Question

A number is chosen randomly among the first 100 natural numbers. Then the probability
that the number chosen is multiple of 7, will be
7
(A)
7
(B)
(D)
50
13
is
15
29​

Answer 1

number of sample space = no. of ways in which one number can be chosen from 100 natural numbers

= 100C1 = 100

there are 14 multiples of 7 from 1 to 100

7,14,21,....98

no. of ways of choosing one number from this = 14

probability = 14/100

= 7/50

Answer 2

Answer :

P = 7/50

Solution :

Firstly ,

Let's find the number of multiples of 7 between 1 and 100 using concept of AP .

We know that ,

The smallest and the largest multiple between 1 and 100 are 7 and 98 respectively .

Thus ,

First term , a = 7

Last term , l = 98

Common difference , d = 7

Number of terms , n = ?

We have ,

=> l = 98

=> a + (n - 1)d = 98

=> 7 + (n - 1)7 = 98

=> 7 + 7n - 7 = 98

=> 7n = 98

=> n = 98/7

=> n = 14

Hence ,

The number of multiples of 7 between 1 and 100 is 14 .

Here ,

We need to find the probability of getting a multiple of 7 when one number between 1 and 100 is chosen at random .

Thus ,

No. of favourable outcomes = 14

No. of total outcomes = 100

Also ,

We know that ,

$Probability \: , \: P = \frac{No. \: of \: favourable \: outcomes}{No. \:of \: total \: outcomes }$

Thus ,

=> P = 14/100

=> P = 7/50

Hence , P = 7/50 .