Question

2. A number is selected from the first 100 natural numbers.
The probability that the number is divisible by 5 is
Touth​

Answer 1

Step-by-step explanation:

I am assuming that the first 100 natural numbers mean integers in the range [1,100]

Let A denote the set of numbers in the given range that are even, Let B denote the set of numbers in the given range that are divisible by 5

The total count of even numbers in the said range is 50 . Hence, if any number is selected at random, the probability that it is even =50100=12

The total count of numbers divisible by 5 in the said range is 20 . Hence, if any number is selected at random, the probability that it is even =20100=15

We need to calculate P(A∪B)

By the principle of inclusion and exclusion, we have

P(A∪B)=P(A)+P(B)−P(A∩B)

P(A∩B) is the probability that a number is both even and divisible by 5 . By common reasoning, we see that these are the numbers divisible by 10 . We have 10 of them in the said range. Therefore, P(A∩B)=10100=110

∴P(A∪B)=12+15−110=35

Answer 2

Answer:

natural number is selected =100

the possible outcomes which are divisible by 5 are :-5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95 and 100

total numbers of possible outcomes which are divisible by 5= 20

the probability of numbers which are divisible by 5 = 20/100