Discussion
A game consists of spinning an arrow around a stationary disk as shown below. When the arrow comes to rest, there are eight equally likely outcomes. It could come to rest in any one of the sectors numbered 1, 2, 3, 4, 5, 6, 7 or 8 as shown.
Two such disks are used in a game where their arrows are independently spun.
What is the probability that the sum of the numbers on the resulting sectors upon spinning the two disks is equal to 8 after the arrows come to rest?
Answer: D
The probability is 7/64.
Each disk stops on one of 8 equally likely numbers, independently, so there are \(8 \times 8 = 64\) equally likely pairs. The pairs adding to 8 are
(1, 7), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (7, 1)
that is, 7 pairs; (8, 0) is impossible because no sector is numbered 0. The probability is therefore 7/64.
3/32, which is 6/64, leaves out the pair (4, 4); 1/16, which is 4/64, counts each unordered pair such as {3, 5} only once; and 5/64 also undercounts the seven ordered pairs.
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