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# I want to select 5 integers at random without the replacement from integers 1 through 12 inclusive?

Find the probability that at least one of the integers is even. Report answer using 3 decimal places......Im lost!! HELP!!

### 3 Answers

- Joe JLv 41 decade agoFavorite Answer
first you need to know how many ways to select 5 integers.

Based on the description, you care about combinations, not permutations so the total combinations is:

C(12, 5)

= 12! / (5! (12-5)!)

= 12!/(5! 7!)

= 12 * 11 * 10 * 9 * 8 / (5 * 4 * 3 * 2 * 1)

= 792

Now finding the number of combinations with one or more even integers is not very easy. However, finding the exact opposite is pretty straight forward. That would be combinations with no even integers. There's 6 odd intergers from 1 -12, and you need to choose 5. (note: whenever doing probability and combinatorics, always see if finding the the opposite is easier. it often is)

C(6, 5)

= 6! / (5! * (6-5)!)

= 6! / 5!1!

= 6

so there are 792 combinations, but only 6 have only odd numbers. So cominations with 1 or more even umbers is:

792 - 6 = 786

and the probability is:

786/792 = 99.2%

- 1 decade ago
Well I assume without the replacement from integers means once an integer has been selected, it cannot be selected again.

To solve this problem you need to figure out what the odds are that all the numbers you draw will be odd. To do this break it down into 5 different events.

The probability that the first number is odd is 6/12 or .5.

The probability that the second number is odd given the first number is odd is 5/11 or .454545 repeating.

The probability that the third number is odd given the first and second numbers are odd is 4/10 or .4.

The probability that the fourth number is odd given all the others so far are odd is 3/9 or .3333 repeating.

Lastly, the fifth number's probability of being odd is 2/8 or .25.

So to find the probability that you'll get all odds you multiply all these events' probabilities....so .5 * .454545 * .4 *.33333 * .25 = .00758

To find the probability that any number you draw is even you subtract that number from 1, which gives you 99.242%

- Anonymous1 decade ago
P(at least on even number) = 1 - P(no even number) = 1- 6C5

I think.....................