SOLUTION: Each switch in a system of six switches can be either on or off. A power surge randomly resets the six switches. What is the probability that this random setting for the system is

Question 588541: Each switch in a system of six switches can be either on or off. A power surge randomly resets the six switches. What is the probability that this random setting for the system is one in which the first two switches are both in the on position?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
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.

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