Question

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

Answer 1

`(2^(n-1))/(5^n)(2n+3)`

Explanation:

We have given the coin is 0.6 biased means
`p=0.6=(6)/(10)=(3)/(5)`

q =1-p
`=1-(3)/(5)=(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(3)/(5)^0(2)/(5)^(n-0)+^nC_1(3)/(5)^1(2)/(5)^(n-1)=(2^(n-1))/(5^n)(2n+3)`