Question

Calculate the probability of landing at most 1 head when a 0.6-biased coin is tossed

Answer 1

Answer:

$\frac{2^{n-1}}{5^n}(2n+3)$

Step-by-step explanation:

We have given the coin is 0.6 biased means $p=0.6=\frac{6}{10}=\frac{3}{5}$

q =1-p$=1-\frac{3}{5}=\frac{2}{5}$

We have to find the at most one head means sum of probability of 0 head and 1 head

So $p=p(x=0)+p(x=1)$

From binomial distribution the probability is given by $p=^nC_rp^rq^{n-r}$

So probability $p=^nC_0\frac{3}{5}^0\frac{2}{5}^{n-0}+^nC_1\frac{3}{5}^1\frac{2}{5}^{n-1}=\frac{2^{n-1}}{5^n}(2n+3)$