SOLUTION: We randomly select 4 prime numbers without replacement from the first 17 prime numbers. What is the probability that the sum of the four selected numbers is odd? Express answer in

Question 391079: We randomly select 4 prime numbers without replacement from the first 17 prime numbers. What is the probability that the sum of the four selected numbers is odd? Express answer in common fraction.

I did this problem as
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,61

(1/17 * 1/16 * 1/15 * 1/14) * 9
I multiplied by I could form sum as odd.
Please correct me. I am not myself sure with the approach.

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We select 5 numbers at random, with replacement, from the set of integers from 1 to 400 inclusive. What is the probability that the product of 5 numbers is even? Answer should be in common fraction.

I guessed my answer:
(1/400)^5

Please guide me in the above problems.
Thanks
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
There are 17C4 ways to choose four prime numbers without replacement. Suppose that 2 was not one of the numbers chosen. Then, any set of four numbers would result in an even sum (since the sum of four odd numbers is always even). Therefore, 2 must be one of the numbers, and there are 17C3 ways to choose the other three. Therefore the answer is (17C3)/(17C4) = 2/7.

For the second one, you should ask yourself "does (1/400)^5 make sense?" The number you guessed is an extremely minuscule number, which means it's almost impossible for the product to be even. Instead, find the probability that the product is odd, then subtract it from 1 (to find the probability that the number is even).
For the product to be odd, all 5 numbers must be odd. There are 200 odd numbers out of 400, so the probability of each number being odd is 1/2 (since we have replacement). Therefore the probability that the product is odd is (1/2)^5 = 1/32. Subtracting this from 1, we see that the probability of the product being even is 31/32.
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