Question

If a fair coin is tossed $10$ times, find the probability of

(i) Exactly six heads.
(ii) Atleast six heads.

Answer 1

Let $X$ denote the number of heads in the experiment of $10$ trials.

$\therefore X$ is binomial distribution
$n=10;P=\frac{1}{2}\Rightarrow q=1-p=1-\frac{1}{2}=\frac{1}{2}$

$\therefore P(X=x){=}^n{C}_x{q}^{n-x}{p}^x;x=0,1,2,...n$

$P(X=x){=}^{10}{C}_x{\left(\frac{1}{2}\right)}^{10-x}{\left(\frac{1}{2}\right)}^x{=}^{10}{C}_x{\left(\frac{1}{2}\right)}^{10}$

(i) $P(X=6){=}^{10}{C}_6{\left(\frac{1}{2}\right)}^{10}=\frac{10!}{4!(10-4)!}\cdot \frac{1}{2^{10}}=\frac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}\cdot \frac{1}{1024}=\frac{105}{512}$

(ii) $P$ (atleast six heads) $=P(X\ge 6)=P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)$

${=}^{10}{C}_6{\left(\frac{1}{2}\right)}^{10}{+}^{10}{C}_7{\left(\frac{1}{2}\right)}^{10}{+}^{10}{C}_8{\left(\frac{1}{2}\right)}^{10}{+}^{10}{C}_9{\left(\frac{1}{2}\right)}^{10}{+}^{10}{C}_{10}{\left(\frac{1}{2}\right)}^{10}$

$=\frac{1}{2^{10}}(\frac{10!}{6!4!}+\frac{10!}{7!3!}+\frac{10!}{8!2!}+\frac{10!}{9!1!}+1)=\frac{193}{512}$.