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Statistics · The binomial distribution (discrete variable)

Binomial Coefficients and Counting

The yearbook club needs a committee of $3$ students chosen from $12$ volunteers. On a pop quiz with $10$ multiple-choice questions and $4$ choices each, you guess on every question, and you want the chance of getting exactly $3$ right. Both problems ask the same thing: in how many ways can you pick $r$ items out of $n$? The tool that answers it is the binomial coefficient.

Factorials

Counting problems quickly produce long products like $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$. We give them a short name.

Computing 5 factorial and simplifying 8 factorial divided by 6 factorial by canceling common factors 5! = 5·4·3·2·1 = 120 8!/6! = 8·7·6!/6! = 56 the common factors cancel

The trick for fractions of factorials: write the bigger factorial out only until it reaches the smaller one, then cancel.

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Calculate and simplify the value of the following expressions without using a calculator, applying the properties of factorials: a) $\frac{8!}{6!}$ b) $\binom{7}{2}$ c)…

Calculate and simplify the value of the following expressions without using a calculator, applying the properties of factorials:

  1. a) $\frac{8!}{6!}$
  1. b) $\binom{7}{2}$
  1. c) $5! - 4!$
Solve

Binomial coefficients

Back to the yearbook committee. Suppose the order of choosing does not matter: {Ava, Ben, Cruz} is the same committee as {Cruz, Ava, Ben}.

Why this formula? If order did matter (president, vice president, treasurer), there would be $n(n-1)\cdots(n-r+1)$ ordered lists. Each committee of $r$ people shows up $r!$ times in those lists, once for every way to reorder it. Dividing by $r!$ removes the repeats.

The six committees of 2 students that can be chosen from 4 students A, B, C and D Choose 2 of 4 students: A, B, C, D A, BA, CA, D B, CB, DC, D C(4,2) = 4·3/(2·1) = 6 committees
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To organize the senior class graduation party, an events committee made up of $3$ students must be chosen. If there are $20$ student volunteers in the class, in how many…

To organize the senior class graduation party, an events committee made up of $3$ students must be chosen. If there are $20$ student volunteers in the class, in how many different ways can this group be chosen? Use the formula for binomial coefficients:

$$\binom{n}{r} = \frac{n!}{r!(n-r)!}$$ Students (n = 20) Committee (r = 3) C₂₀,₃ = 1140
Solve

Symmetry and Pascal's rule

Choosing $2$ students for a committee is the same as choosing the $n-2$ students who are left out.

When $r$ is close to $n$, switch to $n-r$ first. It saves a lot of arithmetic.

Pascal's triangle from row 0 to row 5 showing that 4 plus 6 equals 10 n = 0n = 1n = 2 n = 3n = 4n = 5 1 11 121 1331 14641 15101051 C(4,1) + C(4,2) = C(5,2): 4 + 6 = 10

Row $n$ of the triangle lists $\binom{n}{0}, \binom{n}{1}, \dots, \binom{n}{n}$. The symmetry shows up as the left-right mirror image of every row.

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In the section on binomial coefficients, there is a symmetry property which states that $\binom{n}{r} = \binom{n}{n-r}$. Verify this property by calculating the values…

In the section on binomial coefficients, there is a symmetry property which states that $\binom{n}{r} = \binom{n}{n-r}$.

Verify this property by calculating the values of $\binom{10}{8}$ and $\binom{10}{2}$. What conclusion do you draw regarding the calculation effort when $r$ is a number close to $n$?

Solve
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Prove, using the definition of a binomial coefficient using factorials, the following identity (known as Pascal's rule or Stifel's property): $$\binom{n}{k} +…

Prove, using the definition of a binomial coefficient using factorials, the following identity (known as Pascal's rule or Stifel's property):

$$\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}$$

Apply this result to calculate the value of $\displaystyle \binom{10}{4} + \binom{10}{3}$ without calculating each term separately.

