TS EAMCET 04 May 2019 Shift 2 Solved Paper

Number divisible by 4 from first 30 natural number is 7
Number divisible by 7 from first 30 natural number is 4
Number divisible by both 4 and 7 from the first 30 number is 1
Therefore, the required probability is,
P‌‌=‌

7

30

+‌

4

30

−‌

1

30

‌‌=‌

10

30

‌‌=‌

1

3