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.

• An easier way to do the first: Our player either makes or does not make any given shot, two possibilities. Hence the total number of outcomes is $2^{10}=1024$. – lulu Feb 1 '16 at 18:35
• To clarify: the way you have phrased the question, I assume you are asking for the probability that our player makes the $i^{th}$ free throw, I mean, that's what it says. But in your discussion you appear to be asking for the probability that he makes exactly $i$ shots which is completely different. My posted solution addresses the first question, not the second. – lulu Feb 1 '16 at 18:58

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: $$P(A_j)=\frac 37-\frac {16}{21}(.3)^j$$

Here are results from a simulation in R of 100,000 ten-throw sessions, baed on @lulu's analysis.

The simulation is based on a matrix MAT with $m = 100,000$ rows and $n = 10$ columns. At the end of the simulation, the matrix has a 1 in cell $(i,j)$ if the $j$th throw in the $i$th session was a success. The $m$-vector x contains the number of successes in each of the $m$ sessions. The table shows the simulated distribution of the number of successes per ten-throw session, and this is plotted in the figure.

 m = 10^5;  n = 10
MAT = matrix(0, nrow=m, ncol=n)  # all 0's to start
MAT[,1] = rbinom(m, 1, .2)       # fill first column
for(i in 2:n) {
MAT[,i] = rbinom(m, 1, (MAT[,i-1]+1)*.3) }
x = rowSums(MAT)
round(table(x)/m, 3)
##  x
##     0     1     2     3     4     5     6     7     8     9    10
## 0.032 0.083 0.135 0.176 0.187 0.159 0.115 0.068 0.033 0.011 0.002
plot(table(x)/m, type="h", lwd=2, ylab="PDF", main="Distribution of Successes")
abline(h=0, col="darkgreen")


As a reality check, we confirm that the key conditional structure is properly captured by the program:

(a) If throw #2 (in col 2) is a success, then throw #3 (in col 3) should be a success with probability 0.6.

 mean(MAT[,3][MAT[,2]==1])  # vec of results on 3rd given S on 2nd
## 0.6014073               # 'mean' is proportion of S's


(b) If throw #2 (in col 2) is a failure, then throw #3 (in col 3) should be a success with probability 0.3.

 mean(MAT[,3][MAT[,2]==0])
## 0.3002676


Finally, according to @lulu's analysis, by the tenth throw the proportion of successes ought to be approaching $3/7 = 0.4285714.$

 throw.10 = MAT[,10]
mean(throw.10)
## 0.42746


Simulating 100-throw sessions, I got 3/7 to three place accuracy for the 100th throw.

I realize this is not a substitute for a formal mathematical solution to your problem, but it does a pretty good job of confirming @lulu's analysis, and may give you something against which to compare your analytical solution.