0
$\begingroup$

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?

$\endgroup$
1
0
$\begingroup$

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 $

For more info see this question.

A precise answer thanks to Henry in the comments

$P(X\ge 10)=1-P(X\le 9) = 1- \sum\limits_{k=0}^9{100 \choose k}0.6^k0.4^{100-k}\approx 1 - 1.256\times 10^{-26}$

$\endgroup$
1
  • 1
    $\begingroup$ 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 $\endgroup$ – Henry May 16 '20 at 9:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.