Coin flip probability. At least 2 tails from 3 flips? Ok, so I'm a parent helping my 12yo with a 7th grade math question. We couldn't figure out the right way to get the answer for this one...
Q: A coin is flipped 3 times. What is the probability of getting at least 2 tails?
I thought the answer would be 1/2 x 1/2 which would equal 1/4 with the third flip not mattering, but that's not correct. Listing the outcomes (H being heads and T being tails... HHH, HHT, HTH, HTT, THH, THT, TTH, TTT), it's clear that 1/2 the outcomes result in at least 2 tails. So, is there a way to figure this out mathematically as a function of fractions, with each coin toss being a 1/2 probability? Is listing the outcomes and counting from there the only way?
Thanks in advance!
 A: One option is to, as you did here, just list out all the possibilities.  Another way of thinking about things:  The condition "at least two tails" is equivalent to "more tails than heads".  If you flip three coins, you'll be in exactly one of two situations:


*

*More tails than heads.

*More heads than tails.  


Which of these two situations (if any) is more likely?  What does that imply about the probability of each situation?  
A follow up question you and your 12yo might want to think about: If I flip $4$ coins, the probability of getting more tails than heads is not $\frac{1}{2}$.  Why can't I just use the same argument as I did for $3$ coins?  
A: Let's take case of two tails:
First flip gets you tail then in second it doesn't but in third it does. P = 0.5(tails)*0.5(heads)*0.5(tails)
Or there can be an another case
Where first flip gets you tail then second flip also gets you tail but third doesn't . P=0.5(tails)*0.5(tails)*0.5(heads)
Or 
First flip gets you head second gets you tail and third also gets you tail. P=0.5(heads)*0.5(tails)*0.5(tails)
Now take the case of three tails:
First gets you tails and second and third also gets you the same. P=0.5(tails)*0.5(tails)*0.5(tails)
Now add them all P= 0.125+0.125+0.125+0.125=0.5
So the probability is 0.5
You could have use combinations to count how many ways are there to arrange but for a 12 year child that would be too much ( if he is not Sheldon Cooper :P)
