Probability Problem

2 numbers out of 16 numbers can be chosen in 120 ways 16c2=120

The table it symmetrical:

1 2 3 4 are same with 13 14 15 16 ( Top Row & Bottom Row )

and

5 6 7 8 are same with 9 10 11 12 ( Middle Rows )

1 and 4 will be a rectangle with 2, 5 and 3,8 = 4 total
2 and 3 will be a rectangle with 1,6,3 and 2,7,4 = 6 total

also you should keep in mind that these rectangles overlap!

so 6+4=10

5 and 8 will be a rectangle with 1,6,9 and 4,7,12 = 6 total
6 and 7 will be a rectangle with 2,5,7,10 and 3,6,11,8 = 8 total


so 8+6=14

Since the rest of the matrix is simetrical and we overlap all the rectangles:

10+14=24 total

24/120 = 1/5

It takes time to explain and write it but this can be solved under 2 minutes.

Similar approach

Picking 2 numbers out 16 16C2 = 120

LISTING OUT THE POSSIBILITES BY BRUTE FORCE APPROACH
1:2,1:5,2:3,2:6,3:4,3:7,4:8,5:6,5:9,6:7,6:10,7:8,7:11,8:12,9:10,9:13,10:11,10:14,11:12,11:15,12:16,13:14,14:15,15:16

=24

So 24/120 = 1/5

Friendly suggestion :If we are not able to find the shortcut method we are better off going wiht the brute force approach instead of wasting time trying to come up with a shorcut.

cramya wrote:Similar approach

Picking 2 numbers out 16 16C2 = 120

LISTING OUT THE POSSIBILITES BY BRUTE FORCE APPROACH
1:2,1:5,2:3,2:6,3:4,3:7,4:8,5:6,5:9,6:7,6:10,7:8,7:11,8:12,9:10,9:13,10:11,10:14,11:12,11:15,12:16,13:14,14:15,15:16

=24

So 24/120 = 1/5

Friendly suggestion :If we are not able to find the shortcut method we are better off going wiht the brute force approach instead of wasting time trying to come up with a shorcut.

Cramya,

Tell us more about Brute Force Approach master!

I am all ears!

Serioulsy the master stuff creeps me out Jus kidding!

I was just saying that listing out the possibiliteis works sometimes (may not work if there are 1000's of possibilities) In this problem it just happenned to work wihtin the 2 minutes we have to solve a problem!

Strictly my opinion. I think ur approach is good but not all minds are alike!! Whatever works!

I did it this way using the attached image:

Two numbers can form a rectangle, only if one of their sides is common. There are three methods of choosing such numbers. Each RED number if chosen first can have 2 numbers which will help it make a rectangle. Similarly each BLUE number can have 3 such numbers and each BLACK number can have 4 such numbers.


From the table

Total probability = probability when the first selection is a Red number + probability when the first selection is a Blue number + probability when the first selection is a Black number

= (4/16*2/15)+(8/16*3/15)+(4/16*4/15)
=1/30+1/10+1/15
= (1+3+2)/30=6/30=1/5



For ease of understanding:

probability when the first selection is a Red number
if first number is a RED one, then there are only TWO numbers which if chosen next can make a rectangle eg. 1 with 2, 5 & 4 with 3,8 & 13 with 9,14 & 16 with 12, 15
ie probability of choosing method 1 = Probability of choosing a RED no from sixteen numbers * Probability of choosing the two numbers from fifteen which will form the rectangle
= (4/16)*(2/15)


probability when the first selection is a Blue number
if first number is a BLUE one, then there are only THREE numbers which if chosen next can make a rectangle eg. 2 with 1,3,6 & 3 with 2,7,4 etc.
= (8/16*3/15)

probability when the first selection is a Black number
if first number is a BLACK one, then there are only FOUR numbers which if chosen next can make a rectangle eg. 6 with 2,5,10,7 etc
= (4/16*4/15)

to get the final probability, just add up these three probabilities.

All of our methods are taking FOREVER!!

I need to know the GMAT way of solving this.

No of rectangles: 16C2
No of rectangles per row: 3
No of rows: 4
Flip the square to double the combinations:
total: 3*4*2 = 24
Probability: 24/120

brianoduor wrote:No of rectangles: 16C2
No of rectangles per row: 3
No of rows: 4
Flip the square to double the combinations:
total: 3*4*2 = 24
Probability: 24/120

NASTY!!!!

Now we are talking! you are the man! This is how it is done ladies and gentleman!