Question

8 coins are tossed simultaneously then what is the probability of geting atleast 6 heads?

Answer 1

This is a binomial distribution with $n=8$ and $p=1/2$
$Pr(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$

$Pr(X=6)=\binom{8}{6}(1/2)^6(1-1/2)^{8-6}=\frac{7}{64}$
$Pr(X=7)=\binom{8}{7}(1/2)^7(1-1/2)^{8-7}=\frac{1}{32}$
$Pr(X=8)=\binom{8}{8}(1/2)^8(1-1/2)^{8-8}=\frac{1}{256}$

$Pr(X\ge{6})=Pr(X=6)+Pr(X=7)+Pr(X=8)$

$=\frac{7}{64}+\frac{1}{32}+\frac{1}{256}$

$=\boxed{\frac{37}{256}}$

Answer 2

P(H)= 37/256

This is the probability. You may take P>5