Question

A fair coin is tossed 8 times, find the probability of

(i) exactly 5 heads (ii) at least six heads (iii) at most six heads.

Answer 1

Let X denote the number of heads obtained when a fair is tossed 8 times.

Now, X is a binomial distribution with n = 8, $p=\frac{1}{2}$ and $q=1-\frac{1}{2}=\frac{1}{2}$.

$\therefore \mathrm{P}\left(X=r\right){=}^8{\mathrm{C}}_r{\left(\frac{1}{2}\right)}^{8-r}{\left(\frac{1}{2}\right)}^r{=}^8{\mathrm{C}}_r{\left(\frac{1}{2}\right)}^8,r=0,1,2,...,8$

(i) Probability of getting exactly 5 heads = $\mathrm{P}\left(X=5\right){=}^8{\mathrm{C}}_5{\left(\frac{1}{2}\right)}^8=\frac{56}{256}=\frac{7}{32}$

(ii) Probability of getting atleast 6 heads
$$\begin{gathered} =\mathrm{P}\left(X\ge 6\right) \\ =\mathrm{P}\left(X=6\right)+\mathrm{P}\left(\mathrm{X}=7\right)+\mathrm{P}\left(\mathrm{X}=8\right) \\ {=}^8{\mathrm{C}}_6{\left(\frac{1}{2}\right)}^8{+}^8{\mathrm{C}}_7{\left(\frac{1}{2}\right)}^8{+}^8{\mathrm{C}}_8{\left(\frac{1}{2}\right)}^8 \\ =\left(28+8+1\right)\times \frac{1}{256} \\ =\frac{37}{256} \end{gathered}$$
(iii) Probability of getting at most 6 heads
$$\begin{gathered} =\mathrm{P}\left(X\le 6\right) \\ =1-\left[\mathrm{P}\left(\mathrm{X}=7\right)+\mathrm{P}\left(\mathrm{X}=8\right)\right] \\ =1-\left[{}^8{\mathrm{C}}_7{\left(\frac{1}{2}\right)}^8{+}^8{\mathrm{C}}_8{\left(\frac{1}{2}\right)}^8\right] \\ =1-\left(\frac{8}{256}+\frac{1}{256}\right) \\ =1-\frac{9}{256} \\ =\frac{247}{256} \end{gathered}$$