# Within the interval {0,1} we randomly choose two numbers: x and y and Determine the probability that the number 5x+y is divisible by 3

my approach was like this ,

Assuming 𝑛 ∈ ℕ∗,two numbers are drawn randomly. Lets call the first one 𝑥 and the second 𝑦. Furthermore we learn that both are drawn from the uniform distribution on {0,1}. (Actually the distribution is not given, but I chose a simple one to not complicate matters). Because the two events are supposed to be independent (which I also assume), we can draw all possibly combinations as a pair of numbers (𝑥,𝑦) that is (0,1) & (1,0),

We need to establish the probability the number 5x+y is divisible by 3.

case1: when (𝑥,𝑦) = (0,1) : we get 5x+y = 1

1 is not divisible by 3.

case2: when (𝑥,𝑦) = (1,0) : we get 5x+y = 5

5 is not divisible by 3.

so the probability that 5x+y within interval {0,1} is divisible by 3 is zero.

I am definitely not convinced by my approach . kindly suggest.

You've got the right idea, but you're missing half the cases. Since $$x$$ and $$y$$ are independent, they can take the same value; we're sampling with replacement, if you like. As well as $$(0,1)$$ and $$(1,0)$$, $$(x,y)$$ can also be equal to $$(0,0)$$ or $$(1,1)$$. Add those two cases, and you should get the answer you're looking for.