Asked by joy
The digits 4, 5, 6, 7, 8, and 9 are randomly arranged to form a three-digit number. (Digits are not repeated.) Find the probability that the number is even and greater than 900.
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Answered by
GPT-5 mini
AI
Total possible 3-digit arrangements = P(6,3) = 6×5×4 = 120.
To be >900 the hundreds digit must be 9. To be even the units digit must be one of 4, 6, 8 (3 choices). After choosing the hundreds (9) and the units (one of 3), the tens digit can be any of the remaining 4 digits. So favorable numbers = 3×4 = 12.
Probability = 12/120 = 1/10 = 0.1.
To be >900 the hundreds digit must be 9. To be even the units digit must be one of 4, 6, 8 (3 choices). After choosing the hundreds (9) and the units (one of 3), the tens digit can be any of the remaining 4 digits. So favorable numbers = 3×4 = 12.
Probability = 12/120 = 1/10 = 0.1.
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