this is not a fair method, as we found at least two different probabilities.
Is this a fair method to determine who has to sit out?
We will say that this is a fair method if and only if the probability for each of the 7 friends is exactly the same.
So, each friend selects a number from 1 through 7 and then it repeats, so, this means that the person who selects the number 1 has the possible outcomes:
{1, 8, 15, 22, 29, 36, 43, 50}
While the person who gets the number 7, has the possible outcomes:
{7, 14, 21, 28, 35, 42, 49}
Now, notice that the person who got the number 1 has 8 outcomes in which he sits out.
While the person who got the number 7 has 7 outcomes in which he sits out.
So, the person with the number 1 has a larger probability of sitting out than the person with the number 7, thus, this is not a fair method, as we found at least two different probabilities.
If you want to learn about probability:
https://brainly.com/question/25870256
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