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:

P (F_{1} = score) P (F_{2} = score) P (F_{3} = miss) P (F_{4} = miss) = (0.9) (0.9) (0.1) (0.1) = 0.0081

where $F_{i}$ denotes the outcome of the $i$ th free throw.

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(a)

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

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(b)

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

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(c)

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

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(d)

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

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Subsection 3.4.1 Binomial Probabilities

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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:

P (X = k) = C (n, k) p^{k} (1 - p)^{n - k} .

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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.

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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.

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(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.

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(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.

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(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.

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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.

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(a)

What is the probability that she makes exactly 2 free throws, that is, $P (X = 2) ?$

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(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:


    
        
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numbaskets = rep(0, 1000)
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freethrows=4
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p=0.9
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for(i in 1:freethrows){
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  numbaskets=numbaskets+sample(c(1, 0), 1000, replace = TRUE, prob = c(p, 1-p)) 
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}
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hist(numbaskets)
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print(freethrows)

    
    
    
    
        
            
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(c)

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


    
        
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length(which(numbaskets==2))

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

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(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?

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(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:


    
        
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numbasketsTM = rep(0, 1000)
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freethrows=FIXME
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p=FIXME
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for(i in 1:freethrows){
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  numbasketsTM=numbasketsTM+sample(c(1, 0), 1000, replace = TRUE, prob = c(p, 1-p)) 
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}
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hist(numbasketsTM)
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print(freethrows)

    
    
    
    
        
            
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(f)

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


    
        
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length(which(numbasketsTM==FIXME))

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

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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


    
        
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dbinom(4, 5, 0.65)

    
    
    
    
        
            
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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.


    
        
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dbinom(FIXME, FIXME, FIXME)

    
    
    
    
        
            
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(a)

Compute $P (X = k)$ for $k = 8, 9$ and $10 .$

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(b)

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

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(c)

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

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(d)

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

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(e)

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

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(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).

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(g)

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


    
        
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pbinom(3, 10, 0.6, lower.tail = TRUE)

    
    
    
    
        
            
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(h)

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


    
        
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pbinom(7, 10, 0.6, lower.tail = FALSE)

    
    
    
    
        
            
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(i)

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

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Subsection 3.4.2 Expectation and Variance

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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.

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(a)

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

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(b)

Compute $P (X = 0),$ $P (X = 1),$ $P (X = 2),$ $P (X = 3),$ $P (X = 4) .$

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(c)

Use Remark 2.4.4 to compute $E (X) .$ How does this value compare to (a)?

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(d)

Use Remark 2.4.5 to compute $V a r (X) .$

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(e)

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


    
        
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mean(numbaskets)

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

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(f)

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


    
        
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var(numbaskets)

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

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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.

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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 $i .$ Note that the probability distribution of each $X_{i}$ would be:

\begin{array}{ccc} k & 0 & 1 \\ P (X_{i} = k) & 0.1 & 0.9 \end{array}

and that

X = X_{1} + X_{2} + X_{3} + X_{4}

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

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(a)

Use Remark 2.4.4 to compute $E (X_{i}) .$

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(b)

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

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(c)

Use Remark 2.4.5 to compute $V a r (X_{i}) .$

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(d)

Use Remark 2.5.1 to compute $V a r E (X) .$ How does this value compare to what you found in Activity 3.4.5 (d)?

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Remark 3.4.5.

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

E (X) = n p, V a r (X) = n p (1 - p), σ_{X} = \sqrt{n p (1 - p)} .

This only applies to binomial random variables.