Your mind will explode #4

Let $X_n$ be the number of shots made in $n$ consecutive attempts. We know that $\displaystyle a_n=\frac{X_n}{n}$ . Since $\displaystyle a_{10}=0.4=\frac{2}{5}$ we have that

$\displaystyle a_{10}=\frac{2}{5}=\frac{X_{10}}{10} \ \Longrightarrow \ X_{10}=4$

Let $\textbf{t}=(t_1,...,t_{10})$ the trials vector such that

$t_i=\begin{cases}1, & \text{if he/she makes a shot} \\ 0, & \text{otherwise} \end{cases}, \ i \in[0,10]$

resulting in $\displaystyle X_n=\sum_{i=1}^n t_i$ . We also know that $a_n \leq 0.4$ for $n \in [1,9]$ , so

$\displaystyle a_n=\frac{X_n}{n} \leq \frac{2}{5} \ \Longrightarrow \ X_{n}\leq \Big\lfloor\frac{2n}{5}\Big\rfloor$

and

$X_1=0, \ X_2=0, \\ X_3 \leq 1, \ X_4 \leq 1, \\ X_5 \leq 2, \ X_6 \leq 2, \ X_7 \leq 2, \\ X_8 \leq 3, \ X_9 \leq 3$

and eventually $X_{10}=4$ , as we previously said. Let's rewrite $\textbf{t}=(0 \ 0 \ | \ t_3, \ t_4 \ | \ t_5, \ t_6, \ t_7 \ | \ t_8, \ t_9 \ | \ 1)$ and put the remaning three $1s$ in $\textbf{t}$ , following the boundaries previously written. I'll use the notation $|\cdot|\cdot|\cdot|$ to indicate the number $(\#)$ of $1s$ will appear between the bars, i.e. the number of $1s$ assigned to the $t_i$ between the bars (the rest will be $0s$ ), and for each of them we'll count the number of possible combination. Hence

$\#( | 1 | 1 | 1 | )=2 \cdot 3 \cdot 2=12 \\ \#( | 1 | 0 | 2 | )=2 \\ \#( | 0 | 2 | 1 | )=2 \cdot 3=6 \\ \#( | 0 | 1 | 2 | )=3$

Summing up $2+12+6+3=23$ , so that the final probability is

$\displaystyle \mathbb{P}(a_{10}=0.4 \wedge a_n \leq 0.4, \ 1 \leq n \leq 9)=23\Big(\frac{2}{5}\Big)^4 \Big(1-\frac{2}{5}\Big)^6=\frac{268272}{9765625}=\frac{2^4 \cdot 3^6 \cdot 23}{5^{10}}$

Eventually

$(p+q+r+s)(a+b+c)=(2+3+23+5)(4+6+10)=\boxed{660}$