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# Probability in Divisibility | AMC-10A, 2003 | Problem 15

Try this beautiful problem from AMC 10A, 2003 based on Probability in Divisibility.

## Probability in Divisibility - AMC-10A, 2003- Problem 15

What is the probability that an integer in the set $${1,2,3,…,100}$$ is divisible by $$2$$ and not divisible by $$3$$?

• $$\frac {33}{100}$$
• $$\frac{1}{6}$$
• $$\frac{17}{50}$$
• $$\frac{1}{2}$$
• $$\frac{18}{25}$$

### Key Concepts

Number system

Probability

divisibility

Answer: $$\frac{17}{50}$$

AMC-10A (2003) Problem 15

Pre College Mathematics

## Try with Hints

There are total number of integers are $$100$$.and numer of integers divisible by $$2$$ is $$\frac{100}{2}$$=$$50$$. Now we have to find out divisible by $$2$$ and not divisible by $$3$$. so at first we have to find out the numbers of integers which are divisible by $$2$$ and $$3$$ both.......

can you finish the problem........

To be divisible by both $$2$$ and $$3$$, a number must be divisible by the lcm of $$(2,3)=6$$.

Therefore numbers of integers which are divisible by $$6$$=$$\frac{100}{6}=16$$ (between $$1$$ & $$100$$)

can you finish the problem........

Therefore the number of integers which are divisible by $$2$$ and not divisible by $$3$$= $$50 - 16=34$$.

So require probability= $$\frac{34}{100}=\frac{17}{50}$$

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Try this beautiful problem from AMC 10A, 2003 based on Probability in Divisibility.

## Probability in Divisibility - AMC-10A, 2003- Problem 15

What is the probability that an integer in the set $${1,2,3,…,100}$$ is divisible by $$2$$ and not divisible by $$3$$?

• $$\frac {33}{100}$$
• $$\frac{1}{6}$$
• $$\frac{17}{50}$$
• $$\frac{1}{2}$$
• $$\frac{18}{25}$$

### Key Concepts

Number system

Probability

divisibility

Answer: $$\frac{17}{50}$$

AMC-10A (2003) Problem 15

Pre College Mathematics

## Try with Hints

There are total number of integers are $$100$$.and numer of integers divisible by $$2$$ is $$\frac{100}{2}$$=$$50$$. Now we have to find out divisible by $$2$$ and not divisible by $$3$$. so at first we have to find out the numbers of integers which are divisible by $$2$$ and $$3$$ both.......

can you finish the problem........

To be divisible by both $$2$$ and $$3$$, a number must be divisible by the lcm of $$(2,3)=6$$.

Therefore numbers of integers which are divisible by $$6$$=$$\frac{100}{6}=16$$ (between $$1$$ & $$100$$)

can you finish the problem........

Therefore the number of integers which are divisible by $$2$$ and not divisible by $$3$$= $$50 - 16=34$$.

So require probability= $$\frac{34}{100}=\frac{17}{50}$$

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