Solve

Counting with combinations

Many counting problems split a choice into groups. Two rules do the work. The Fundamental Counting Principle says that if one choice can be made in $a$ ways and another in $b$ ways, together they can be made in $a \cdot b$ ways. For "at least" or "at most" problems, you add the separate cases.

Choosing 1 of 8 sophomores and 2 of 6 juniors gives 8 times 15 equals 120 committees 1 of 8 sophomores C(8,1) = 8 × 2 of 6 juniors C(6,2) = 15 8 × 15 = 120 committees

When order does matter, count arrangements instead. Arranging all $n$ different items in a row can be done in $n!$ ways. If each of $r$ spots can reuse any of $m$ options, there are $m^r$ possibilities.

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In a high-performance sports center, they want to select a group of 4 athletes to represent the United States in an international relay competition. If there are 12…

In a high-performance sports center, they want to select a group of 4 athletes to represent the United States in an international relay competition. If there are 12 athletes available at the center (7 men and 5 women), calculate:

  1. a) The total number of different ways in which the team of 4 people can be chosen.
  2. b) In how many of those combinations would the team be formed by exactly 2 men and 2 women?
  3. c) Use the properties of binomial coefficients to prove, without calculating the numerical value, that:
$$\binom{12}{4} = \binom{11}{3} + \binom{11}{4}$$ Athletes (12) 7 Men A 5 Women Team (4) E.g.: 2 Men + 2 Women C₁₂,₄ C₇,₂ C₅,₂ × C₁₂,₄ = C₁₁,₃ (with A) + C₁₁,₄ (without A)
Solve
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Solve the following questions related to combinatorics and factorials: a) Find the value of $n$ in the following equation: $$\binom{n}{2} = 21$$ b) Simplify the…

Solve the following questions related to combinatorics and factorials:

  1. a) Find the value of $n$ in the following equation:
$$\binom{n}{2} = 21$$
  1. b) Simplify the following expression, leaving the result in terms of $n$:
$$\frac{(n+1)! + n!}{(n-1)!}$$
  1. c) How many 5-digit numbers can be formed using only the digits $\{1, 2, 3, 4, 5\}$ if no digit is allowed to be repeated? How would the result change if we allow repetitions?
Solve
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In the mathematics department of a high school, they want to form a graduation committee. There are $12$ Science students and $10$ Social Sciences students. a) In how…

In the mathematics department of a high school, they want to form a graduation committee. There are $12$ Science students and $10$ Social Sciences students.

  1. a) In how many ways can a committee of $6$ people be chosen if there must be at least $4$ Social Sciences students? Use the properties of binomial coefficients to simplify the manual calculations.
  1. b) If the selection process for each member were independent and the probability of choosing a Social Sciences student were the actual proportion in the group ($10/22$), what would be the probability that in a committee of $6$ people chosen with replacement there would be exactly $3$ of each subject area?
Solve

Discrete and continuous variables

Before using counting in probability, we need to know what kind of random variable we are counting.

A discrete variable shown as separate dots at whole numbers and a continuous variable shown as an unbroken interval on a number line discrete 0, 1, 2, 3, 4… (counted) continuous any value in an interval (measured)

A binomial random variable counts successes, so it is always discrete.

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Before studying the binomial distribution, it is essential to distinguish the type of variable. Classify the following random variables as discrete or continuous: 1. The…

Before studying the binomial distribution, it is essential to distinguish the type of variable. Classify the following random variables as discrete or continuous:

  1. The number of "likes" an Instagram post receives in an hour.

  2. The exact time it takes a student to get to school.

  3. The number of people who test positive on an antigen test in a sample of $50$ individuals.

  4. The height of the players on the US national basketball team.

Solve

Is it binomial?

Not every count of "successes" is binomial. A situation is a binomial setting when all four conditions of BINS hold.

The four BINS conditions for a binomial setting: binary, independent, number of trials fixed, same probability BINS Binary: success or failure Independent trials Number of trials is fixed Same probability p on every trial

If the setting is binomial with $n$ trials and success probability $p$, we write $X \sim B(n, p)$, where $X$ is the number of successes. (AP problems may write "binomial with $n$ and $p$.") A common shorthand is $q = 1 - p$ for the probability of failure.

