4.6.6 problem 89

(a)

Since $\text{E}\left(N_C\right)=115p_C$, $\text{Var}\left(N_C\right)={\sum }_{k=0}^{115}{k}^2{p}_C^k-{\left(115p_C\right)}^2$.

(b)

Let $I_j$ be the indicator random variable that CATCAT starts at position $j$. Then, the expected number of CATCAT is $\text{E}\left(X\right)=110{\left(p_C{p}_A{p}_T\right)}^2$.

(c)

In a sequence of length $6$, the desired options are CATxxx, xxxCAT. Thus, $P\left(\text{at least one CAT}\right)=2\left(p_C{p}_A{p}_T\right(1-p_C{p}_A{p}_T\left)\right)+{\left(p_C{p}_A{p}_T\right)}^2$.