Suppose a coin is tossed which has a probability $q$ of returning heads.
Now there are $n$ people who are guessing the outcome of this toss. These people choose independently of each other and have a probability $p$ of choosing heads.
Each toss divides the group of $n$ people into $\mathbb{H}$ and $\mathbb{T}$ where $\mathbb{H}$ are those who have choosen heads and $\mathbb{T}$ for tails.
Let $H:=|\mathbb{H}|$ and is clearly $BinomialDistribution(n,p)$ and similarly $|\mathbb{T}|=n-|\mathbb{H}|=n-H$ is clearly $BinomialDistribution(n,1-p)$.
My question is what is the distribution of the random variable representing number of people with correct choice, i.e. $B\times H+(1-B)(n-H)$ where $B:=1_{\{\text{toss returned head}\}}$.
Note: $B$ is just $BernoulliDistribution(q)=BinomialDistribution(1,q)$.
Let $X$ be the number of people who correctly chose the outcome of the coin toss. Let $Y$ a bernoulli random variable with $Y=1$ corresponding to coin coming up heads. $$ X\mid Y=1\sim \text{Bin}(n, p) $$ and $$ X\mid Y=0\sim \text{Bin}(n, 1-p) $$ so by the law of total probability $$ \begin{align} P(X=k)&=P(X=k\mid Y=1)P(Y=1)+P(X=k\mid Y=0)P(Y=0)\\ &=qP(X=k\mid Y=1)+(1-q)P(X=k\mid Y=0) \end{align} $$ for $0\leq k\leq n$ and the distribution of $X$ is a mixture of binomial distributions.