A biased coin is tossed $10$ times with probability of obtaining head as $\frac{1}{3}$. Let $Y$ denote the total number of tails observed. What is the expectation $E(5Y+2)$ ?
It seems a binomial distribution with $n=10, p=\frac{1}{3}$.
The binomial distribution of $Y$ is $f_i(y)=\binom{10}{i}(1-p)^ip^{n-i},~i=0,1,2,\cdots,10$
The expectation of $y$ is $E(Y)=\sum_{i=0}^{10} i \cdot f_i(y)=\sum_{i=0}^{10} i \cdot (1-p)^ip^{n-i}=\frac{10}{3}~(=np)$.
Thus expectation of $E(5Y+2)=5E(Y)+2=\frac{50}{3}+2=\frac{56}{3}.$
But the answer is given $\frac{100}{3}$. I don't see how.
Your $p$ should be $2/3$. Here, success is defined as getting a $T$, the probability of which is $\frac23$. Therefore $np = \frac{20}{3}$. You can finish the rest.