Have a very good idea but would love some help / confirmation. Thank you!

A box contains 3 dice. Die 1 is biased such that a "1" is tossed with probability 0.75. The remaining probability is distributed equally among the other 5 outcomes. Die 2 is fair, and thus, is equally likely to come up any of the numbers "1" to "6". Finally, die 3 has "1"'s on all sides. A die is randomly selected from the box and is tossed. Let's define the following events: Let Di = die i is selected (i = 1, 2, 3). Let Of = the jth toss comes up as a "1". When answering the questions below, express your solution in terms of the events defined above. a. If the first toss is a "1", what is the probability that die 3 was tossed? b. Now, suppose the same die is tossed again, and another "1" is tossed. What is the probability that die 3 was tossed? C. Suppose that we kept tossing the same die, and a "1" kept appearing each time, what would you expect the conditional probability of flipping die 3 given that we continue to see "1"'s converge to? Explain. d. Now, consider a sequence of 4 tosses of the same die, suppose that we observe the sequence (1, 1, 1, 6). Determine the probability that we tossed die i, for i = 1, 2, 3, so you will be calculating three probabilities. Note: You can define S4 to be the event that a "6" is tossed on the 4th toss. e. Now let's start over again. All 3 dice are in the box again. One die is drawn randomly from the box. Suppose that we observe a "1" on every toss in a sequence of n tosses of this die. What is the minimum value of n such that the probability that we tossed die 3 is greater than 0.95