Question 588541
6 switches with each switch either being on or off.
if the probability of a switch being on or off is the same, then the probability that the first 2 switches would be on and the other 4 switches would be off would be .5^6.
if you are just looking for the probability that the first 2 switches are on and don't care whether the other switches are on or off, then the probability that the first 2 switches would be on is .5^2.
we'll assume 3 switches for simplicity.
the number of possible conditions for the 3 switches is 2^3 = 8
those possible conditions are:
on on on
on on off
on off on
on off off
off on on
off on off
off off on
off off off
the probability for a switch being on or off is the same, namely 1/2 (.5).
this means that 50% of the time the switch will be on and 50% the switch will be off if you tested the condition of the switch an infinite number of times.
the probability that the first 2 switches would be on is therefore .5*.5 = .25
look at all the possible conditions of the 3 switches and you'll see that the on on condition for the first 2 switches occurs 2 times out of 8 which is the same as 25% of the time which is the same as .25.
look at all the possible conditons of the 3 switches and you'll see that the on on off condition for all 3 switches (first 2 are on and third is off) occurs 1 time out of 8 which is the same as .125.
the probability that the first 2 are on and the third is off is .5^3 which is also equal to .125.
bottom line is the formula works for 3 switches and there is no reason to believe the same formula won't work for 6 switches.
results are:
probability that the first 2 switches would be in the on position regardless of the position of the last 4 switches is .5^2.
probability that the first 2 switches would be in the on position and the other 4 switches would be in the off position would be .5^6.