**T5.11. Dice Game Probability Exercise**Your teacher has created a "fair" dice game. Here's how it works:- Your teacher rolls a fair eight-sided die.- You roll a fair six-sided die.- Each player rolls once.- The player with the higher number wins.- In case of a tie, neither player wins.The table below shows the possible outcomes:| | **Teacher Rolls** | ||-----------|:------------------:|------------|| **You Roll** | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 || **1** | | | | | | | | || **2** | | | | | | | | || **3** | | | | | | | | || **4** | | | | | | | | || **5** | | | | | | | | || **6** | | | | | | | | |**Questions:**(a) Let $ A $ be the event "your teacher wins." Find $ P(A) $.(b) Let $ B $ be the event "you get a 3 on your first roll." Find $ P(A \cup B) $.(c) Are events $ A $ and $ B $ independent? Justify your answer.

The total possible outcomes are, Teacher Rolls You Roll 1 2 3 4 5 6 7 8 1 (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (1, 7) (1, 8) 2 (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (2, 7) (2, 8) 3 (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (3, 7) (3, 8) 4 (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (4, 7) (4, 8) 5 (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (5, 7) (5, 8) 6 (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) (6, 7) (6, 8)(a) The event A be "Your teacher wins". The possible outcomes are, Teacher Rolls You Roll 1 2 3 4 5 6 7 8 1 (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (1, 7) (1, 8) 2 (2, 3) (2, 4) (2, 5) (2, 6) (2, 7) (2, 8) 3 (3, 4) (3, 5) (3, 6) (3, 7) (3, 8) 4 (4, 5) (4, 6) (4, 7) (4, 8) 5 (5, 6) (5, 7) (5, 8) 6 (6, 7) (6, 8) The probability PA is, PA=nATotal=2748=0.5625 Therefore, the probability PA is 0.5625.(b) The event B be "you get a 3 on your first roll". The possible outcomes are. (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (3, 7) (3, 8) The probability PB is, PB=nBTotal=848=16=0.1667 The probability PB is 0.1667. The possible outcomes for A∩B is, (3, 4) (3, 5) (3, 6) (3, 7) (3, 8) The probability PA∪B is, PA∪B=PA+PB-PA∩B=2748+848-548=27+8-548=3048=0.625 Therefore, the probability PA∪B is 0.625.(c) When events are independent happening of one event does not influence happening of another event. When events are independent if PA|B=PA or PB|A=PB. The probability PA|B is, PA|B=PA∩BPB=548848=58=0.625≠0.5625≠PA Therefore, the events A and B are not independent.