the probability of getting a number between 1 and 100 which is divisible by 7 is ___?

the first number b/w 1 and 100 divisible by 7 is __ 7__

the last number b/w 1 and 100 divisible by7 is* 98*

a_{n} = a + ( n- 1) d

98 = 7 + 7n - 7

98 / 7= n

n= 14

favorable outcomes=14

probability=14/98

= 1/7

- -2

Smallest number divisible by seven = 7

Largest number divisible by seven = 98

Here a = 7 , d = 7 an = 9

an = a + (n-1)d

Substituting the value

98 = 7 + (n-1)7

98 - 7 = (n-1)7

91 = (n - 1)7

91/7 = (n-1)

(n-1) = 13

n = 14

Hence the number of favourable outcomes = 14

Total number of outcomes = 100

Probability of getting a number between 1 and 100 is 14/100 or 7/50

Hope its correct

- 0

** well.. the ques itself says "between 1 & 100" **

**this simply implies that a number (n) which is **

**1<n<100.**

**where n is an integer(as per the ques.)**

**so total possible outcomes = 98**

**but the total possible outcomes would have been 100 only and only if the question said - "find the probability of getting a number divisible by 7 in first 100 natural numbers."**

**hope it helps.. **

**cheers.. :)**

- 9