Question

Numbers written from 3 to 101 and one number is selected .Find the probability of getting

a) a single digit b) a two digit number c) a number divisible by 3 and 5 d) a prime number less than 30 e) a perfect square f) a perfect cube​

Answer 1

Step-by-step explanation:

Given Numbers written from 3 to 101 and one number is selected .Find the probability of getting a) a single digit b) a two digit number c) a number divisible by 3 and 5 d) a prime number less than 30 e) a perfect square f) a perfect cube

• So the numbers are written from 3 to 101 and one number is selected.
• So probability = Favourable outcomes / total number of outcomes
• Total numbers written will be 99
• So we need to find single digit, so it will be 3,4,5,6,7,8,9 = 7
• So probability = 7 / 99
• Two digit numbers will be from 10 to 99 = 90
• So probability = 90 / 99
• Numbers divisible by 3 and 5 = multiples of 3 = 32 / 99
• Also multiples of 5 = 20 / 99
• Prime number less than 30 = 3,5,7,11,13,17,29,23,29 = 9
• So probability = 9 / 99
• Perfect square = 4,9,16,25,36,49,64,81,100 = 9
• So probability = 9/99
• Perfect cube = 8,27,64 = 3
• So probability = 3/99

Reference link will be

https://brainly.in/question/5372704

Answer 2

Answer:

hope it helps

Step-by-step explanation:

o the numbers are written from 3 to 101 and one number is selected.

So probability = Favourable outcomes / total number of outcomes

Total numbers written will be 99

So we need to find single digit, so it will be 3,4,5,6,7,8,9 = 7

So probability = 7 / 99

Two digit numbers will be from 10 to 99 = 90

So probability = 90 / 99

Numbers divisible by 3 and 5 = multiples of 3 = 32 / 99

Also multiples of 5 = 20 / 99

Prime number less than 30 = 3,5,7,11,13,17,29,23,29 = 9

So probability = 9 / 99

Perfect square = 4,9,16,25,36,49,64,81,100 = 9

So probability = 9/99

Perfect cube = 8,27,64 = 3

So probability = 3/99