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A study states that $70\%$ of American teenagers use TikTok daily. If we randomly select a group of $8$ teenagers from a high school: a) Identify the parameters $n$ and…

A study states that $70\%$ of American teenagers use TikTok daily. If we randomly select a group of $8$ teenagers from a high school:

  1. a) Identify the parameters $n$ and $p$ of the binomial distribution that this situation follows.
  2. b) Define the random variable $X$ associated with this experiment.
  3. c) What is the probability that exactly all $8$ teenagers use the app? (Express only the setup of the formula).
0.1 0.2 0.3 P(X=k) 0 1 2 3 4 5 6 7 8 Teenagers who use TikTok (k) P(X=8) = ?
Solve
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A young entrepreneur has launched a sustainable fashion app. According to her market research, $35\%$ of users who download the app end up making a purchase in the first…

A young entrepreneur has launched a sustainable fashion app. According to her market research, $35\%$ of users who download the app end up making a purchase in the first week. If 10 new independent users have registered today:

  1. a) Define the random variable $X$ and justify whether it follows a binomial distribution, indicating its parameters.
  2. b) Calculate the probability that exactly 4 of these users make a purchase.
  3. c) What is the probability that, at most, 2 users buy in the first week?
Solve

The binomial probability formula

Here is where the binomial coefficient pays off. Flip a fair coin $4$ times and ask for exactly $2$ heads. The sequences that work are HHTT, HTHT, HTTH, THHT, THTH, TTHH. There are $6$ of them, which is $\binom{4}{2}$. Each one has probability $0.5^4$.

The tree below shows $3$ trials with success probability $p$ and failure probability $q = 1 - p$. The three leaves with exactly $2$ successes are highlighted.

Tree diagram of three trials, each a success S or a failure F, with the probability of each of the eight outcomes SSS p³ SSF p²q SFS p²q SFF pq² FSS p²q FSF pq² FFS pq² FFF q³ Upper branch = success S (probability p) Lower branch = failure F (probability q = 1 − p)

Exactly $2$ successes can happen in $3$ ways, each with probability $p^2 q$, so $P(X=2) = 3\,p^2 q = \binom{3}{2} p^2 q$.

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Knowing that for a binomial distribution $B(n, p)$, the probability of obtaining $k$ successes is: $$P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}$$ Calculate the…

Knowing that for a binomial distribution $B(n, p)$, the probability of obtaining $k$ successes is:

$$P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}$$

Calculate the probability of obtaining exactly $2$ heads when flipping a fair coin $4$ times.

Solve
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A content creator on TikTok knows that the probability of one of their videos going "viral" (exceeding 100,000 views) is 0.05. If they decide to post one video a day for…

A content creator on TikTok knows that the probability of one of their videos going "viral" (exceeding 100,000 views) is 0.05. If they decide to post one video a day for a week (7 days):

  1. a) What is the probability that none of their videos go viral?
  2. b) Calculate the probability that at least 2 videos are viral.
  3. c) Represent graphically using a tree diagram the possibilities of success (V) and failure (N) for the first 3 days and assign the probabilities to each branch.
Day 1 Day 2 Day 3 Result Probability 0.05 0.95 0.05 0.95 0.05 0.95 0.05 0.95 0.05 0.95 0.05 0.95 0.05 0.95 V N V N V N V N V N V N V N VVV VVN VNV VNN NVV NVN NNV NNN 0.05³ 0.05²·0.95 0.05²·0.95 0.05·0.95² 0.05²·0.95 0.05·0.95² 0.05·0.95² 0.95³ V: Viral Video (0.05) N: Not Viral (0.95)
Solve

At least, at most, and the complement

Questions rarely ask for one exact value. First translate the words into an inequality.

