Probability two distinct integers less than $100$'s product is divisible by $3$

Question

I have been working on a problem that I feel I used the right approach for, but my answer is wrong.

The problem statement is

What is the probability that the product of two distinct positive integers less than $100$ is divisible by $3$?

I have tried a pairing technique, where $3$ and every other number other than $3$ are joined, creating $99$ possibilities. Then, I extended this logic to $6$ and other multiples of $3$ up to $99$, and found the successes to be $67+68+69+70+\dots+98+99$ out of $9900$ choices.

Is there something wrong with my method, or did I possibly make a computational error? Any help is appreciated!

Answer 1

The positive integers less than $100$ are the numbers $1-99$. $33$ of them are divisible by $3$, hence $66$ are not.

There are $$\binom{66}{2}=2145$$ pairs of integers whose product is NOT divisble by $3$ out of $$\binom{99}{2}=4851$$ possible pairs. Hence, $2706$ out of $4851$ pairs lead to a number divisible by $3$, so the probabilty is $$\frac{82}{147}=0.55782$$

Answer 2

For the product to not be divisible by three, both integers would have to not be divisible by three.

Since there are $99$ integers to choose from, $66$ of which are not divisible by three, and the numbers must be distinct, the probability of picking two numbers both not divisible by three would be $\frac{66}{99}\cdot\frac{65}{98}=0.44218$.

Therefore, the probability that the product is divisible by three would be $.55782$.

Answer 3

Since the resulting number is divisible by $3$ if either or both of the chosen numbers are divisible by $3,$ it makes sense to initially work out the probability of the product not being divisible by 3, which requires that both of the distinct integers are also not divisible by three.

As written, there are $66$ (out of $99$) integers in range not divisible by $3$. The probability of choosing two of these is thus:

$$ \frac{\binom{66}{2}}{\binom{99}{2}} = \frac{\frac{66\cdot65}{2\cdot 1}}{\frac{99\cdot 98}{2\cdot 1}} = \frac{66\cdot65}{99\cdot 98}= \frac{2\cdot65}{3\cdot 98}= \frac{65}{147}$$

giving the required probability of divisibility by $3$ as $$1- \frac{65}{147}= \fbox{$\frac{82}{147}$}$$

Answer 4

There are $(1+2+3+4+5+6+7+8+9+10)\times 99=5445$ numbers divisible by 3 in the products of two positive integers less than 100.

The possible products are $99\times 99$ so the probability is $p=\dfrac{5}{9}$