Math problem challenge

[chuckywang]

Given that five vertices of a cube are selected, what is the probability that one face has all its vertices selected?

 

[dug777]

5/8+4/7+3/6+2/5+1/4 i think (or should the +'s be x's?)

 

[Mark R]

3/7?

My solution (may include spoilers)

 

[Kelemvor]

Pretty low. There's 12 total and you have to get just the right 4...

 

[leftyman]

um..11?

 

[dug777]

Originally posted by: FrankyJunior
Pretty low. There's 12 total and you have to get just the right 4...

oops..well repeat what i did but with 12 instead of 8...and find out whether it is + or x

 

[Fraggable]

2

 

[Mark R]

Pretty low. There's 12 total and you have to get just the right 4...

12?

 

[dug777]

Originally posted by: Mark R

Pretty low. There's 12 total and you have to get just the right 4...

12?

i thought there were 8 myself...

 

[Mark R]

i thought there were 8 myself...

This is, admittedly, an unusual occurrence. But you were correct.

 

[dug777]

Originally posted by: Mark R

i thought there were 8 myself...

This is, admittedly, an unusual occurrence. But you were correct.

did my initial maths look right then? i really can't remember whether i should have used + or x tho'...

 

[Mark R]

Originally posted by: dug777
did my initial maths look right then? i really can't remember whether i should have used + or x tho'...

Sadly not.

You should have used x rather than +. Even then you'd only have worked out the probability of selected a particular combination of 5 of 8 vertices, and not the number of possible combinations of vertices that meet the requirement.

 

[bonkers325]

nvm

 

[Confused]

42

 

[dug777]

Originally posted by: Mark R

Originally posted by: dug777
did my initial maths look right then? i really can't remember whether i should have used + or x tho'...

Sadly not.

You should have used x rather than +. Even then you'd only have worked out the probability of selected a particular combination of 5 of 8 vertices, and not the number of possible combinations of vertices that meet the requirement.

drat...it's been a long time since i did applicable maths

 

[akubi]

24/56 = 0.428571429

 

[charlieg6]

Depends on if it's any face or a particular face. For any face: You can pick any first corner, so the probability starts with 1. Now you have 1 corner attached to 3 faces, so you can pick any corner attached to those three faces (6/7 probability). Picking the second corner you've narrowed it two faces, so you have 4 out of 6 vertices left to choose from. Finally, you must complete the face, which is a 1/5 probability.

6/7 * 4/6 * 1/5 = .11428 (11.42%)

If it's a predetermined face that has to be selected it's much easier:
4/8 * 3/7 * 2/6 * 1/5 = 0.01428 (1.4%)

Edit: I'm wrong, misread the question that you can select 5 (I only did four). I think it changes my answers to
22.856% and 2.857% respectively.

 

[CreativeTom]

ELEVENTYBILLION

 

[glugglug]

6 side * 4 potential other vertices / 8C5 = 4 * 6 / (8 * 7 * 6 / 3 / 2) = 24/56 = 3/7.

 

[bonkers325]

Originally posted by: chuckywang
Given that five vertices of a cube are selected, what is the probability that one face has all its vertices selected?

there are 6 faces of a cube, 4 vertices to a face. 8 vertices in total

probability that a face w/ 4 of its vertices selected is 1/6

the probability that one face will have its verticies selected is

(1/6) / [ (8!) / (5!3!) ]

0.167 / [ (8*7*6*5*4*3*2*1) / (5*4*3*2*1 * 3*2*1) ]

0.167 / [ 40320 / 720 ]

0.167 / 56

= 0.003 or .3%

 

[chuckywang]

Geez, there's a lot of answers here...