Table of Contents
What is the probability that a randomly selected 2 digit number would be a multiple of 3 but not of 5?
Probability that a two-digit number is a multiple of 3 and not a multiple of 5 = 30-6 = 2490= 415.
How many two-digit numbers which are multiples of 3?
2 digit numbers which are multiples of 3: 12, 15, 18, 21, …, 90, 93, 96, 99. Consider this A.P. : a = 12, d = 3, 99 = 12 + (n- 1) (3) => n – 1 = 87/3 = 29 => n = 30. There are 30 multiples of 3 in all 2 digit numbers.
What is the smallest positive 2 digit whole number divisible by 3 and such that the sum of its digits is 9?
The number we are looking for is 18 and the answer is D. It is divisible by 3 and the sum of its digits is equal to 9 and it is the smallest and positive whole number with such properties.
What is the probability that a two digit?
There are 90 two digit numbers(99–9). Out of this there are 6 numbers divisible by 15(15, 30, 45, 60, 75, 90), which are also divisible by 5. Therefore, the favorable cases are 30–6=24. Hence, the required probability is 24/90 = 4/15.
What is the probability of choosing a two-digit number at random?
Since there are 90 two-digit numbers in all (how do I know this?), the probability of choosing a two-digit number ‘at random’ (with the meaning described above) is 30 90 = 1 3. 8 clever moves when you have $1,000 in the bank.
How many 2-digit positive integers are there?
2-digit positive integers are 10, 11, 12.., 98, 99. These are 90 numbers out of which one number can be chosen in 90 ways. Total number of elementary events = 90.
How many two-digit integers are multiples of 3?
Find the probability that a randomly chosen two-digit integer is a multiple of 3. 2-digit positive integers are 10, 11, 12.., 98, 99. These are 90 numbers out of which one number can be chosen in 90 ways. Total number of elementary events = 90. Out of these 90 numbers, 30 numbers (12, 15, 18., 96, 99) are multiples of 3.
What are the two digit numbers that are divisible by 3?
With that out of the way, the two digit numbers that are divisible by 3 are 12, 15, 18, …, 99, of which there are 30 (how do I know this?). Since there are 90 two-digit numbers in all (how do I know this?), the probability of choosing a two-digit number ‘at random’ (with the meaning described above) is 30 90 = 1 3.