Tom says that he needs 6 rolls to obtain each possible outcome on a 6-sided die. On the fourth roll, he rolls his second "3”. Tom says that the die is loaded and that each outcome is not equally likely. Is Tom correct here? If you think Tom is incorrect, how many rolls should Tom make until he sees each number occurring about 1/6 of the time? You must show all of your work to receive credit.

I need the last part as I know that he is incorrect

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Step-by-step explanation:

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let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3

then the equation is (x-x₁)(x-x₂) = 0

if you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.

step-by-step explanation:

you can see that a=1 (there is just an x²)

b = (-5 -2√7 + 2√7 - 5)/3 = -10/3

c = -2√7)/3 ) * (-5+2√7)/3) = -1/3

so your answer is 3x^2 – 10x – 1 = 0