Probability Binomial Coin Flip

Question

Question: How many times must a fair coin be tossed so that the probability of at least one head exceeds 0.9?

I have no clue what to do. From what I have tried:

I might have to consider nCr $(1/2)^r(1/2)^{n-r} > 0.9$ but i dont see how that takes me anywhere?

Answer 1

Let's consider a single coin toss. The probability of not gettings heads is $0.5$. This means the probability of not getting heads $n$ times in a row is $0.5^n$.

Say that after $m$ coin tosses, the chance of getting heads at least once is $0.9$. This means that the chance of getting $m$ tails in a row is less than $0.1$, so: $$0.1>0.5^m$$ One could find the lowest such $m$ by simply trying out some possibilities, or taking the logarithm base $0.5$ of both sides: $$3.32\approx\log_{0.5}0.1>m$$ So the answer is $4$ times.

Answer 2

Pr(atleast 1 heads) + Pr(no heads) = 1

Pr(atleast 1 heads) = 1 - Pr(no heads) > 0.9

Pr(no heads) < 0.1

Let number of coins be n. So probability that one coin is tails(not heads) is 0.5.

For n coins this probability is $0.5^n$

Thus, we have

$0.5^n$<0.1

Taking log on both sides

n*log(0.5)< log(0.1)

(as log(0.5) is negative quantity, sign of inequality changes)

n>3.32

n=4