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Section 3.3 The Binomial Formula

OpenIntro: Binomial Distribution video
Figure 3.3.1 Binomial Distribution video

Suppose we randomly selected four individuals to participate in the “shock” study. What is the chance exactly one of them will be a success? Let's call the four people Allen (\(A\)), Brittany (\(B\)), Caroline (\(C\)), and Damian (\(D\)) for convenience. Also, suppose 35% of people are successes as in the previous version of this example.

Solution

Let's consider a scenario where one person refuses:

\begin{align*} \amp P(A=\text{refuse}, B=\text{shock}, C=\text{shock}, D=\text{shock} )\\ \amp = P(A=\text{refuse} )\ P(B=\text{shock} )\ P(C=\text{shock} )\ P( =\text{shock} )\\ \amp = (0.35) (0.65) (0.65) (0.65) = (0.35)^1 (0.65)^3 = 0.096 \end{align*}

But there are three other scenarios: Brittany, Caroline, or Damian could have been the one to refuse. In each of these cases, the probability is again \((0.35)^1(0.65)^3\text{.}\) These four scenarios exhaust all the possible ways that exactly one of these four people could refuse to administer the most severe shock, so the total probability is \(4\times(0.35)^1(0.65)^3 = 0.38\text{.}\)

Verify that the scenario where Brittany is the only one to refuse to give the most severe shock has probability \((0.35)^1(0.65)^3\text{.}\)  1 

\begin{gather*} P(A=\text{shock} ,\text{ } B=\text{refuse} ,\text{ } C=\text{shock} ,\text{ } D=\text{shock} )\\ = (0.65)(0.35)(0.65)(0.65) = (0.35)^1(0.65)^3 \end{gather*}

Subsection 3.3.1 Understanding the formula

To solve the scenario outlined in Example 3.3.2 we use what is called the Binomal Formula. The binomal formula gives the probability of having \(k\) successes in \(n\) independent trials where probability of an individual success in one trial is \(p\) (in Example 3.3.2, \(n=4\text{,}\) \(k=1\text{,}\) \(p=0.35\)). In order to develop this formula, we reexamine each part of the example.

There were four individuals who could have been the one to refuse, and each of these four scenarios had the same probability. Thus, we could identify the final probability as

\begin{gather} \text { [#of scenarios] } \times P(\text{ single scenario } )\label{binomial_in_words}\tag{3.3.1} \end{gather}

The first component of this equation is the number of ways to arrange the \(k=1\) successes among the \(n=4\) trials. The second component is the probability of any of the four (equally probable) scenarios.

Consider \(P(\)single scenario\()\) under the general case of \(k\) successes and \(n-k\) failures in the \(n\) trials. In any such scenario, we apply the Multiplication Rule for independent events:

\begin{gather*} p^k(1-p)^{n-k} \end{gather*}

This is our general formula for \(P(\)single scenario\()\text{.}\)

Secondly, we introduce a general formula for the number of ways to choose \(k\) successes in \(n\) trials, i.e. arrange \(k\) successes and \(n-k\) failures:

\begin{gather*} {n\choose k} = \frac{n!}{k!(n-k)!} \end{gather*}

The quantity \({n\choose k}\) is read n choose k. 2 Other notation for \(n\) choose \(k\) includes \(_nC_k\text{,}\) \(C_n^k\text{,}\) and \(C(n,k)\text{.}\) The exclamation point notation (e.g. \(k!\)) denotes a factorial expression.

\begin{gather*} 0! = 1\\ 1! = 1\\ 2! = 2\times1 = 2\\ 3! = 3\times2\times1 = 6\\ 4! = 4\times3\times2\times1 = 24\\ \vdots\\ n! = n\times(n-1)\times...\times3\times2\times1 \end{gather*}

Using the formula, we can compute the number of ways to choose \(k=1\) successes in \(n=4\) trials:

\begin{gather*} {4 \choose 1} = \frac{4!}{1!(4-1)!} = \frac{4!}{1!3!} = \frac{4\times3\times2\times1}{(1)(3\times2\times1)} = 4 \end{gather*}

This result is exactly what we found by carefully thinking of each possible scenario in Example 3.3.2.

Substituting \(n\) choose \(k\) for the number of scenarios and \(p^k(1-p)^{n-k}\) for the single scenario probability in Equation (3.3.1) yields the general binomial formula.

Binomial formula

Suppose the probability of a single trial being a success is \(p\text{.}\) Then the probability of observing exactly \(k\) successes in \(n\) independent trials is given by

\begin{gather} {n\choose k}p^k(1-p)^{n-k} = \frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}\label{binomial_formula}\tag{3.3.2} \end{gather}

Subsection 3.3.2 When and how to apply the formula

TIP: Is it binomial? Four conditions to check.

(1) The trials are independent.

(2) The number of trials, \(n\text{,}\) is fixed.

