Probability of 3 Heads in 10 Coin Flips What’s the probability of getting 3 heads and 7 tails if one flips a fair coin 10 times. I just can’t figure out how to model this correctly.
 A: Your question is related to the binomial distribution.
You do $n = 10$ trials. The probability of one successful trial is $p = \frac{1}{2}$. You want $k = 3$ successes and $n - k = 7$ failures. The probability is:
$$
\binom{n}{k} p^k (1-p)^{n-k} = \binom{10}{3} \cdot \left(\dfrac{1}{2}\right)^{3} \cdot \left(\dfrac{1}{2}\right)^{7} = \dfrac{15}{128}
$$
One way to understand this formula: You want $k$ successes (probability: $p^k$) and $n-k$ failures (probability: $(1-p)^{n-k}$). The successes can occur anywhere in the trials, and there are $\binom{n}{k}$ to arrange $k$ successes in $n$ trials.
A: If you want them in that order then $\dfrac{1}{2^{10}} =\dfrac{1}{1024}$
If order does not matter then ${10 \choose 3}\times \dfrac{1}{2^{10}}=\dfrac{120}{1024} =\dfrac{15}{128}$
A: We build a mathematical model of the experiment.  Write H for head and T for tail.  Record the results of the tosses as a string of length $10$, made up of the letters H and/or T.  So for example the string HHHTTHHTHT means that we got a head, then a head, then a head, then a tail, and so on.
There are $2^{10}$ such strings of length $10$.  This is because we have $2$ choices for the first letter, and for every such choice we have $2$ choices for the second letter, and for every choice of the first two letters, we have $2$ choices for the third letter, and so on.
Because we assume that the coin is fair, and that the result we get on say the first $6$ tosses does not affect the probability of getting a head on the $7$-th toss, each of these $2^{10}$ ($1024$) strings is equally likely. Since the probabilities must add up to $1$, each string has probability $\frac{1}{2^{10}}$.
So for example the outcome HHHHHHHHHH is just as likely as the outcome HTTHHTHTHT. This may have an intuitively implausible feel, but it fits in very well with experiments.
Now let us assume that we will be happy only if we get exactly $3$ heads.  To find the probability we will be happy, we count the number of strings that will make us happy. Suppose there are $k$ such strings. Then the probability we will be happy is $\frac{k}{2^{10}}$.
Now we need to find $k$. So we need to count the number of strings that have exactly $3$ H's.  To do this, we find the number of ways to choose where the H's will occur. So we must choose $3$ places (from the $10$ available) for the H's to be. 
We can choose $3$ objects from $10$ in $\binom{10}{3}$ ways. This number is called also by various other names, such as $C_3^{10}$, or ${}_{10}C_3$, or $C(10,3)$, and there are other names too. It is called a binomial coefficient, because it is the coefficient of $x^3$ when the expression $(1+x)^{10}$ is expanded.  
There is a useful formula for the binomial coefficients. In general
$$\binom{n}{r}=\frac{n!}{r!(n-r)!}.$$
In particular, $\binom{10}{3}=\frac{10!}{3!7!}$. This turns out to be $120$. So the probability of exactly $3$ heads in $10$ tosses is
$\frac{120}{1024}$.
Remark: The idea can be substantially generalized. If we toss a coin $n$ times, and the probability of a head on any toss is $p$ (which need not be equal to $1/2$, the coin could be unfair), then the probability of exactly $k$ heads is
$$\binom{n}{k}p^k(1-p)^{n-k}.$$
This probability model is called the Binomial distribution. It is of great practical importance, since it underlies all simple yes/no polling. 
A: You need to model this with a binomial distribution, with $n=10$ and $p=0.5$.
A: You are looking for 
$$\frac{\text{Number of Relevant Outcomes}}{\text{Number ofTotal Outcomes}}.$$
The number of total outcomes is $2^{10}$. The number of relevant outcomes is the number of ways you can get exactly three heads in a string of 10 coin flips, or ${10}\choose{3}$. So the answer is 
$$\frac{{10}\choose{3}}{2^{10}}.$$
A: If this is in a continuum and you just need the probability of 3 of 10, then the practical answer is shown on the grey line here.  I see it is about 15%.  Other tips on Statistical Ideas web log, including treatment of any broader trials prior to the selection (e.g., 10 in this case).
A: Using Pascal's Triangle:
Total of elements in row 10 = 1024
Choose element 4 of row 10 (this represents the number of ways three heads can occur): 120
Probability = 120/1024
A: What I would do is to first find the total # of outcomes.
$$\text{Outcomes}=2^{10}=1024$$
Then you have to arrange 3 heads and 7 tails within 10 open slots.
We can name our Heads: $H(A)$, $H(B)$, and $H(C)$. Now, you can arrange the heads in such ways that $H(A)$ will essentially have 10 open slots, $H(B)$ will have 9 open slots and $H(C)$ will have 8 open slots. In other words there are: $$10 \times 9 \times 8 = 720$$ ways to arrange these heads. However, this accounts for the fact that all heads are different. Therefore the number of positions in which $H(x)$ can be arranged in is $$3 \times 2 \times 1 = 6,$$ and so there are: $$\frac{720}{6} = 120 $$
scenarios of which 3 heads and 7 tails are given. Therefore
$$P= \frac{120}{1024} = \frac{15}{128} \approx 11.7\%.\tag{getting $3$ heads and $7$ tails}$$ 
A: I think the easiest way to understand this problem is to do some rephrasing. Once we do that you can see that it is a simple combination problem.
Original question: What is the probability of getting only three heads with 10 coin flips?
There are 2 possibilities for each coin flip and 10 flips so the total number of outcomes is $2^{10}=1024.$
Now we want to know the total number of outcomes that result in only 3 heads with 10 coin flips. This is our desired outcome. This is where I find rephrasing very helpful. 
Rephrase to: There are 10 canidates named Flip1, Flip2, Flip3,...Flip 10 running for the 3 council positions called 'Heads'. How many different groups of council members could there be if they all had eqaul chance of winning
. 
We can quickly see that this is a combination problem because all three positions are the same and we can't distinguish them apart. The number of combinations is:
$10!/((10-3)!*3!) = 120$ combinations
Therefore the total probability is $120/1024 = 15/128$
I hope this helps people see the similarity between this question and the more common combination problem format.
