Answer:
[tex]\frac{2^{n-1}}{5^n}(2n+3)[/tex]
Step-by-step explanation:
We have given the coin is 0.6 biased means [tex]p=0.6=\frac{6}{10}=\frac{3}{5}[/tex]
q =1-p[tex]=1-\frac{3}{5}=\frac{2}{5}[/tex]
We have to find the at most one head means sum of probability of 0 head and 1 head
So [tex]p=p(x=0)+p(x=1)[/tex]
From binomial distribution the probability is given by [tex]p=^nC_rp^rq^{n-r}[/tex]
So probability [tex]p=^nC_0\frac{3}{5}^0\frac{2}{5}^{n-0}+^nC_1\frac{3}{5}^1\frac{2}{5}^{n-1}=\frac{2^{n-1}}{5^n}(2n+3)[/tex]