Binomial distribution: notes and practice questions
- Describes the probability of successes in independent trials of a binary event (success/failure) with probability of success .
Probability mass function:
where .
- Mean: , Variance: .
How it is examined
Paper 2 only, in practice, because the guidance says probabilities are found with technology. Justifying that the binomial is an appropriate model (fixed , independent trials, constant , two outcomes) is a legitimate `Explain` part. needs and the boundary is where the errors are. 4 to 7 marks.
with and . The probability function itself is not given, which matches the guidance that probabilities come from technology.
- Binomial distribution.
- Mean and variance of the binomial distribution.
Not required: formal proof of mean and variance.
Linking questions
- Enrichment: hypothesis testing using the binomial distribution. Note that hypothesis testing is not on the AA syllabus at either level.
Practice questions
22 questions · 12 medium · 10 hardQuestion 1
MediumPaper 2 · calculator7 marksA manufacturer produces electronic components. It is known that the probability that a randomly selected component is defective is 0.03. A quality control inspector takes a random sample of 40 components from a large batch.
(a) Find the probability that there is at least one defective component in the sample.
(b) Given that there is at least one defective component in the sample, find the probability that there are at most three defective components.
Consider the complementary event. What is the probability that there are NO defective components?
This is a conditional probability problem. Remember the formula . Identify events A and B correctly.
Question 2
HardPaper 2 · calculator16 marks(a) An electronics factory produces two types of resistors: Type A and Type B.
The resistance, (in Ohms), of Type A resistors is normally distributed with a mean of 120 Ohms and a standard deviation of 5 Ohms.
Find the probability that a randomly selected Type A resistor has a resistance less than 115 Ohms.
(b) In a random selection of 10 Type A resistors, find the probability that exactly 3 have a resistance less than 115 Ohms.
(c) The resistance, (in Ohms), of Type B resistors is normally distributed with a mean of 135 Ohms and a standard deviation of 7 Ohms.
Each day, 70% of the resistors produced are Type A, and 30% are Type B.
On a particular day, a resistor is randomly selected from all those produced at the factory.
Let represent 'Type A resistor' and represent 'Type B resistor'.
(i) Find the probability that the randomly selected resistor has a resistance less than 115 Ohms.
(ii) Given that a randomly selected resistor has a resistance less than 115 Ohms, find the probability that it is a Type A resistor.
(d) The machine that makes the Type A resistors is adjusted so that the mean resistance of the Type A resistors remains the same, but their standard deviation changes to Ohms. The machine that makes the Type B resistors is not adjusted. The probability that the resistance of a randomly selected resistor from these machines is now less than 115 Ohms is 0.18.
Find the value of .
Use the normal cumulative distribution function (CDF) on your GDC. Remember to input the lower bound, upper bound, mean, and standard deviation.
This is a binomial probability problem. Identify the number of trials, the number of successes, and the probability of success from part (a).
You need to consider both types of resistors. Calculate the probability for Type B resistors first, then use the law of total probability, taking into account the proportions of each type.
This is a conditional probability problem, often solved using Bayes' theorem. You need the probability of a Type A resistor having low resistance and the overall probability of a low resistance resistor.
Set up an equation for the new total probability, similar to part (c.i). You'll need to solve for the new probability , then use the inverse normal function to find the z-score, and finally calculate the new standard deviation .
Question 3
MediumPaper 2 · calculator15 marksThe delivery times, minutes, for packages from a logistics hub to a regional distribution center can be modelled by a normal distribution with a mean of 120 minutes and a standard deviation of minutes.
Given that 3% of the delivery times are longer than 135 minutes, find the value of .
Find the probability that a randomly selected package will have a delivery time of more than 130 minutes.
Given that a package delivery takes longer than 130 minutes, find the probability that it takes less than 135 minutes.
On a particular day, there are 80 packages scheduled for delivery from the hub.
