TITLE: Counting outcomes for coin tosses
QUESTION [6 upvotes]: Don't laugh, this is a dumb question, but my brain just doesn't work mathematically.  A question in my math class says 

A coin is tossed 4 times.  Compute the probability of at least 2 tails
  occurring.

OK, so I know I figure out how many total events are in the sample, then figure out how many possible ways at least 2 tails are occurring, and divide.  My problem is, I can NEVER seem to figure out how many total events there are!  I start with HHHH, HHHT, HHTH, HTHH, and so on, but I always get lost somewhere along the way, miss an event, and never get them all.  My book says there are 16 different possibilities.  Is there a better way of figuring out how many different events could happen??

REPLY [2 votes]: Partition the possible results into three sets: $T_2$ (exactly two tails occur), $T_{>2}$ (more than two tails occur), and $T_{<2}$ (fewer than two tails occur). These sets are exclusive and exhaustive, so $P(T_2)+P(T_{>2})+P(T_{<2})=1$. Furthermore, $P(T_{>2})=P(H_{<2})$, and by symmetry, $P(T_2)+2P(T_{<2})=1$, so $P(T_{2})=1-2P(T_{<2})$ It’s not hard to find the probability P(T_2), which is the probability that the sequence of throws is an arrangement of $2$ $H$’s and $2$ $T$’s, of which there are $4\choose2$ many, giving a probability of ${4\choose2} / 2^4=\frac38$, so $P(T_{<2})=\frac{1-\frac38}2=\frac5{16}$, and therefore the probability of having not fewer than $2$ tails is $1-\frac5{16}=\frac{11}{16}$.