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?
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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
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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
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?
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.
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 \(86/90\approx 0.956\)
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