Words Inequality
at least 3 $X \ge 3$
more than 3 $X > 3$, which is $X \ge 4$
at most 3 $X \le 3$
fewer than 3 $X < 3$, which is $X \le 2$
between 2 and 4, inclusive $2 \le X \le 4$

To find the probability, add the individual terms $P(X=k)$. If that means adding many terms, use the complement: the probabilities of all values $0, 1, \dots, n$ add to $1$, so

$$P(X \ge k) = 1 - P(X \le k-1)$$

The most common case is "at least one": $P(X \ge 1) = 1 - P(X = 0) = 1 - (1-p)^n$.

A bar of total probability 1 split into the small piece P(X=0) equal to 0.178 and the large piece P(X at least 1) equal to 0.822 P(X = 0) = 0.178 P(X ≥ 1) = 1 − 0.178 = 0.822 whole bar = total probability 1
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A multiple-choice exam consists of $5$ true or false questions. If a student answers at random (i.e., $p=0,5$): a) What is the probability that they do not get any…

A multiple-choice exam consists of $5$ true or false questions. If a student answers at random (i.e., $p=0,5$):

  1. a) What is the probability that they do not get any questions right ($P(X=0)$)?
  2. b) Using the complementary event, set up how you would calculate the probability that they get at least one question right.
Exam: 5 questions (Independent events) F₁ F₂ F₃ F₄ F₅ 0.5 0.5 0.5 0.5 P(X=0) = 0.5 × 0.5 × 0.5 × 0.5 × 0.5 = 0.5⁵ = 0.03125 0.1 0.2 0.3 Probability 0 1 2 3 4 5 Number of correct answers (X) P(X=0) P(X ≥ 1) = 1 - P(X=0)
Solve
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It is known that $15\%$ of the containers produced by a plastic recycling plant do not meet the required purity standards. A quality inspector randomly selects 8…

It is known that $15\%$ of the containers produced by a plastic recycling plant do not meet the required purity standards. A quality inspector randomly selects 8 containers for analysis.

  1. a) Calculate the probability that all the selected containers are suitable (meet the standards).
  2. b) Calculate the probability that at least one of the containers is defective.
  3. c) What is the probability that there are more than 6 suitable containers in the sample?
0.1 0.2 0.3 0.4 Probability Defective Suitable 0 1 2 3 4 5 6 7 8 8 7 6 5 4 3 2 1 0 0.27 0.38 0.24 0.08 0.02 n = 8 p = 0.15 (Defective) q = 0.85 (Suitable) a) All suitable c) > 6 suitable b) At least 1 defective
Solve
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A delivery app has a probability of delivery delay of $p = 0,15$. A delivery driver makes $20$ deliveries in a night shift. a) What is the probability that they have…

A delivery app has a probability of delivery delay of $p = 0,15$. A delivery driver makes $20$ deliveries in a night shift.

  1. a) What is the probability that they have exactly $3$ delays? Express the binomial coefficient as a product of fractions before calculating.
  1. b) If the delivery driver receives a penalty if they have more than $4$ delays, what is the probability of being penalized? (Set up the calculation using the complementary event and justify why it is more efficient in this case).
0,25 0,20 0,15 0,10 0,05 0,00 Probability P(X=k) 0 1 2 3 4 5 6 7 8 ... Number of delays (k) Penalty Limit a) P(X=3) Sin penalización: P(X ≤ 4) (5 terms to calculate) b) Penalización: P(X > 4) (16 terms to calculate) 1 - P(X ≤ 4) Problem data X ~ B(20, 0.15) n = 20 p = 0.15
Solve

Mean and standard deviation

Look at the distribution of heads in $4$ fair coin flips. The counts $\binom{4}{k} = 1, 4, 6, 4, 1$ give the bar heights, out of $16$ total sequences.

Bar graph of the binomial distribution B(4, 0.5) with probabilities 1/16, 4/16, 6/16, 4/16, 1/16 and the mean at 2 1/164/166/164/161/16 01234 number of heads mean μ = 2

The distribution balances at $2$ heads, which is $4 \cdot 0.5$. This pattern holds for every binomial.