Find the expected number of packages that will have a delivery time of more than 130 minutes.
Find the probability that more than 8 of the packages on this particular day will have a delivery time of more than 130 minutes.
Use the inverse normal distribution function on your GDC to find the z-score corresponding to the given percentile. Remember that if 3% are longer than 135 minutes, then 97% are shorter than 135 minutes.
Use the normal cumulative distribution function (CDF) on your GDC. Remember to use the standard deviation found in part (a).
This is a conditional probability problem. Recall the formula . Here, event A is 'delivery takes less than 135 minutes' and event B is 'delivery takes longer than 130 minutes'.
This involves a binomial distribution. The expected value for a binomial distribution is given by , where is the number of trials and is the probability of success for a single trial (from part (b) ).
Use the binomial cumulative distribution function (CDF) on your GDC. Remember that 'more than 8' means , which can be calculated as .
Question 4
HardPaper 2 · calculator18 marks(a) A new automated coffee machine is programmed to dispense coffee. The volume of coffee dispensed, ml, is normally distributed with a mean of 200 ml and a standard deviation of ml.
On 15% of occasions, the machine dispenses more than 210 ml of coffee.
Find the value of .
(b) On a randomly selected occasion, find the probability that the machine dispenses more than 205 ml of coffee.
(c) The machine is considered to have 'over-filled' a cup if it dispenses more than 215 ml of coffee. Seven customers order coffee. Assume the volume dispensed for each customer is independent.
Find the probability that at least one of these seven coffees is over-filled.
(d) Given that at least one of the seven coffees is over-filled, find the probability that exactly two of them are over-filled.
(e) The café serves 25 customers in an hour. The machine requires maintenance if it over-fills more than 3 coffees during that hour. So far, 18 customers have been served, and the machine has over-filled 2 coffees.
Find the probability that the machine will NOT require maintenance by the end of the hour.
For a normal distribution, you can use the inverse normal function on your GDC to find the z-score corresponding to a given probability. Remember the formula for the z-score: .
Use the standard deviation found in part (a) and the normal distribution function on your GDC to find the probability.
First, calculate the probability of a single coffee being over-filled. Then, consider a binomial distribution for the number of over-filled coffees among the seven customers. 'At least one' often implies using the complementary probability.
This is a conditional probability problem. Remember the formula . Here, event A is 'exactly two over-filled' and event B is 'at least one over-filled'. What is the intersection of these two events?
Determine how many more customers will be served and how many more over-fills are allowed to avoid maintenance. Then, use the binomial distribution for the remaining trials.
Question 5
MediumPaper 2 · calculator15 marks(a) The lifespan of a new type of battery, hours, can be modelled by a normal distribution with a mean of 1200 hours and a standard deviation of hours.
Given that 5% of the batteries last longer than 1280 hours, find the value of .
(b) Find the probability that a randomly selected battery will have a lifespan of less than 1150 hours.
(c) Given that a battery lasts longer than 1150 hours, find the probability that it lasts less than 1250 hours.
(d) A batch of 500 batteries is produced. Find the expected number of batteries that will have a lifespan of less than 1150 hours.
(e) Find the probability that more than 85 of the batteries in this batch will have a lifespan of less than 1150 hours.
For part (a), use the inverse normal function on your GDC to find the z-score corresponding to the given percentile. Then, use the z-score formula to solve for . Remember that 5% lasting longer means 95% last less than that value.
For part (b), use the normal cumulative distribution function (CDF) on your GDC with the mean and standard deviation found in part (a). You need to find .
For part (c), this is a conditional probability problem. Recall the formula . Here, A is 'lasts less than 1250 hours' and B is 'lasts longer than 1150 hours'. So you need to find and .
For part (d), this involves a binomial distribution. The number of trials is the batch size, and the probability of success is the probability calculated in part (b). The expected number of successes in a binomial distribution is given by .
