Question

An unbiased coin is tossed 8 times . Find by using binomial distribution , the probability of getting atleast 3 heads .

Answer 1

$p=$ probability of getting head in one toss $=\frac{1}{2}$

$q=(1-p)=\frac{1}{2}$

$x=$ no. of heads

$\therefore p(x\ge 3)$

$=1-p(x\le 2)$

$=1-p(x=0)-p(x=1)-p(x=2)$

Now $p(x=0)={}^8{C}_0{\left(\frac{1}{2}\right)}^0{\left(\frac{1}{2}\right)}^8$

$={\left(\frac{1}{2}\right)}^8$

$p(x=1)={}^8{C}_1{\left(\frac{1}{2}\right)}^1{\left(\frac{1}{2}\right)}^7$

$=8\times {\left(\frac{1}{2}\right)}^8$

$p(x=2)={}^8{C}_2{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^6$

$=\frac{8!}{6!\times 2!}\times \frac{1}{2^8}$

$=56\times {\left(\frac{1}{2}\right)}^8$

$\therefore p(x\ge 3)=1-{\left(\frac{1}{2}\right)}^8-8\times {\left(\frac{1}{2}\right)}^8-56{\left(\frac{1}{2}\right)}^8$

$=1-\frac{65}{2^8}$.