Gwydion wrote:
Hi, could you please tell me the mistake in the following logic?
First person entering pushes one button, it doesn't matter which: p=1
Second person pushes a button that has not been pressed before. Since one is already pushed, only 5 remain: p=5/6
Same logic for third person: p=4/6
.
.
.
This leaves us with The probability of all pushing a different button of: 1*5/6*4/6*4/6*2/6*1/6 or 5!/6^5
Where's the mistake?
Expanding on Gwydion's post - it is true that p=1, but because the answers all have 6 in them, the first person should be written as p=6/6 to make it easier to see the answer:
First person walks in and can push any button (6/6)
Probability that second person will press any of the remaining 5 buttons (5/6)
Probability that third person will press any of the remaining 4 buttons (4/6)
Probability that fourth person will press any of the remaining 3 buttons (3/6)
Probability that fifth person will press either of the remaining 2 buttons (2/6)
Probability that sixth person will press the remaining button (1/6)
\(\frac{6}{6} * \frac{5}{6} * \frac{4}{6} * \frac{3}{6} * \frac{2}{6} * \frac{1}{6} = \frac{6*5*4*3*2*1}{6*6*6*6*6*6} = \frac{6!}{6^6}\)
Answer is A