For part (e), you need to calculate for the binomial distribution . Remember that , and your GDC can compute using binomial CDF.
Question 6
HardPaper 2 · calculator18 marksIn a large university, 200 students were surveyed. Of those, 120 were undergraduates (U) and the rest postgraduates (P).
Each student in the survey was asked whether they preferred quiet zones (Q) or collaborative areas (C) for studying. It was found that 75 of the undergraduates preferred quiet zones. The total number of students who preferred collaborative areas was 100. This information is shown in the following table.
| Quiet Zones (Q) | Collaborative Areas (C) | Total | |
|---|---|---|---|
| Undergraduates (U) | 75 | p | 120 |
| Postgraduates (P) | x | 55 | 80 |
| Total | q | 100 | 200 |
Find the value of
;
.
Three students are chosen at random from those surveyed. Find the probability that all three are postgraduates.
Given that , find the value of .
A student is chosen at random from those surveyed. Write down the probability that they are a postgraduate who prefers quiet zones.
Determine if the events P (Postgraduate) and Q (prefers Quiet Zones) are independent. Justify your answer.
It can be assumed that the survey results are representative of the university population. Ten students from the university are chosen at random. Find the probability that at least five of them prefer quiet zones.
Use the row total for undergraduates and the number of undergraduates preferring quiet zones to find .
Use the grand total and the total number of students preferring collaborative areas to find .
Remember that once a student is chosen, they are not replaced. This affects the total number of students and postgraduates for subsequent selections.
Recall the formula for conditional probability: . In this case, .
This is a direct probability from the completed table. Look for the cell representing postgraduates who prefer quiet zones and divide by the total number of students.
Two events A and B are independent if or if . Calculate these probabilities using your table values.
This scenario involves a fixed number of trials (10 students) and a probability of success (preferring quiet zones) for each trial. Consider which probability distribution is appropriate.
Question 7
MediumPaper 2 · calculator7 marksLiam and Chloe are participating in an archery tournament.
The scores, points, achieved by Liam on a target can be modelled by a normal distribution with mean 85 and standard deviation 4.
The scores, points, achieved by Chloe on a target can be modelled by a normal distribution with mean 88 and standard deviation 3.
In the first round of the tournament, each competitor takes six shots. To qualify for the next round, a competitor must achieve at least one score of 90 points or greater in the first round.
Find the probability that only one of Liam or Chloe qualifies for the next round of the tournament.
First, calculate the probability of a single successful shot for each competitor using the normal distribution. Then, use the binomial distribution to find the probability that each competitor qualifies for the round. Finally, combine these probabilities to find the chance that only one qualifies.
Question 8
HardPaper 2 · calculator16 marksThe lifespan, (in hours), of a certain type of LED light bulb is modelled by a normal distribution with mean and standard deviation .
It is known that and .
Find the probability that a randomly selected light bulb has a lifespan between hours and hours.
Find the value of and the value of .
A manufacturer tests a batch of randomly selected light bulbs. Any bulb with a lifespan greater than hours is considered a 'long-life' bulb. Lifespans of bulbs are independent of each other.
Find the probability that exactly bulbs in the batch are 'long-life' bulbs.
Given that fewer than bulbs are 'long-life' bulbs, find the probability that exactly bulbs are 'long-life' bulbs.
In another factory, a different type of LED light bulb is produced. The lifespan of these bulbs, (in hours), is normally distributed with a mean of hours. The interquartile range (IQR) for these bulbs is hours.
Find the value of the standard deviation, , for this type of bulb.
Recall that the sum of probabilities for all possible outcomes in a continuous distribution is 1. Consider the regions defined by the given probabilities.
Use the inverse normal function to find the z-scores corresponding to the given probabilities. Then set up a system of two linear equations using the formula and solve for and .
This scenario involves a fixed number of trials (bulbs), two possible outcomes ('long-life' or not), and independent trials. This suggests a binomial distribution.
