Seven students are called on at random on two separate occasions. One of the students has been called on for the first occasion. What is probably that this student will NOT be called on the second time?

The correct way to solve this (1/7) for the first time and (1/7) for the second time. The chances of the student being called on twice is (1/49), so 1-(1/49) is 48/49.

The student has a 48/49 chance of NOT being called on twice.

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However, I was wondering why the following solution doesn't work--if you were to use (1/7) for the first time BUT (6/7) the second time to represent the chances of another student being called on. (1/7)*(6/7) = 6/49.

So the first time, the student has a 1/7 chance of being called on. The second time, he has a 6/7 chance of NOT being called on. Why can't we use this to establish the chances of him being called on only once?