The digits 1, 2, 3, 4, 5, 6, and 7 are randomly arranged to form a? three-digit number.? (digits are not? repeated.) find the probability that the number is even and greater than 700.

Question

Molly Holloway

Mathematics

20 May, 05:24

The digits 1, 2, 3, 4, 5, 6, and 7 are randomly arranged to form a? three-digit number.? (digits are not? repeated.) find the probability that the number is even and greater than 700.

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Lincoln · Answer 1

Nice twist of similar questions.

There are 7*6*5=210 possible 3-digit numbers formed by digits 1-7 without repetition.

For numbers greater than 700, the first digit must be 7.

Of the 30, the last digit must be a 2,4, or 6 to be an even number, that leaves 3 possibilities for the last digit.

The middle digit can be any number of the remaining 5 digits.

So the number of qualifying candidates would be

1*5*3=15 (1 for the first digit, 5 choices for the second, and 3 choices for the last digit).

Probability of randomly choosing a qualifying candidate (>700 & even) is therefore 15/210=1/14