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In a binomial distribution $X \sim B(100; 0,2)$, calculate: a) The mean or expected value ($\mu = n \cdot p$). b) The variance ($\sigma^2 = n \cdot p \cdot q$). c) The…

In a binomial distribution $X \sim B(100; 0,2)$, calculate:

  1. a) The mean or expected value ($\mu = n \cdot p$).
  2. b) The variance ($\sigma^2 = n \cdot p \cdot q$).
  3. c) The standard deviation ($\sigma = \sqrt{n \cdot p \cdot q}$).

Briefly explain what the value of the mean signifies in this context.

Solve
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A multiple-choice exam consists of 12 questions, each with 4 answer options of which only one is correct. A student who has not studied at all decides to answer all the…

A multiple-choice exam consists of 12 questions, each with 4 answer options of which only one is correct. A student who has not studied at all decides to answer all the questions at random.

  1. a) Calculate the expected number of correct answers (mean) and the standard deviation of this distribution.
  2. b) What is the probability that the student passes the exam (obtains at least 6 correct answers)?
  3. c) If the exam had 100 questions, what parameters would the resulting binomial distribution have? (Do not perform the probability calculations).
Solve
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In a vaccination campaign at a high school, it has been observed that $10\%$ of the students experience some mild side effect (such as a sore arm). If a class with 25…

In a vaccination campaign at a high school, it has been observed that $10\%$ of the students experience some mild side effect (such as a sore arm). If a class with 25 vaccinated students is selected:

  1. a) What is the mean number of students expected to have side effects?
  2. b) Calculate the probability that the number of students with side effects is less than the mean.
  3. c) What is the probability that exactly 3 students have side effects?
Solve

Finding unknown $n$ or $p$

Sometimes the unknown is a parameter, not a probability.

Using the mean. If the mean must be at most some value, solve $np \le$ that value. For $p = 0.04$ and a mean of at most $2$: $0.04\,n \le 2$, so $n \le 50$.

Using the mean and the variance. Divide the variance by the mean: $\dfrac{\sigma^2}{\mu} = \dfrac{np(1-p)}{np} = 1 - p$. Then find $p$, then $n = \mu / p$.

Finding the smallest $n$. If each trial has a small success probability, how many trials until success is likely? "No success in $n$ trials" has probability $(1-p)^n$. Solve an inequality in the exponent with logarithms.

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A courier company uses drones to deliver parcels in rural areas. It has been proven that the probability of a drone having a precision error upon landing is 0.12. If a…

A courier company uses drones to deliver parcels in rural areas. It has been proven that the probability of a drone having a precision error upon landing is 0.12. If a drone makes 15 landings in a day:

  1. a) Calculate the probability that it is successful in all landings.
  2. b) Calculate the probability that it has errors in more than 2 landings.
  3. c) If the company wants the mean number of errors per day to be at most 1, how many landings should the drone make per day following the same error rate?
Solve
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In an e-sports tournament, the probability of a professional "League of Legends" player achieving a "Pentakill" in a competitive match is $p = 0,008$. A team decides to…

In an e-sports tournament, the probability of a professional "League of Legends" player achieving a "Pentakill" in a competitive match is $p = 0,008$. A team decides to participate in a league consisting of a certain number of matches $n$. It is known that the probability of the team not achieving any "Pentakills" in the entire league is less than $0,6$.

  1. a) Determine, using logarithms, the minimum number of matches the team must play for this condition to be met.
  1. b) If they finally play $120$ matches, calculate the probability that they achieve exactly two "Pentakills", expressing the final result to four decimal places. To calculate the binomial coefficient, use the factorial formula: $\displaystyle \binom{n}{k} = \frac{n!}{k!(n-k)!}$.
Limit P(X=0) < 0.6 0.0 0.2 0.4 0.6 Probability P(X=k) 0 1 2 3 4 5 Number of Pentakills (k) P(X=0) P(X=2) = ? X ~ B(120, 0.008) n=120 matches
Solve
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A well-known social network is analyzing the behavior of its "stories". It has been proven that the distribution of the number of users who interact with an ad follows a…

A well-known social network is analyzing the behavior of its "stories". It has been proven that the distribution of the number of users who interact with an ad follows a binomial model. In a sample of users, it is known that the mean number of interactions is $\mu = 12$ and the variance is $\sigma^2 = 7,2$.

