Answer to question 10

The outcomes are all the different ways we can draw a ticket from F, replace it, then draw another from F.

To solve the problem, we need to identify each of the three tickets in F.  Let's call one of those with a 0, "0_A", and the other one with 0, "0_B".  The third one, which has a 1, we will call "1".

The outcomes then are all possible combinations of the two tickets--in order--which are 0_A followed by any of 0_A, 0_B, or 1; 0_B followed by any of 0_A, 0_B, or 1; or 1 followed by any of 0_A, 0_B, or 1.  That's nine combinations.

The box for F # F therefore has nine tickets, one for each of the nine combinations.