A fair spinner is divided into 8 congruent sectors numbered 1 through 8. A player spins th… — ACT

A fair spinner is divided into $8$ congruent sectors numbered $1$ through $8$ . A player spins the spinner twice in succession. How many of the $64$ possible outcomes result in a sum of the two spins that is strictly greater than $12$ ?

Answer: 10 outcomes

Answer

There are 10 outcomes that yield a sum strictly greater than 12.

Systematically listing all ordered pairs

(x, y)

from

\{1, 2, \dots, 8\} \times \{1, 2, \dots, 8\}

such that

x + y > 12

yields 4 outcomes for sum 13, 3 outcomes for sum 14, 2 outcomes for sum 15, and 1 outcome for sum 16, totaling 10 valid outcomes.

Step-by-Step Solution

Determine the acceptable sums for the two spins

The possible sums strictly greater than 12 are 13, 14, 15, and 16.

Since each spin has a maximum value of 8, the maximum possible sum is 8 + 8 = 16.

Count the ordered pairs (spin 1, spin 2) for each valid sum

4 outcomes for sum 13, 3 outcomes for sum 14, 2 outcomes for sum 15, and 1 outcome for sum 16.

First and second spins are ordered, so (5,8) and (8,5) represent distinct outcomes.

Sum the outcome counts across all valid cases

4 + 3 + 2 + 1 = 10 outcomes.

The sets of outcomes for distinct sums are mutually exclusive.

Key Concept

Basic Probability and Counting Sample Space Outcomes

Estimated Time:1m 15s

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