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What's the probability of getting 3 heads and 7 tails is one flips a fair coin 10 times. I just can't figure out how to model this correctly.

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In how many ways can you order the string $HHHTTTTTT$?. – Pedro Tamaroff May 30 '12 at 22:42

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.

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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.

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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


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@Newbie1923 This approach is good here since all outcomes are equally likely. But if the coin is unfair not all outcomes will be equally likely, so you will have to do the binomial model instead. – rschwieb May 30 '12 at 22:58

You need to model this with a binomial distribution, with $n=10$ and $p=0.5$.

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Perhaps you could add a bit to this answer? – mixedmath Jun 2 '12 at 9:48

If you want them in that order then $\dfrac{1}{2^{10}} =\dfrac{1}{1024}$

If order does not matter then ${10 \choose 3}\times \dfrac{1}{2^{10}}=\dfrac{120}{1024} =\dfrac{15}{128}$

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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

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Hi The Bobster, please use appropriate formatting as described in the ChatJax tutorial to your right of the page. – Don Larynx Nov 14 '13 at 5:40
This answer provides only information that all of the other answers also provided - and the other answers are both older by a year and a half and much more complete. – Did Nov 14 '13 at 6:03

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).

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