A basketball player makes 80% of her free throws attempts. Find the average number of free throw attempts needed in order to make 3 free throws
in a row.

Question

A basketball player makes 80% of her free throws attempts. Find the average number of free throw attempts needed in order to make 3 free throws 
in a row.

anonymous · Answer

i think you can solve this via $.8\times x=3$

anonymous · Answer

My interpretation:$$
X \sim Binomial[n,p] = Binomial[n,0.80]
$$We want $n$ such that: $$
E[X] = np = n(0.8) = 3
$$

anonymous · Answer

However $n$ has to be an integer, so you might only be able to get an average slightly less or over.

highschoolmom2010 · Answer

im lost on this to be honest

anonymous · Answer

Do you know what the binomial probability distribution is?

anonymous · Answer

@wio didn't we write the same thing?

anonymous · Answer

$$.8x=3\iff x=\frac{8}{.3}$$

anonymous · Answer

I don't get your reasoning

anonymous · Answer

it is exactly what you wrote, cept i used $x$ and you used $n$ unless i misread your answer

ybarrap · Answer

Yes, both of yours are identical, it's the mean of the binomial random variable

tkhunny · Answer

Notice how the three made free throws are not necessarily consecutive.

highschoolmom2010 · Answer

yep im still lost

ybarrap · Answer

@highschoolmom2010 you are given that the chance of a successful free throw is 80%. This means that the chance of failing is 20%. There are two probability distributions you should be considering: Binomial and Geometric.

$\underline{Binomial}$

When you have a random process where there are only 2 options, success or failure, we can use what is called the "binomial distribution". This same distribution can predict the number of heads in 5 tosses (for example), because in each attempt there are only two options, heads or tails.

So whenever you encounter a random process that can only take on 2 values, think binomial.

For example, if you wanted to know chance of making only 1 free throws in three tries, this would be $0.80\times0.20\times0.20$ and since the free throw could have been made in either the 1st, 2nd or third attempt, we multiply this number times 3.

If you can model a process by the binomial distribution, calculating the average number of successes in $n$ tries is extremely easy. It is just the probability of success times the number of attempts: $0.85\times n$

$\underline{Geometric}$

There is another distribution type that take on two values, success and failure. That is the geometric distribution. This is the one where you want to find the number of $consecutive$ unsuccessful attempts before you get a single successful one.

The thing that should trigger you to use the geometric distribution is something like, "consecutive" or "in a row." The binomial models the chance of getting 3 free throws out of say 5, but not necessarily "in order."

Your problem seems to be one of the geometric type. But we need to modify the parameters a little to make it "fit" the geometric.

Take the tossing of a coin. The probability, we call it $p$, of a head is 1/2. So the average number of tosses you need to get a head is $\cfrac{1}{p}$. So, we need to toss the coin 2 times, $on~average$ before we get a head.

So for your problem, we need to find p. 

We will call a success "three consecutive free throws" and a failure otherwise.

The chance of 3 consecutive free throws is $0.80\times0.80\times0.80=0.80^3$.

So, $p=0.80^3$. 

And the average number of throws before we get three free throws in a row is $\cfrac{1}{p}=\cfrac{1}{0.80^3}\approx1.95$. Each of these represents 3 free throw attempts, so we multiply by 3 to get the average number of free throw attempts to get 3 in a row: $\cfrac{3}{0.80^3}\approx6$.

For the 1st 2 throws, there is 0 chance of 3 in a row. So, when we computed 1.95 above, it is obvious that that could not be our final answer. It is not until the 3rd free-throw  attempt, that "success" is even possible. So, chance of "success" starts on the 3rd free throw attempt, but on average we need 6 tries to get 3 successful free throws $in~ a~ row$.

If it were certain that we would make a free throw, then p=1 and 1/p = 1 is the average number of throws needed to get three in a row. But then we need to multiply by 3 to get the actual number of attempts. 

You can experiment with different values of p to see if the averages you get makes sense. For example,  if each free throw had a 50% chance of success then you would need 24 attempts to get 3 in a row, on average. But with p = 90%, then you would need only between 4 and 5 attempts, on average, which is less than 6 (for your problem), where p= 80%. 

I hope this helps.

highschoolmom2010 · Answer

@ybarrap yes that helpes

ybarrap · Answer

One last thing that I thought would be interesting to note.

If you compare the number of attempts needed to get 3 successful free throws, regardless of order, we can use the binomial and that would be about 4 attempts, on average (because $n\times.80=3\implies n=3.2$)

This makes sense since you would expect that it would be easier (and thus less attempts required) to make 3 free throws in any order, versus 3 free throws consecutively. 

This is like asking, which would take less time , to toss the coin until you get 3 heads in a row (geometric) or to toss the coin until you get 3 heads (binomial). Try it if you don't know -- toss a coin several times, chances are that you will get 3 heads in any order more frequently versus 3 heads in a row. Remember we are talking about averages, so you would have to try many times.

So 4 (binomial average) < 6 (geometric average), which is what we would expect using this argument.