Question

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

Answer 1

Finding the probability that the coin lands on heads exactly $4$ times.

Given,

A coin is flipped $10$ times.

The probability of getting a head when a coin is tossed is,

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

The probability of getting a tail when a coin is tossed is,

$$\begin{gathered} \mathrm{q}=1-\mathrm{p} \\ =\frac{1}{2} \end{gathered}$$

The probability of getting exactly $r$ heads in $n$ tosses is given by,

$={}^{n}C_r{\left(p\right)}^4{\left(q\right)}^{n-r}$

Substituting the values $p,q,n=10$ and $r=4$ in the above formula, we get

$$\begin{gathered} \therefore \mathrm{Probability}={}^{10}\mathrm{C}_4{\left(\mathrm{p}\right)}^4{\left(\mathrm{q}\right)}^{10-4} \\ =\frac{10!}{4!\left(10-4\right)!}{\left(\frac{1}{2}\right)}^4{\left(\frac{1}{2}\right)}^{10-4} \\ =\frac{10!}{6!4!}\times \frac{1}{16}\times \frac{1}{64} \\ =\frac{10\times 9\times 8\times 7}{4\times 3\times 2\times 16\times 64} \\ =\frac{5\times 3\times 7}{4\times 2\times 64} \\ =\frac{105}{512} \end{gathered}$$

Therefore, the required probability is $\frac{105}{512}$.