Misunderstanding basic probability question I appear to be misunderstanding a basic probability concept. The question is: you flip four coins. At least two are tails. What is the probability that exactly three are tails? 
I know the answer isn't 1/2, but I don't know why that's so. Isn't the probability of just getting 1 tail in the remaining two coins 1/2?
Thanks
 A: Count like this:  When you flip 4 coins, there are 16 possible outcomes.  List them and cross off all the cases which do not have at least two tails.  That leaves 11 possibilites.  Of the 11, how many have exactly 3 tails?  $4$.  So the answer is $4/11.$
A: I believe the source of your confusion is arising from the fact that the specific tosses on which the two tails took place are not specified. So if we compute the conditional probability which is being described, letting $A$ be the event that we get at least $2$ tails, and $B$ be the event that we get exactly $3$ tails, we get: $$P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{P(B)}{P(A)}=\frac{4(1/2)^{4}}{(6+4+1)(1/2)^{4}}=\frac{4}{11}.$$ You can see quite clearly the importance of the number of ways these different sequences can happen: there are $6$ sequences of heads and tails that result in $2$ tails, but only $4$ sequences of heads and tails that result in $1$ tail, and this results in the probability being smaller than your original attempt.
A: "At least two are tails" does not specify which coins are tails – or heads for that matter. The $\frac12$ answer assumes that two specific coins are tails first, but either or both may be heads.
