Three distinct integers are selected at random between 1 and 2016, inc

Question

Three distinct integers are selected at random between 1 and 2016, inclusive (without replacement). Which of the following is a correct statement about the probability p that the product of the three integers is odd?

(A) p < 1/8
(B) p = 1/8
(C) 1/8 < p < 1/3
(D) p = 1/3
(E) p > 1/3

(A) p < 1/8
(B) p = 1/8
(C) 1/8 < p < 1/3
(D) p = 1/3
(E) p > 1/3

TornDilemma · Answer

No of Odd Integers between 1 and 2016=1008

Probability that thier product is odd= \frac{1008C3}{2016C3}

\frac{1008*1007*1006}{2016*2015*2014}=\frac{1006}{2*2*2015} .

Now,  \frac{1}{2015}<\frac{1}{2012} 
So, \frac{1006}{2*2*2015}<\frac{1006}{2*2*2012}  or  p<\frac{1}{8} .

IMO, Option A.

greenisthecolor · Answer

I took way too much time to solve this. Could someone tell me a quick way to solve this?

nick1816 · Answer

Product of integers is odd, when all of them are odd numbers.
There are 1008 odd numbers from 1 to 2016

the probability p that the product of the three integers is odd= ( \frac{1008}{2016} )*( \frac{1007}{2015} )*( \frac{1006}{2014} )
p= \frac{(1008*1007*1006)}{(2016*2015*2014)} 
p={ \frac{1}{4} }*( \frac{1006}{2015} )

\frac{1006}{2015} < \frac{1007.5}{2015} 
Hence p=  \frac{1}{4*2.xyz} 
p<1/8

IanStewart · Answer

The probability our first selection is odd is 1/2. But if we're doing things without replacement, once we remove that first odd number, the probability the next is odd will be slightly less than 1/2, because now less than half of our numbers are odd. And the probability the third is odd again is even smaller than that. So the probability all three are odd will be less than (1/2)(1/2)(1/2) = 1/8 and A is the answer.

aliakberza · Answer

I tried to solve and reduce the fractions too, but then realized that should there have been replacement (i.e. numbers would be brought back to 2016 after each selection) the probability would have been exactly 1/2 * 1/2 * 1/2 = 1/8. Now as a general rule of thumb when the same number is added or subtracted to/from both the numerator and denominator, the resulting fraction will always have a greater or lower value respectively. Because there is no replacement each time a number is selected, we know that the absolute value of the fraction is decreasing after each selection. If with replacement we know the probability would have been 1/8, then without replacement it should be p<\frac{1}{8} . Hope this helps.

IanStewart · Answer

That is true only when the overall value of the fraction is less than 1 (and when everything is positive). If you instead start with a fraction like 3/2, and you add 1 to the numerator and denominator, you'll see that the value drops. And if any numbers might be negative, the rules are a lot more complicated.

aliakberza · Answer

Correct. Thanks for the input. IanStewart

ScottTargetTestPrep · Answer

In order for the product of the three integer to be odd, each integer must be odd. There are 1008 odd integers from 1 to 2016, inclusive, so P(odd and odd and odd) = 1008/2016 x 1007/2015 x 1006/2014.

We see that 1008/2016 is exactly 1/2. Although 1007/2015 and 1006/2014 are not exactly 1/2, they are very close to 1/2. Therefore, the probability of selecting 3 odd numbers is approximately ½ x ½ x ½ = ⅛. Finally, because 1007/2015 and 1006/2014 are each slightly less than ½, the probability sought will be slightly less than ⅛.

Answer: A

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

Three distinct integers are selected at random between 1 and 2016, inc

Prep Toolkit

Top 5 GMAT Debrief Videos

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

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Three distinct integers are selected at random between 1 and 2016, inc

Prep Toolkit

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

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards