Question

Hey can you please answer this?

Each time you click a toggle switch, the switch either turns from off to on or
from on to off. Suppose that you start with three toggle switches with one of
them on and two of them off. On each move you randomly select one of the
three switches and click it. Let m and n be relatively prime positive integers
so that
m/n
is the probability that after four such clicks, one switch will be on
and two of them will be off. Find m + n.

Answer 1

i don't think this will be easy, and i think it will take some time

Answer 2

did you try anything? for example, if all the moves are on one switch, then there will be one on and two off right?

Answer 3

likewise if three are on the first switch (the on one) and the other one is not , then there are still two off and one on

Answer 4

if two are on the first switch, then the next two must be on one, otherwise you get
on on on

Answer 5

after going through some cases, maybe it would be easier to find the probability that there is one off and two on. seem less likely

Answer 6

meant the other way around, sorry
one off, two on

Answer 7

i found out that its impossible that after 1 click and 3 clicks, it is impossible to have the same combination as the first time

Answer 8

if all four on one switch, one on and two off
if three on the on switch, then still one off and two on
if two from the on switch, then you have on off off and so the next two have to be from different off switches
that probability is \((\frac{1}{3})^4\) two from the first, one each from the second

Answer 9

also found that after 2 clicks, it is 1/3 probability to have them good

Answer 10

to summerize so far, all four from any one switch, one on and two off
three from first , one on and two off
two from first only get one off and two on if they are from different switches \(p=(\frac{1}{3})^4\)
what about one from the first switch?

Answer 11

one on the first switch, then the other two have to come from different switches to get all off

Answer 12

probability of that is also \(p=(\frac{1}{3})^4\)

Answer 13

so far only two ways not to end up with one on and two off, both with probability \((\frac{1}{3})^4\)

Answer 14

how about none on the first switch?

Answer 15

then must be three on one and one on the other, othewise still one on and two off

Answer 16

probability of that is \(2\times (\frac{1}{3})^4\)

Answer 17

i think that is it unless i miscounted

Answer 18

so although i would not bet on it, looks like total ways to get them not one on and two off is
\[4\times (\frac{1}{3})^4=\frac{4}{81}\] and so probability of one on and two off is
\[\frac{77}{81}\]

Answer 19

this may certainly be wrong, but if you check through the cases i think that is the method needed

Answer 20

im currently checking that out, thanks for you help :)