Maria has a spinner with 2 sections labeled a, 3 sections labeled b, 2 sections labeled c, 4 sections labeled d, and 1 section labeled e. all 12 sections are the same size.
she spins it 20 times with outcomes as follows:

outcome
number of times

a
4

b
6

c
3

d
5

e
2

a) find the theoretical probability that the spinner lands on d.
b) find the theoretical probability that the spinner lands on a.
c) find the theoretical probability that the spinner lands on d or a.
d) find the experimental probability that the spinner lands on b.
e) find the experimental probability that the spinner lands on e.

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

ab will be equal to

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step-by-step explanation:

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