# If you flip a fair coin 10 times, what is the probability that it lands on heads exactly 4 times?

$\setminus \frac{105}{{2}^{9}}$

#### Explanation:

The probability of getting a head in a single toss

$p = \frac{1}{2}$

The probability of not getting a head in a single toss

$q = 1 - \frac{1}{2} = \frac{1}{2}$

Now, using Binomial theorem of probability,

The probability of getting exactly $r = 4$ heads in total $n = 10$ tosses

${=}^{10} {C}_{4} {\left(\frac{1}{2}\right)}^{4} {\left(\frac{1}{2}\right)}^{10 - 4}$

=\frac{10\times9\times8\times7}{4!}\frac{1}{2^4}\frac{1}{2^6}

$= \setminus \frac{{2}^{4} \setminus \cdot 9 \setminus \cdot 35}{24 \left({2}^{10}\right)}$

$= \setminus \frac{105}{{2}^{9}}$

Jun 26, 2018

The probability is approximately 20.51%.

#### Explanation:

This question uses the binomial distribution.

Let $X$ be the number of heads in 10 tosses.
Then $X$ is distributed as "Bin"(n=10," "p=1/2).

The probability of $X$ being 4 is therefore

${\text{P"(X=4)=}}_{10} {C}_{4} {\left(\frac{1}{2}\right)}^{4} {\left(1 - \frac{1}{2}\right)}^{10 - 4}$

$\textcolor{w h i t e}{\text{P} \left(X = 4\right)} = 210 \left(\frac{1}{16}\right) {\left(\frac{1}{2}\right)}^{6}$

$\textcolor{w h i t e}{\text{P} \left(X = 4\right)} = 210 \left(\frac{1}{16}\right) \left(\frac{1}{64}\right)$

$\textcolor{w h i t e}{\text{P} \left(X = 4\right)} = \frac{210}{1024} = \frac{105}{512} \approx 0.2051$