What is the probability that the product of two randomly chosen (distinct) numbers between $1$ and $100$ (inclusive) is divisible by $3$?
Now, what I am doing is to find out the number of ways of selecting a multiple of $3$ out of $100$ and then selecting any of the other remaining numbers. This can be done in $^{33}C_1 \times ^{99}C_1$ ways. The whole process of selecting two numbers can be carried out in $^{100}C_2$ ways. That gives the probability as: $ \dfrac { 33 } { 100 } $. However this answer is wrong. What's wrong in what I am doing?
The given solution considers two separate cases. First case, you select both the numbers as multiples of $3$ and in the other case, you select one multiple and one non-multiple of $3$. Why do that?
Best Answer
Can you tell how many times the pair (3,6) was considered in your solution?
Once, choosing a multiple of 3 like 3 and then one of the remaining numbers, 6. Once again, choosing a multiple of 3 like 6 and then one of the remaining numbers, 3.
In you solution, the pair (3,6), just like all pairs with both numbers divisible by 3, was considered twice. That's why the solution split the problem in these cases.