If a 3-digit integer is selected at random from the integers 100 through 199, inclusive, what is the probability that the first digit and the last digit of the integer are each equal to one more than the middle digit?

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Solution

The correct option is D
Total numbers from 100 to 199 = 100;
Only number which satisfies the given conditions is 121;
Probability = $\frac{1}{100}$

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If a 3-digit integer is selected at random from the integers 100 through 199, inclusive, what is the probability that the first digit and the last digit of the integer are each equal to one more than the middle digit?

The correct option is D Total numbers from 100 to 199 = 100; Only number which satisfies the given conditions is 121; Probability = 1100

The correct option is D
Total numbers from 100 to 199 = 100;
Only number which satisfies the given conditions is 121;
Probability = $\frac{1}{100}$