Question

A natural number is chosen at random from amongst the first 300. what is the probability that the number, so chosen number is divisible by 3 or 5?

Answer 1

Heya

All the numbers divisible by 3 till 300 are

3,6,9,....300

They form an arithmetic progression.

an = a +(n-1)d

300 = 3 + (n-1)3

(300-3)/3 = n-1

297/3 = n-1

99 +1 = 100 = n

There are 100 numbers divisible by 3.

The numbers divisible by 5 are

5,10,15....300

They form an arithmetic progression.

300 = 5+(n-1)5

(300-5)/5 = n-1

295/5 = 59 = n-1

n = 60

Numbers divisible by both 3 and 5 are divisible by 15

15,30,......300

300 = 15+(n-1)15

285/15 = n-1

19+1 =n

n =20

No. Of numbers divisible by 3 or 5 are

No. Of Numbers divisible by 3 + no. Of numbers divisible by 5 - no. Of numbers divisible by 15

[ This is because if we count the numbers divisible by 3 and 5 individually, few numbers which are common in both of them are counted twice.

For example, while counting the numbers divisible by 3, we even count the numbers 15,30,... (Which are divisible by 5 also)

While counting the numbers divisible by 5, we count 15,30...(which are divisible by 3 also) again. We count them twice. So we subtract it]

= 100+60-20

= 140

Probability of picking a number divisible by 3 or 5 is

= 140/300

= 7/15

Hope it helps! ^_^