(3) Each trial outcome can be classified as a success or failure.

(4) The probability of a success, \(p\text{,}\) is the same for each trial.

What is the probability that 3 of 8 randomly selected students will refuse to administer the worst shock, i.e. 5 of 8 will?

Answer

We would like to apply the binomial model, so we check our conditions. The number of trials is fixed (\(n=8\)) (condition 2) and each trial outcome can be classified as a success or failure (condition 3). Because the sample is random, the trials are independent (condition 1) and the probability of a success is the same for each trial (condition 4).

In the outcome of interest, there are \(k=3\) successes in \(n=8\) trials, and the probability of a success is \(p=0.35\text{.}\) So the probability that 3 of 8 will refuse is given by

\begin{align*} { 8 \choose 3}(0.35)^3(1-0.35)^{8-3} \amp =\amp \frac{8!}{3!(8-3)!}(0.35)^3(1-0.35)^{8-3}\\ \amp = \frac{8!}{3!5!}(0.35)^3(0.65)^5 \end{align*}

Dealing with the factorial part:

\begin{gather*} \frac{8!}{3!5!} = \frac{8\times7\times6\times5\times4\times3\times2\times1}{(3\times2\times1)(5\times4\times3\times2\times1)} = \frac{8\times7\times6}{3\times2\times1} = 56 \end{gather*}

Using \((0.35)^3(0.65)^5 \approx 0.005\text{,}\) the final probability is about \(56*0.005 = 0.28\text{.}\)

TIP: Computing binomial probabilities

The first step in using the binomial model is to check that the model is appropriate. If it is, the next step is to identify \(n\text{,}\) \(p\text{,}\) and \(k\text{.}\) The final step is to apply the formulas and interpret the results.

The probability that a random smoker will develop a severe lung condition in his or her lifetime is about \(0.3\text{.}\) If you have 4 friends who smoke, are the conditions for the binomial model satisfied? 3 One possible answer: if the friends know each other, then the independence assumption is probably not satisfied. For example, acquaintances may have similar smoking habits.

Suppose these four friends do not know each other and we can treat them as if they were a random sample from the population. Is the binomial model appropriate? What is the probability that (a) none of them will develop a severe lung condition? (b) One will develop a severe lung condition? (c) That no more than one will develop a severe lung condition? 4 To check if the binomial model is appropriate, we must verify the conditions. (i) Since we are supposing we can treat the friends as a random sample, they are independent. (ii) We have a fixed number of trials (\(n=4\)). (iii) Each outcome is a success or failure. (iv) The probability of a success is the same for each trials since the individuals are like a random sample (\(p=0.3\) if we say a “success” is someone getting a lung condition, a morbid choice). Compute parts (a) and (b) from the binomial formula in Equation (3.3.2): \(P(0) = {4 \choose 0} (0.3)^0 (0.7)^4 = 1\times1\times0.7^4 = 0.2401\text{,}\) \(P(1) = {4 \choose 1} (0.3)^1(0.7)^{3} = 0.4116\text{.}\) Note: \(0!=1\text{,}\) . Part (c) can be computed as the sum of parts (a) and (b): \(P(0) + P(1) = 0.2401 + 0.4116 = 0.6517\text{.}\) That is, there is about a 65% chance that no more than one of your four smoking friends will develop a severe lung condition.

What is the probability that at least 2 of your 4 smoking friends will develop a severe lung condition in their lifetimes? 5 The complement (no more than one will develop a severe lung condition) as computed in Guided Practice 3.3.6 as 0.6517, so we compute one minus this value: 0.3483.

Suppose you have 7 friends who are smokers and they can be treated as a random sample of smokers. What is the probability that at most 2 of your 7 friends will develop a severe lung condition. 6 \(P(\)0, 1, or 2 develop severe lung condition\() = P(k=0) + P(k=1)+P(k=2) = 0.6471\text{.}\)

Below we consider the first term in the binomial probability, \(n\) choose \(k\) under some special scenarios.

Why is it true that \({n \choose 0}=1\) and \({n \choose n}=1\) for any \(n\text{?}\)

 7 Frame these expressions into words. How many different ways are there to arrange 0 successes and \(n\) failures in \(n\) trials? (1 way.) How many different ways are there to arrange \(n\) successes and 0 failures in \(n\) trials? (1 way.)

How many ways can you arrange one success and \(n-1\) failures in \(n\) trials? How many ways can you arrange \(n-1\) successes and one failure in \(n\) trials? 8 One success and \(n-1\) failures: there are exactly \(n\) unique places we can put the success, so there are \(n\) ways to arrange one success and \(n-1\) failures. A similar argument is used for the second question. Mathematically, we show these results by verifying the following two equations:

\begin{gather*} {n \choose 1} = n, \qquad {n \choose n-1} = n \end{gather*}

There are 13 marbles in a bag. 4 are blue and 9 are red. Randomly draw 5 marbles without replacement. Find the probability you get exactly 3 blue marbles.

