Question

Assume that we have m coins. We toss each one of them n times. The probability of heads showing up for each coin isp. What’s the probability of getting all heads for at least one coin?

Answer 1

Answer:

$1-(1-p^n)^m$

Step-by-step explanation:

For a coin, the probability of head showing in a single toss is $p$.

$P(H)=p$

Its complement, the probability of not head is

$P(\Sim H)=1-p$

This is a binomial distribution. In $n$ tosses, the probability of having all heads (i.e. $n$ heads) is

$P(\text{all heads})=\binom{n}{n}p^n(1-p)^0=p^n$

Let's call this value $a$.

For $m$ coins, we determine the probability of at least 1 coin showing all heads by first finding its complement i.e. the probability of no coin showing all heads. This is also a binomial distribution.

$P(\text{no coin showing all heads})=\binom{m}{0}a^0(1-a)^m=(1-a)^m$

$P(\text{at least 1 coin showing all heads})=1-P(\text{no coin showing all heads})$

$P(\text{no coin showing all heads})=1-(1-a)^m=1-(1-p^n)^m$