## Loser66 one year ago 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

1. Loser66

@oldrin.bataku

2. anonymous

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

3. Loser66

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

4. anonymous

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

5. Loser66

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?

6. Loser66

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.

7. anonymous

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 $$86/90\approx 0.956$$

8. Loser66

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