There are 10 coins and each one shows heads with probability p, independently of each of the other coins. Each coin which shows head is tossed again, and stop if shows tail. What is the probability mass function of the number of heads after the second round of tosses?
And what is the expected number of heads after the second round of tosses?
My attempt：
Let Y be the number of heads after the first tosses ~ Bin(10,p). Thus the probability that 0≤k≤10 tosses show heads in the first round is given by $P(Y=k)=\binom{10}{k} p^k(1−p)^{10−k}$
Let X be the number of heads from the second tosses.$P(X=m)=\binom{10}{m} (p^2)^m(1−p^2)^{10−m}$
Here I am confused about the question since the question is asking for the number of heads after the second round of tosses, does it mean the sum of two number? i.e. <= 20 heads total? How can I do it then? And how to calculate the expected value
The question states that each coin which shows head in the first toss is tossed again, and stop if shows tail. We need to find probability mass function of the number of heads after the second round of tosses. So I interpret it as number of heads that show after the second round of tosses and not total number of heads from two tosses.
As only coins that show $H$ in the first toss are tossed again, if $X$ is the number of heads after end of two tosses, probability mass function of $X$ is given by,
$ \displaystyle P(X = m) = \sum\limits_{k=m}^{10} \binom{10}{k} p^k(1−p)^{10−k} \binom{k}{m} p^m(1−p)^{k-m}$
$ \displaystyle = \sum\limits_{k=m}^{10} \cfrac{10!}{(10-k)! \ m! \ (k-m)!} \cdot p^{m+k}(1−p)^{10−m} $
where $m \in (0, 10)$.
Expected number of heads after two tosses
$E(X) = \sum\limits_{m=0}^{10} m \cdot p(X = m) $