The professor asked us to imagine a scenario where we have a basketball player who isn't good at shooting free throws. He makes his first free throw with probability $0.2$ After the first free throw, he makes a free throw with probability $0.6$ if he made the preceding one, and he makes a free throw with probability $0.3$ if he missed the preceding one. During practice, he throws $10$ free throws.
What is the total number of outcomes? Let $A_i$ be the event that he makes the $i$th free throw. Find $P(A_i)$ for $i=2, 3, 4$.
I figured out the total number of possible outcomes using binomial coefficients ${n \choose r} = \frac{n!}{(n-r)!r!} $. The number of possible ways outcomes are as follows:
\begin{eqnarray}
\text{Possible ways he will make 10 free throws} & = & {10 \choose 10} = 1 \\
\text{Possible ways he will make 9 free throws} & = & {10 \choose 9} = 10 \\
\text{Possible ways he will make 8 free throws} & = & {10 \choose 8} = 45 \\
\text{Possible ways he will make 7 free throws} & = & {10 \choose 7} = 120 \\
\text{Possible ways he will make 6 free throws} & = & {10 \choose 6} = 210 \\
\text{Possible ways he will make 5 free throws} & = & {10 \choose 5} = 252 \\
\text{Possible ways he will make 4 free throws} & = & {10 \choose 4} = 210 \\
\text{Possible ways he will make 3 free throws} & = & {10 \choose 3} = 120 \\
\text{Possible ways he will make 2 free throws} & = & {10 \choose 2} = 45 \\
\text{Possible ways he will make 1 free throws} & = & {10 \choose 1} = 10 \\
\text{Possible ways he will make 0 free throws} & = & {10 \choose 0} = 1 \\
\end{eqnarray}
These outcomes add up to $1024$. So that's done.
Now, if $B_j$ represents the event where this basketball player made $j$ number of free throws. Then $B_{10}$ has $1$ outcome, $B_{9}$ has $10$ outcomes, $B_{8}$ has $45$ outcomes, and so on until we get to $B_{0}$ which has $1$ outcome. From each of these I need to figure out how many outcomes in each $B_j$ made the $i$th free throw. These outcomes would then go into $A_i$.
Now, this is where I am a little overwhelmed. How would I find the number of outcomes in each $A_i$ without having to conduct it tediously. From there, how would I take into consideration the free throw probabilities $0.2, 0.6, 0.3$. I assume this has to do something with conditional probability, but since I am new to probability, I am not sure.
I am terribly sorry for the long post. I am truly grateful for all the help and advice you give me. Thank You for your time, take care, and have a wonderful day.
Best Answer
For the second part, it is probably easiest to proceed recursively.
We are told $P(A_1)=.2$.
Now suppose we have computed $P(A_{j-1})$ and want to compute $P(A_j)$. Look at the first $j-1$ shots. We only care about the last one, which we know to be a hit with probability $P(A_{j-1})$ (and therefore a miss with probability $ 1-P(A_{j-1}))$ It follows that $$P(A_j)=.6\times P(A_{j-1})+ .3\times (1-P(A_{j-1}))=.3 \times P(A_{j-1})+.3 = .3\times (P(A_{j-1})+1)$$
In this way it is very easy to compute all the $P(A_j)$.
Note: it wasn't part of the question, but it is interesting to note that in the limit as the number of free throws gets large, the probability goes to $\frac 37$; this follows quickly from the recursion. Indeed, the recursion can be solved in closed form by standard methods. Doing so, we get: $$P(A_j)=\frac 37-\frac {16}{21}(.3)^j$$