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

egmat · Answer

Hi reruma,

The key that unlocks this whole problem is to flip it around. A product of three numbers is  even  unless  every  one of them is odd. So instead of chasing the  9/10 , work with the complement:

P(all three odd) =  1  -  9/10  =  1/10

Step 1 - count the odds.  Among the first n positive integers, the number of odd ones is about half. So we need:

C(odds, 3) / C(n, 3) =  1/10

Here's the part most people miss:  the count of odds depends on whether n is even or odd , so you have to check both. That's exactly why the question says "sum of the  possible  values."

Step 2 - n even.  Say n =  2m , so there are m odds.

C(m, 3) / C(2m, 3) = (m-2) / (4(2m-1)) =  1/10

Solving:  10 (m-2) =  4 (2m-1) -  2m = 16  - m =  8 , so  n = 16 .

Check: C( 8 , 3 ) / C( 16 , 3 ) =  56 / 560  =  1/10 . ✓

Step 3 - n odd.  Say n =  2m-1 , again m odds. Setting the ratio to  1/10  and simplifying leads to m2 -  2 m -  3  =  0  - (m-3)(m+1) =  0  - m =  3 , so  n = 5 .

Check: C( 3 , 3 ) / C( 5 , 3 ) =  1/10 . ✓

Step 4 - add them.  Possible values are  5  and  16 , so the sum is  21  - answer  E .

The trap:  if you only solve the even case, you get n =  16  and grab  B . The whole point of "sum of possible values" is that there's a second value hiding in the odd-n case - always test both parities when the number of odds isn't fixed.

Answer: E

HarshavardhanR · Answer

P (even product) = 9/10 (given)

Then,

P (odd product) = 1 - 9/10 = 1/10.

So, saying that the P (even product) = 9/10 is the same as saying that the P (odd product) = 1/10.

When will the product of 3 integers be even?

If even one of the 3 integers is even.

When will the product of 3 integers NOT be even i.e., be odd?
 
If not even one of the 3 integers is even.

-----

Case 1:  n is an even number. Say n = 2k

So,
- # odd numbers = k
- # even numbers = k

Given:

\frac{kC3}{nC3}  =  \frac{kC3}{2kC3}  =  \frac{1}{10}

Simplifying this, we get -> 5k - 10 = 4k - 2

=> k = 8

Therefore, one possible value of n = 2k = 16

Case 2:  n is an odd number. Say n = 2k + 1. Then, there will be k even numbers and k+1 odd numbers.

For example: 1,2,3,4,5,6,7 -> # odd = 4, # even = 3. In a case like this, there will be one more odd number.

Given:

\frac{(k+1)C3}{(2k+1)C3}  = 1/10

Simplifying this, we get -> 5 k^2  - 5 = 4 k^2  - 1

=> k = 2 (k cannot be -2).

Therefore, another possible value of n = 2k+1 = 5.

----

Thus,

The  sum of the possible values of n = 16 + 5 =   21 . Choice E.

---
Harsha

If three of first n positive integers are chosen at random

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