Answer

Because the probability of success \(p\) is not the same for each trial, we cannot use the binomial formula. However, we can use the same logic to arrive at the following answer.

\begin{align*} P(x = 3) \amp = (\text{ # of combinations with 3 blue } )\times P(\text{3 blue and 2 red in a specific order} )\\ \amp ={5\choose 3}\times P(\text{ RRRBB } )\\ \amp = {5\choose 3}\left(\frac{4}{13}\times \frac{3}{12}\times \frac{2}{11} \times \frac{9}{10} \times \frac{8}{9}\right)\\ \amp = 0.1112 \end{align*}

Subsection 3.3.3 Calculator: binomial probabilities

Needs video TI-83/84: Computing the binomial coefficient, \({n\choose k}\)

Use MATH, PRB, nCr to evaluate \(n\) choose \(r\text{.}\) Here \(r\) and \(k\) are different letters for the same quantity.

  1. Type the value of n.

  2. Select MATH.

  3. Right arrow to PRB.

  4. Choose 3:nCr.

  5. Type the value of k.

  6. Hit ENTER.

Example: 5 nCr 3 means 5 choose 3.

Casio fx-9750GII: Computing the binomial coefficient, \({n\choose k}\)

  1. Navigate to the RUN-MAT section (hit MENU, then hit 1).

  2. Enter a value for \(n\text{.}\)

  3. Go to CATALOG (hit buttons SHIFT and then 7).

  4. Type C (hit the ln button), then navigate down to the bolded C and hit EXE.

  5. Enter the value of \(k\text{.}\) Example of what it should look like: 7C3.

  6. Hit EXE.

TI-84: Computing the binomial formula, \(P(X = k)={n\choose k}p^k(1-p)^{n-k}\)

Use 2ND VARS, binompdf to evaluate the probability of exactly \(k\) occurrences out of \(n\) independent trials of an event with probability \(p\text{.}\)

  1. Select 2ND VARS (i.e. DISTR)

  2. Choose A:binompdf (use the down arrow to scroll down).

  3. Let trials be \(n\text{.}\)

  4. Let p be \(p\)

  5. Let x value be \(k\text{.}\)

  6. Select Paste and hit ENTER.

TI-83: Do step 1, choose 0:binompdf, then enter \(n\text{,}\) \(p\text{,}\) and \(k\) separated by commas:

binompdf(n, p, k). Then hit ENTER.

TI-84: Computing \(P(X \le k)= {n\choose 0}p^0(1-p)^{n-0} + ... + {n\choose k}p^k(1-p)^{n-k}\)

Use 2ND VARS, binomcdf to evaluate the cumulative probability of at most \(k\) occurrences out of \(n\) independent trials of an event with probability \(p\text{.}\)

  1. Select 2ND VARS (i.e. DISTR)

  2. Choose B:binomcdf (use the down arrow).

  3. Let trials be \(n\text{.}\)

  4. Let p be \(p\)

  5. Let x value be \(k\text{.}\)

  6. Select Paste and hit ENTER.

TI-83: Do steps 1-2, then enter the values for \(n\text{,}\) \(p\text{,}\) and \(k\) separated by commas as follows: binomcdf(n,p,k). Then hit ENTER.

Casio fx-9750GII: Binomial calculations

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  1. Navigate to STAT (MENU, then hit 2).

  2. Select DIST (F5), and then BINM (F5).

  3. Choose whether to calculate the binomial distribution for a specific number of successes, \(P(X = k)\text{,}\) or for a range \(P(X \leq k)\) of values (0 successes, 1 success, ..., \(k\) successes).

    • For a specific number of successes, choose Bpd (F1).

    • To consider the range 0, 1, ..., \(k\) successes, choose Bcd(F1).

  4. If needed, set Data to Variable (Var option, which is F2).

  5. Enter the value for x (\(k\)), Numtrial (\(n\)), and p (probability of a success).

  6. Hit EXE.

Find the number of ways of arranging 3 blue marbles and 2 red marbles. 9 Use \(n\) = 5 and \(k\) = 3 to get 10.

There are 13 marbles in a bag. 4 are blue and 9 are red. Randomly draw 5 marbles with replacement. Find the probability you get exactly 3 blue marbles. 10 Use \(n\) = 5, \(p\) = 4/13, and \(x\) (\(k\)) = 3 to get 0.1396.

There are 13 marbles in a bag. 4 are blue and 9 are red. Randomly draw 5 marbles with replacement. Find the probability you get at most 3 blue marbles (i.e. less than or equal to 3 blue marbles). 11 Use \(n = 5\text{,}\) \(p = 4/13\text{,}\) and \(x = 3\) to get 0.9662.