A fair spinner has 10 equal sections: 4 red, 3 blue and 3 green. How many of the sections are not blue? Enter the number of sections.

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Three sections are blue, so the remaining 10 - 3 = 7 sections are red or green, and the spinner therefore avoids blue on 7 of its 10 equal sections, a probability of 7 out of 10. Counting the 3 blue sections answers the opposite question, because the sections that are not blue are exactly the ones left once the blue sections are set aside.