cup B)=P\left(A\right)+P\left(B\right)=\frac{8}{25}+\frac{2}{25}=\frac{10}{25}=\frac{2}{5}$

Hence, probability of getting a number divisible by $3$ or $11$ is $\frac{2}{5}$.