# What is the probability of at least N heads in M flips of an unfair coin?

I see many answers about getting an exact number of coin flip outcomes with a fair coin, and some about getting an exact number of coin flip outcomes with an unweighted coin (like this one) but I am curious about a the probability of at least a number of outcomes with an unfair coin.

e.g.

If I have an unfair coin that gets tails with a probability of P what is the probability that I get at least N number of heads in K flips?

For instance, if my coin gets tails 60% of the time, what is the chance that I get at least 10 heads in 100 flips?

• You are looking for the Binomial distribution, if I am not mistaken. May 16, 2020 at 8:19

Since each fair/unfair coin toss is a bernoulli trial with parameter P, the question of is really asking for the Cumulative distribution function of a Binomial distribution.

For instance, if we wanted to know what the chance of getting at least 10 heads in 100 flips where the probability of getting heads is 0.4 ($$P(X \ge 10)$$ where $$X \sim Bin(100,0.4)$$)

Using the formula given in the link we get a number extremely close to 1. Or use a calculator such as this to get

$$P(X \ge 10) > 0.999999$$

$$P(X\ge 10)=1-P(X\le 9) = 1- \sum\limits_{k=0}^9{100 \choose k}0.6^k0.4^{100-k}\approx 1 - 1.256\times 10^{-26}$$
• More precisely, $P(X\ge 10)=1-P(X\le 9) = 1- \sum\limits_{k=0}^9{100 \choose k}0.6^k0.4^{100-k}\approx 1 - 1.256\times 10^{-26}$, so many more $9$s May 16, 2020 at 9:12