If three of first n positive integers are chosen at random

Question

If three of the first n positive integers are chosen at random, the probability that their product is even is 9/10. What is the sum of the possible values of n?

A) 15
B) 16
C) 17
D) 19
E) 21

reruma · Answer

How to do this problem? Thanks in advance.

kevincan · Answer

Consider two cases for n

kevincan · Answer

We know that the product of the three integers will be even if and only if at least one of them is even. The product is odd if and only if they are all odd: this is easier to work with

If n is even , n=2k for some integer k

The probability that the product of the three numbers is odd is 
 k(k-1)(k-2)/(2k)(2k-1)(2k-2) = (k-2)/4(2k-1)

This is equal to 1/10 if 10(k-2)=4(2k-1) ie k=8 and n = 16

kevincan · Answer

At this point, it’s wise to look at the answer choices

paragw · Answer

And for second case
If n is odd, n = 2k+1 for some integer k
 The probability that the product of the three numbers is odd is 
(k+1)(k)(k-1)/(2k+1)(2k)(2k-1) = 1/10
 (k^2-1)/(2)(4k^2-1) = 1/10
k^2 = 4
k = 2 & n = 5

Hence, 16 + 5 = 21

If three of first n positive integers are chosen at random

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If three of first n positive integers are chosen at random

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615 to 715 in 15 Days (Ria's Strategies)

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