Answer: Okay so im not entirely sure if you need this still or if i am right but this is what ive got.
a) the chance of landing on the letter D is 1/3 because there are 4 sections labeled D and total there is 12. Therefore, 4 is 1/3 of 12 giving me the answer.
b) the chance of landing on the letter A is 1/6 because there are only 2 sections labeled A and total there is 12. Since that is the case, 2 is 1/6 of 12.
c) the chance of landing on the letter A and D is easy to find. You would just add the chances of getting the letter A and the letter D and there is your answer. 1/3 + 1/6= 3/6 or 1/2.
d) the experimental probability of the spinner landing on B is a 6/20 chance because the number of trials are 20, the denominator, and 6 is the event, numerator, which gives 6/20 or 3/10 simplified.
e) the experimental probability of the spinner landing on E is a 2/20 chance because the number of trials is 20, which is the denominator, and 2 is the event, numerator which gives 2/20 or 1/10 simplified
