Counting outcomes for coin tosses 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??  
 A: For every toss you have two different outcomes, there are four tosses, so you have $2\cdot 2 \cdot 2 \cdot 2 = 2^4 = 16$ different outcomes in total.
You could draw a binary tree to visualize the different roads to an outcome. 

For this problem it might be helpful to consider the not interesting events,
here the ones if no tail ($1$ case) and exactly one tail ($4$ cases) is tossed, so there must be $16-5 = 11$ events with at least two tails.
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A: If you want to list all the events, the important thing is to do it systematically so you don't miss any.
For instance, you can count in binary:
$$0000 
\\ 
0001
\\ 
0010  
\\
0011  
\\
0100   
\\
0101  
\\
0110  
\\
0111  
\\
1000  
\\
1001  
\\
1010  
\\
1011   
\\
1100  
\\
1101  
\\
1110  
\\
1111$$  
and then replace each $0$ by an H and each $1$ by a T (or the other way around).
(By the way, at the end, it's a good idea to count your entries to double-check that you didn't miss one.  In this case, the total number of entries should be $2^4=16.)$
A: As mvw stated, there are a total of 16 different outcomes, which you will use as your denominator. 
Now you need to find the total outcomes for at least 2 tails occurring among 4 coin flips. Then divide this by the total amount of outcomes (16).
There are ${^4C_2} + {^4C_3} + {^4C_4} = 6+4+1 = 11$ ways to get at least 2 tails.
Thus $11/16$ or $68.75\%$ chance of at least 2 tails occurring among 4 coin flips

PS: $^nC_r$ is the binomial coefficient, also written as $\binom n r$ or $\frac{n!}{r!(n-r)!}$, and counts ways to select $r$ things from a set of $n$; in this case, the coin tosses which turn up heads.
A: 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 $1-\frac5{16}=\frac{11}{16}$.
