Question

An integer is chosen at random from first hundred natural numbers. The probability that the integer chosen the multiple of 5 is * 1/5 5/10 1/4 None of these

Answer 1

Question -

An integer is chosen at random from the first hundred natural numbers.

The probability that the integer is chosen the multiple of 5 is -

[ A] { 1 / 5 }

[ B ] { 5 / 10 }

[ C ] { 1 / 4 }

[ D ] None of these

Solution -

Here , the first hundred natural numbers means -

=> 1 to 100

Integers that are multiples of 5

=> 5 , 10 , 15 , 100

So ,

This series can be seen as an ap .

a = 5

d = 5

N th term -

=> a + ( n - 1 ) d

=> 5 + 5 ( n - 1 )

=> 5 + 5 n - 5

=> 5n

But , the nth term is 100

=> n = 20 .

Thus , there are 20 integers , between 1 to 100 , which are multiples of 5 .

Now.

Total number of Integers -

=> 100 - 1 + 1

=> 100

So

The probability that the integer is chosen the multiple of 5 -

=> [ Number Of Multiples Of 5 ] / [ Total numbers ]

=>> [ 20 / 100 ]

=>> [ 1 / 5 ]

Hence , Option A is the correct Answer ....

____________

Answer 2

$\huge\sf\pink{Answer}$

☞ Your answer is 1/5

$\rule{110}1$

$\huge\sf\blue{Given}$

✭ Limit = First 100 natural numbers

$\rule{110}1$

$\huge\sf\gray{To \:Find}$

☆ Probability that the chosen integer is a multiple of 5

$\rule{110}1$

$\huge\sf\purple{Steps}$

The first hundred positive integers are,

$\sf \color{red}{\longmapsto S = \bigg\lgroup 1,2,3,4..... 99,100 \bigg\rgroup}$

So the total number of events possible = N(s) = (100)

Let A be the event,

Then the favourable outcomes are,

$\sf \color{aqua}{\leadsto A = \bigg\lgroup 5,10,15,20,25,30.....95,100\bigg\rgroup}$

So the total number of favourable outcomes = N(a) = (20)

➝ P(a) = The probably of getting an integer divisible by 5

➝ $\sf\dfrac{N(a)}{N(s)}$

➝ $\sf {\dfrac{20}{100}}$

➝ $\sf\color{lime}{\dfrac{1}{5}}$

$\rule{170}3$