  1. a) Calculate the sample size $n$ and the probability of interaction $p$ for each user.
  1. b) Once $n$ and $p$ have been obtained, determine the probability that, in a new sample of the same size, the number of interactions is greater than $2$ but less than or equal to $4$.
Solve

Multi-step binomial problems

Harder problems combine the tools above. Three patterns come up often.

The mode. The most likely value of $X \sim B(n, p)$ is the whole number $k$ with $(n+1)p - 1 \le k \le (n+1)p$. For $B(20, 0.4)$: $(21)(0.4) = 8.4$, so $7.4 \le k \le 8.4$ and the mode is $k = 8$. If $(n+1)p$ is itself a whole number, there are two modes.

Conditional probability. "Given that at least $3$" means we restrict to the outcomes $X \ge 3$: $P(A \mid B) = \dfrac{P(A \text{ and } B)}{P(B)}$.

Two stages. First use a binomial (or counting) to find the probability that one item has a property. Then use that probability as the new $p$ in a second binomial, or multiply by the number of items to get an expected count.

Two-stage problem: first the probability that one box is returned, then the expected number of returned boxes among 100 Stage 1: one box of 5 cases X ~ B(5, 0.1) defective P(X ≥ 2) = 0.0815 box is returned → Stage 2: 100 boxes Y ~ B(100, 0.0815) expected returned: 100 × 0.0815 ≈ 8.1
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A smartphone component manufacturing company carries out quality control. Each production batch of a critical part contains $2\%$ defective units. The parts are packed…

A smartphone component manufacturing company carries out quality control. Each production batch of a critical part contains $2\%$ defective units. The parts are packed in boxes of $15$ units. A customer buys a shipment of $100$ boxes.

  1. a) What is the probability that in a randomly chosen box there are at least $2$ defective parts?
  1. b) If the customer returns the box if it contains more than one defective part, how many boxes are expected to be returned out of the $100$ they bought?
Solve
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A study on reading habits in 17-year-old teenagers states that $70\%$ read regularly in digital format. $10$ teenagers are selected at random. a) Calculate the…

A study on reading habits in 17-year-old teenagers states that $70\%$ read regularly in digital format. $10$ teenagers are selected at random.

  1. a) Calculate the probability that the number of teenagers who read digitally is equal to the mode of this binomial distribution. (Note: The mode in a $B(n, p)$ is the integer value $k$ such that $(n+1)p - 1 \leq k \leq (n+1)p$).
  1. b) If it is known that at least $8$ of them read in digital format, what is the probability that all $10$ do so? (Conditional probability).
0.3 0.2 0.1 0 P(X=k) 0 1 2 3 4 5 6 7 8 9 10 Mode X ≥ 8 X = 10 B(10, 0.7)
Solve
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A civil service exam consists of $80$ multiple-choice questions with $4$ options each, of which only one is correct. A candidate answers $12$ questions they did not know…

A civil service exam consists of $80$ multiple-choice questions with $4$ options each, of which only one is correct. A candidate answers $12$ questions they did not know at random.

  1. a) Let $X$ be the variable "number of correct answers in the $12$ questions". Define the parameters of the distribution.
  1. b) Calculate the probability that they have more correct answers than incorrect ones.
  1. c) If they need at least one correct answer to pass that section, calculate the probability that they fail.
Solve

Back to the beginning: Choosing $3$ of $12$ volunteers for the yearbook committee can be done in $\binom{12}{3} = 220$ ways. For the pop quiz, guessing on $10$ questions with $4$ choices gives $X \sim B(10, 0.25)$, and the chance of exactly $3$ correct is $\binom{10}{3}(0.25)^3(0.75)^7 = 120(0.015625)(0.1335) \approx 0.250$. The coefficient $120$ counts the ways the $3$ correct answers can be placed among the $10$ questions.

Summary

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