A number is randomly selected from the first 20 positive integers. What is the probability that it is divisible by 2, 3 or 4? Express your answer as a common fraction.

Guest Aug 1, 2017

#1**+1 **

To figure this problem out, let's first take a look at all the natural numbers up to 20.

The question asks for the probability for a number to be divisible by 2, 3, or 4. However, 2 is a factor of 4, so any number divisible by 4 is already divisible by 4, so there is no need to ever check for divisibility by 2. Also, if you prove a number is divisible by 2, you do not need to check if a number is divisible by 3.Let's take a look:

Natural Numbers from 1-20 | Divisible by 2? | Divisible by 3? | |||||||

1 | |||||||||

2 | ✔ | ||||||||

3 | ✔ | ||||||||

4 | ✔ | ||||||||

5 | |||||||||

6 | ✔ | ✔ | |||||||

7 | |||||||||

8 | ✔ | ||||||||

9 | ✔ | ||||||||

10 | ✔ | ||||||||

11 | |||||||||

12 | ✔ | ✔ | |||||||

13 | |||||||||

14 | ✔ | ||||||||

15 | ✔ | ||||||||

16 | ✔ | ||||||||

17 | |||||||||

18 | ✔ | ✔ | |||||||

19 | |||||||||

20 | ✔ | ||||||||

Now, let's count how many numbers are divisible by either 2 or 3. There are 13 numbers that divisible by either 2 or 3.

Therefore, there is a \(\frac{13}{20}\) chance that a number selected at random from the natural numbers 1 to 20 is divisible by either 2 or 3.

TheXSquaredFactor
Aug 1, 2017

#1**+1 **

Best Answer

To figure this problem out, let's first take a look at all the natural numbers up to 20.

The question asks for the probability for a number to be divisible by 2, 3, or 4. However, 2 is a factor of 4, so any number divisible by 4 is already divisible by 4, so there is no need to ever check for divisibility by 2. Also, if you prove a number is divisible by 2, you do not need to check if a number is divisible by 3.Let's take a look:

Natural Numbers from 1-20 | Divisible by 2? | Divisible by 3? | |||||||

1 | |||||||||

2 | ✔ | ||||||||

3 | ✔ | ||||||||

4 | ✔ | ||||||||

5 | |||||||||

6 | ✔ | ✔ | |||||||

7 | |||||||||

8 | ✔ | ||||||||

9 | ✔ | ||||||||

10 | ✔ | ||||||||

11 | |||||||||

12 | ✔ | ✔ | |||||||

13 | |||||||||

14 | ✔ | ||||||||

15 | ✔ | ||||||||

16 | ✔ | ||||||||

17 | |||||||||

18 | ✔ | ✔ | |||||||

19 | |||||||||

20 | ✔ | ||||||||

Now, let's count how many numbers are divisible by either 2 or 3. There are 13 numbers that divisible by either 2 or 3.

Therefore, there is a \(\frac{13}{20}\) chance that a number selected at random from the natural numbers 1 to 20 is divisible by either 2 or 3.

TheXSquaredFactor
Aug 1, 2017