Binomial Distribution (D4)

Section 3.4 Binomial Distribution (D4)

In this section, we explore the binomial probability distribution .

Exploration 3.4.1. Free Throws.

Suppose a basketball player has a 90% chance shot of making a free-throw, and suppose that the success of each free-throw was independent of each other. If she took 4 shots, the probability that she makes the first two and misses the last two would be:

\begin{equation*} P(F_1=\text{score})P(F_2=\text{score})P(F_3=\text{miss})P(F_4=\text{miss})=(0.9)(0.9)(0.1)(0.1)=0.0081 \end{equation*}

where \(F_i\) denotes the outcome of the \(i\)th free throw.

(a)

What is the probability that she takes 4 free throws, makes the last two shots, but misses the first two?

(b)

What is the probability that she takes 4 free throws, makes the first and third shots, but misses the other two?

(c)

What do we notice about the probabilities found in (a) and (b)?

(d)

Using Definition 3.3.3, how many ways can we select 2 our of 4 free-throws for her to make?

Subsection 3.4.1 Binomial Probabilities

Definition 3.4.1.

Given \(n\) independent trials, each of which has \(p\) probability of success, let \(X\) denote the number of successes attained. We call \(X\) a binomial random variable. The probability that exactly \(k\) successes are attained is:

\begin{equation*} P(X=k)=C(n,k) p^k (1-p)^{n-k}. \end{equation*}

Remark 3.4.2.

The \(C(n,k)\) stems from the number of ways to select \(k\) out of \(n\) trials to be succesful.
The \(p^k\) comes from the probability that the \(k\) selected trials are succesful.
The \((1-p)^{n-k}\) comes from the probability that the remaining \(n-k\) not-selected trials are unsuccesful.

Activity 3.4.2. Snow White.

A binomial random variable must:

Have a fixed number of trials.
Have a fixed probability of success per trial.
Have trials that are independent of each other.

For the following, identify which of the above fail.

(a)

There are 10 apples, 4 are poisoned, Snow White picks 3 at random and eats them, let \(X\) denote the number of poisoned apples she eats.

(b)

There are 10 apples, they are all either poisoned or all not poisoned with a 50/50 chance each. Snow White eats 3 apples and let \(X\) denote the number of poisoned apples she eats.

(c)

There are 10 apples, each has a 10% chance of being poisoned, independently of the others. Snow White eats apples until she has a poisoned apple, let \(X\) denote the number of total apples she eats.

Activity 3.4.3. Free Throws.

Consider the basketball player from Exploration 3.4.1, taking \(n=4\) free throws with a \(p=0.9\) probability of each free-throw being a success. Let \(X\) denote the number of succesful free throws.

(a)

What is the probability that she makes exactly 2 free throws, that is, \(P(X=2)\text{?}\)

(b)

Run the following code to simulate 1000 trials of freethrows=4 free throws taken with p=0.9 probability of success each, and plot a histogram of the number of succesful free throws:

(c)

Run the following code to see how many of these trials resulted in two succesful free throws:

How does this compare to what you found in (a)?

(d)

Suppose her less experienced teammate only has a \(p=0.65\) chance of making a shot. If she takes 5 free throws, what are the chances she makes exactly \(k=3\) free throws?

(e)

Modify and run the following code to simulate 1000 trials of free throws for the teammate and plot a histogram of the number of succesful free throws:

(f)

Modify and run the following code to see how many of these trials resulted in three succesful free throws:

How does this compare to what you found in (a)?

Remark 3.4.3.

One can use Desmos or R to directly compute binomial probabilities. The following compute the probability that a binomial random vairable \(X\) with \(n=5\) trials, each with probability \(p=0.65\) probability of success, has \(k=4\) successes

Activity 3.4.4. Meeting Time.

A department chair calls for a meeting at a certain time. The ten faculty in the department each have a 60% chance of making the meeting. Assume whether or not a professor can make the meeting is independent of the other faculty. Let \(X\) denote the number of professors who can make the meeting.

(a)

Compute \(P(X=k)\) for \(k=8, 9\) and \(10\text{.}\)

(b)

Sum the values you found in (a) to find the probability that at least 8 professors make the meeting (\(P(X\geq 8)\)).

(c)

If less than 4 professors make the meeting, how many professors could have made the meeting?

(d)

Find the probability that less than 4 professors make the meeting.

(e)

In the Desmos above, set Min:8, Max:10 and compare the output to what you found in (a).

(f)

In the Desmos above, set Min: and Max: to values which correspond to “less than 4”. (Less than 4 would mean 4 is not included.) Compare the output to what you found in (d).

(g)

Run the following to compute the probability that 3 or fewer professors would make the meeting:

(h)

Run the following to compute the probability that more than 7 professors would make the meeting:

(i)

What is the probability that at least 4 and at most 7 professors make the meeting?

Subsection 3.4.2 Expectation and Variance

Activity 3.4.5. Free Throws Revisited..

Recall the basketball player from Activity 3.4.3 with a \(p=0.9\) chance of making a free throw. Suppose she takes 4 shots and let \(X\) denote the number of succesful shots.

(a)

Since she makes 90% of her free throws, how many baskets on average would she make with 4 shots?

(b)

Compute \(P(X=0)\text{,}\) \(P(X=1)\text{,}\) \(P(X=2)\text{,}\) \(P(X=3)\text{,}\) \(P(X=4)\text{.}\)

(c)

Use Remark 2.4.4 to compute \(E(X)\text{.}\) How does this value compare to (a)?

(d)

Use Remark 2.4.5 to compute \(Var(X)\text{.}\)

(e)

Run the following to compute the average number of baskets made in the simulation from Activity 3.4.3:

How does this value compare to what you found in (c)?

(f)

Run the following to compute the variance of baskets made in the simulation from Activity 3.4.3:

How does this value compare to what you found in (d)?

Remark 3.4.4.

Because of the regularity of binomial random variables, there is a special way to compute mean and variance of binomial random variables.

Activity 3.4.6. Free Throws Revisited Again..

Recall the basketball player from Activity 3.4.3 with a \(p=0.9\) chance of making a free throw. Suppose she takes 4 shots and let \(X\) denote the number of succesful shots.

Let \(X_i\) denote the number of baskets made by shot \(i\text{.}\) Note that the probability distribution of each \(X_i\) would be:

\begin{equation*} \begin{array}{|c|c|c|} \hline k \amp 0 \amp 1 \\ \hline P(X_i=k) \amp 0.1 \amp 0.9 \\ \hline \end{array} \end{equation*}

and that

\begin{equation*} X=X_1+X_2+X_3+X_4 \end{equation*}

that is, her total number of baskets is equal to the sum of the baskets scored per shot.

(a)

Use Remark 2.4.4 to compute \(E(X_i)\text{.}\)

(b)

Use Remark 2.5.1 to compute \(E(X)\text{.}\) How does this value compare to what you found in Activity 3.4.5 (c)?

(c)

Use Remark 2.4.5 to compute \(Var(X_i)\text{.}\)

(d)

Use Remark 2.5.1 to compute \(VarE(X)\text{.}\) How does this value compare to what you found in Activity 3.4.5 (d)?

Remark 3.4.5.

Given a binomial random variable \(X\) with \(n\) trials and \(p\) probability of success per trial, we have:

\begin{equation*} E(X)=np, Var(X)=np(1-p), \sigma_X=\sqrt{np(1-p)}. \end{equation*}

This only applies to binomial random variables.