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

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

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Answer 1 · 2017-12-14T03:56:04

$12/120=1/10=10%$

Explanation:

There are $5! = 120$ ways to randomly arrange the 5 digits.

To satisfy the conditions, we want the first digit to be 7. We want the last digit to either be 4 or 6. With the 7 forced to be the first digit, we have 4 digits to account for.

To get an even digit, we have a population of 2 and we are choosing 1:

$"^nC_k=(n!)/((k!)(n-k)!)$ with $n="population", k="picks"$

$"^2C_1=(2!)/((1!)(2-1)!)=2! =2$

We can now account for the remaining 3 digits, which is $3! = 6$ .

Therefore, the number of ways to get an even number greater than 700 is:

$2xx6=12$

and the probability of randomly arranging numbers into this arrangement is:

$12/120=1/10=10%$

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

1 Answer

Explanation:

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