Question

A natural number is chosen at random from amongst first 500. What is hte probability that the number so chosen is divisbile by 3 or 5 ?

Answer 1

Since a number is chosen from first 500

$\therefore n\left(S\right)=500$

Let A be the event of choosing number divisble by 3= [3,6,9,...,498]

$\therefore n\left(A\right)=166$ $\left[\begin{array}{c}\because a+(n-1)d=498\\ 3+(n-1)3=498\\ 3n=498\Rightarrow n=166\end{array}\right]$

$P\left(A\right)=\frac{166}{500}$

B be the event of choosing number

[5,10,15,...,500]

n(B)= 100

[$\because$ a+(n-1)d=500]

$\Rightarrow P\left(B\right)=\frac{100}{500}$

[$\because$ 5+(n-1)5=500]

$\left[\begin{array}{c}5n=500\\ n=100\end{array}\right]$

Also, $(A\cap B)=\{15,30,...,495\}$

$\Rightarrow n(A\cap B)=33$

$\left[\begin{array}{c}\because a+(n-1)d=495\\ 15+(n-1)15=495\\ 15n=495\\ n=3\end{array}\right]$

$P(A\cap B)=\frac{33}{500}$

Now required probability = P(a number is divisible by 3 or 5) =$P(A\cup B)$

$\therefore P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

= $\frac{166}{500}+\frac{100}{500}-\frac{33}{500}=\frac{233}{500}$