How many outcomes are there for noting the remainder when a number is divided by 5?

When n is divided by 5, the remainder is 1. So, possible values of n are 1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56, 61, 66, 71, 76, etc. When n is divided by 7, the remainder is 3. So, possible values of n are 3, 10, 17, 24, 31, 38, 45, 52, 59, 66, 73, etc.

What is the probability that a number selected randomly from 1 to 500 is a multiple of 3 or 7?

What is the probability that both the balls are of different colours? Q5. Two men hit at a target with probabilities 1/2 and 1/3 respectively. What is the probability that exactly one of them hits the target?

When a positive integer is divided by 5 What can be the value of the remainder R?

When a positive integer n is divided by 5, the remainder is 2.

How many positive integers not exceeding 100 and are divisible by neither 5 nor 7?

68 numbers
So, there are exactly 68 numbers not exceeding 100 that are not divisible by 5 or by 7.

What is the divisibility rule for 7 11 and 13?

Testing divisibility by 7, 11, and 13 The original number is divisible by 7 (or 11 or 13) if this alternating sum is divisible by 7 (or 11 or 13 respectively). The alternating sum in our example is 963, which is clearly 9*107, and not divisible by 7, 11, or 13.

When a positive integer is divided by 3 in which form can it be returned?

General Discussion. When positive integer n is divided by 3, the remainder is 2; and when positive integer t is divided by 5, the remainder is 3.

How many positive integers are divisible by 35?

There are 100 positive integers not exceeding 100. Now out of these 100 numbers let x be set of integers divisible by 5, y be set of integers divisible by 7 and z be set of integers divisible by both of them therefore divisible by 35.

How many integers are divisible by 5 and/or 7?

See that z is intersection of sets x and y, therefore (N (x)+N (y)-N (z)) will be the number of integers divisible by 5 and/or 7 (we subtracted N (z) because elements of set z were present in both x and y there to avoid counting integers common to both sets we subtracted N (z) There are 100 positive integers not exceeding 100.

How can I understand shortcut tricks on probability?

You can understand shortcut tricks on Probability by these examples. First of all do a practice set on math of any exam. Choose any twenty math problems and write it down on a page. Do first ten maths using basic formula of this math topic. You also need to keep track of timing.