- #1
Biochemgirl2002
- 30
- 1
Question:
A home security device with 10 buttons is disarmed when three different buttons are pushed in the proper sequence. (No button can be pushed twice.) If the correct code is forgotten, what is the probability of disarming this device?
My attempt:
10!/(10-3)! =( 10x9x8x7x6x5x4x3x2x1)/(7x6x5x4x3x2x1)
= 720
therefore there is a 1/720 chance of getting the right combination.
is this correct?
[Moderator's note: Moved from a technical forum and thus no template.]
A home security device with 10 buttons is disarmed when three different buttons are pushed in the proper sequence. (No button can be pushed twice.) If the correct code is forgotten, what is the probability of disarming this device?
My attempt:
10!/(10-3)! =( 10x9x8x7x6x5x4x3x2x1)/(7x6x5x4x3x2x1)
= 720
therefore there is a 1/720 chance of getting the right combination.
is this correct?
[Moderator's note: Moved from a technical forum and thus no template.]