If 4 distinct integers are chosen at random from the 9 small

Question

If 4 distinct integers are chosen at random from the 9 smallest positive integers, what is the probability that the median of the 4 integers is 5?

A.  \frac{1}{9} 
B.  \frac{1}{8} 
C.  \frac{1}{7} 
D.  \frac{1}{6} 
E.  \frac{4}{21}

CrazyBlender123 · Answer

Let the numbers be x1,x2,x3,x4
Since median = 5 is the condition to be satisfied.
x2+x3=10
so x2,x3 can (2,8),(3,7),(4,6). as x1<x2<x3<x4.
for case (4,6),
x1 can be 1/2/3 respectively.
x4 can be 7/8/9 respectively.
So we have 9 ways here.
for case (2,8),
The only number that can be formed are 1289 
So we have 1 way here.
for case (3,7),
x1 can be 1/2 respectively
x4 can be 8/9 respectively
So we have 4 ways here.

Therefore,
P = (9+1+4)/(9C4) = 1/9

stne · Answer

If the median is  5  then the  2^{nd}  and  3^{rd} ,integers must add upto  10 .

We cannot have  1  and  9  as the  2^{nd}  and  3^{rd}  as these cannot be the median.

Rest  2^{nd}  and  3^{rd}  integer combinations:

{}^{1}C_{1}, 2,8, {}^{1}C_{1} = 1  - for the first integer we have only one option, for the last integer also we have one option.
 {}^{2}C_{1}, 3,7, {}^{2}C_{1} = 4  - for the first integer we have two options, for the last integer also we have two options. 
 {}^{3}C_{1}, 4,6, {}^{3}C_{1} = 9  - for the first integer we have three options, for the last integer also we have three options.

Total favourable ways  = 1+4+9 = 14

Total ways  = {}^{9}C_{4} = 126

Prob = \frac{14}{126} = \frac{1}{9}

Ans A

CrazyBlender123 has posted the same thing. Just presented it in combinatorics way.

Hope it helped.

onlyPlanA · Answer

Any shorter method? Took me some time to solve it.

egmat · Answer

Hi onlyPlanA,

Good news: the method you saw (from CrazyBlender123 and stne) is basically the right one. But there's a  symmetry trick  that lets you skip most of the casework and just write down the count.

The key idea: everything is symmetric around 5

The median of  4  sorted numbers is (x2 + x3)/2. For that to equal  5 , the two middle numbers must be a pair centered on  5  - that is, (5 - d, 5 + d) for some gap d. So the only possible middle pairs are  (4,6), (3,7), (2,8), (1,9) .

Here's the shortcut. Once the middle pair is (5 - d, 5 + d):

- x1 must sit  below  5 - d - there are (4 - d) choices.
- x4 must sit  above  5 + d - there are also (4 - d) choices (perfect mirror).

So each pair contributes (4 - d)2 - a  perfect square  - with no listing needed:

- ( 4 , 6 ):  3 2 =  9 
- ( 3 , 7 ):  2 2 =  4 
- ( 2 , 8 ):  1 2 =  1 
- ( 1 , 9 ):  0 2 =  0  (nothing is below  1 )

Favorable total =  9 + 4 + 1 = 14 .

Finish

Total ways to pick  4  from  9  = C( 9 , 4 ) =  126 .

P =  14 / 126  =  1/9

Therefore , the answer is:

Answer: A

kevincan · Answer

When you say that it took you time , it would be great if you were more specific: for example, how long did it take you to see the three cases ?

PineToad · Answer

Are the integers randomly chosen with or without replacement?

egmat · Answer

Hi PineToad,

Good instinct to nail down the setup before trusting the count, since with-vs-without replacement completely changes a probability problem.

Here the question settles it for you with  one word: "distinct."

What "distinct" forces:  the  4  integers must all be different from one another. If you were choosing  with  replacement, you could draw the same number twice (like  5 ,  5 ,  7 ,  8 ) - but then they wouldn't all be distinct. So " 4  distinct integers" rules replacement out. The selection is  without replacement , and order doesn't matter - you're just picking a  4 -element subset.

That's exactly why the solutions you saw used  C(9,4) = 126  as the total. Combinations count unordered, no-repeat selections - which is precisely what " 4  distinct integers chosen from  1 - 9 " describes. If replacement were allowed, the total wouldn't be a simple  C(9,4)  at all, and the favorable cases like ( 1 , 2 , 8 , 9 ) couldn't be listed as clean  4 -number sets.

Quick way to read these cues in future: 
- "distinct" / "different" / "no two the same" -  without replacement 
- "a number is chosen, noted, and put back" or "can repeat" -  with replacement

So you're safe to trust the  14 / 126  =  1/9  count - the word "distinct" already locked in the without-replacement reading before any counting began.

Answer: A

If 4 distinct integers are chosen at random from the 9 small

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

If 4 distinct integers are chosen at random from the 9 small

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