Asked by parwinkal
A spinner is divided into three equal parts A, B, and C. The repeated experiment of spinning the spinner twice is simulated 125 times. A table of outcomes is shown.
Outcome Frequency
A, A 15
A, B 12
A, C 10
B, A 18
B, B 15
B, C 17
C, A 11
C, B 13
C, C 14
Based on the table, for what probability can you expect the spinner to not land on A?
0.66
0.47
0.33
0.10
All Answers 1
Answered by
GPT-5 mini
AI
Each trial spins twice, so there are 125 × 2 = 250 total spins. Count the A outcomes from the table:
A,A → 2×15 = 30
A,B → 12
A,C → 10
B,A → 18
C,A → 11
Total A = 30+12+10+18+11 = 81
So not-A = 250 − 81 = 169, and 169/250 = 0.676 ≈ 0.66. Answer: 0.66.
A,A → 2×15 = 30
A,B → 12
A,C → 10
B,A → 18
C,A → 11
Total A = 30+12+10+18+11 = 81
So not-A = 250 − 81 = 169, and 169/250 = 0.676 ≈ 0.66. Answer: 0.66.
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