Sofia and Tess will each randomly choose one of the 10 integers from 1 to 10. What is the probability that neither integer chosen will be the square of the other?
Please, help

Question

Sofia and Tess will each randomly choose one of the 10 integers from 1 to 10. What is the probability that neither integer chosen will be the square of the other?
Please, help

loser66 · Answer

@oldrin.bataku

anonymous · Answer

here are our possible combinations where one is the square of the other: $(1,1),(2,4),(4,2),(3,9),(9,3)$ so there are $10^2-5=95$ such combinations that are not of this form, meaning there's a $95/10^2=0.95$ probability

loser66 · Answer

I don't get it!! where does 10^2 come from?

anonymous · Answer

there are 10 choices for Sofia, 10 independent choices for Tess, making a total of $10\cdot 10=10^2=100$ possible ways that Sofia and Tess choose

loser66 · Answer

That is they have 2 set of integers?
Is it not that there is just 10 numbers from 1 to 10 and if Sofia pick 1 then 1 will not be on the fool anymore?

loser66 · Answer

I think of the situation: only 2-4 and 3-9 are the forms we have to find the probability of them, then take 1 minus that probability
I meant the probability of choosing 2 or 3 of Sofia is 1/5, and under that condition, the probability of choosing 4 or 9 of Tess is 1/5 and we manipulate more.

anonymous · Answer

if they choose from the same set, then there will only be $10\cdot 9$ possibilities and only $(2,4),(4,2),(3,9),(9,3)$ will be on the table, so $90-4=86$ giving a probability of $86/90\approx 0.956$

loser66 · Answer

I got you. I think the last comment is best fit. Thank you so much. I got it now