This is a conditional probability problem. Remember the formula . Here, event A is 'exactly 25 bulbs are long-life' and event B is 'fewer than 30 bulbs are long-life'.
The interquartile range is the difference between the upper quartile () and the lower quartile (). For a normal distribution, corresponds to the 25th percentile and to the 75th percentile. Use the inverse normal function to find the z-scores for these percentiles.
Question 9
MediumPaper 2 · calculator8 marks(a) The volume, ml, of liquid in bottles filled by a machine can be modelled by a normal distribution with mean ml and standard deviation ml.
A bottle is selected at random.
Find the probability that it contains more than ml.
(b) According to this model, of the bottles contain between ml and ml.
Find the probability that a randomly selected bottle contains less than ml.
(c) Find the value of .
(d) A quality control inspector randomly selects bottles.
Find the probability that exactly of these bottles contain less than ml.
Use your GDC to find the probability for a normal distribution. Remember to use the correct parameters for mean and standard deviation.
Consider the total probability under the curve and the probabilities you already know. The total area under the probability density function is 1.
You need to use the inverse normal function on your GDC. Ensure you use the cumulative probability for .
This is a binomial probability scenario. Identify the number of trials, the number of successes, and the probability of success from previous parts.
Question 10
HardPaper 2 · calculator16 marks(a) The resistance, R ohms, of resistors produced by a factory is normally distributed with a mean of 100 ohms and a standard deviation of 3.5 ohms.
Find the probability that a randomly selected resistor has a resistance less than 98 ohms.
(b) In a random sample of 15 resistors, find the probability that exactly 4 of them have a resistance less than 98 ohms.
(c.i) The capacitance, C microfarads, of capacitors produced by the same factory is normally distributed with a mean of 50 F and a standard deviation of 2.8 F. Each day, 70% of the components produced are resistors and 30% are capacitors.
Find the probability that a randomly selected component has a value less than its respective threshold (i.e., less than 98 ohms for a resistor or less than 47 F for a capacitor).
(c.ii) Given that a randomly selected component has a value less than its respective threshold, find the probability that it is a resistor.
(d) The resistor manufacturing process is adjusted so that the mean resistance remains 100 ohms but its standard deviation changes to ohms. The capacitor manufacturing process is not adjusted. The probability that a randomly selected component from these machines has a value less than its respective threshold is now 0.160.
Find the value of .
Use the normal cumulative distribution function (CDF) on your GDC. Remember to input the lower bound, upper bound, mean, and standard deviation.
This is a binomial probability problem. Identify the number of trials (n), the number of successes (k), and the probability of success (p) from part (a).
First, find the probability that a capacitor has a capacitance less than 47 F. Then, use the law of total probability, considering the proportion of resistors and capacitors produced.
This is a conditional probability problem. Use Bayes' theorem: P(A|B) = P(A and B) / P(B).
Work backwards. Use the new total probability and the unchanged capacitor probability to find the new probability for resistors. Then use the inverse normal function to find the z-score, and finally calculate the new standard deviation.
Question 11
MediumPaper 2 · calculator5 marksA quality control manager inspects a batch of 40 components for defects. The number of defective components, , follows a binomial distribution, . The variance of the number of defective components is known to be 8.4.
(a) Find the possible values of .
The cost of repairing the batch of components, , is given by the formula (in dollars).
(b) Find Var().
Recall the formula for the variance of a binomial distribution: Var() = . You will need to solve a quadratic equation for .
Remember the property for the variance of a linear transformation: Var() = Var().
Question 12
HardPaper 2 · calculator19 marksA new automated manufacturing process produces components. The time, in minutes, taken for a critical assembly step is modelled by a continuous random variable , with a probability density function defined by
Find the exact value of .
Find .
The assembly step is considered "efficient" if it takes less than 1.5 minutes. Each assembly step is independent. Determine the least number of assembly steps required to be 99% sure of at least one efficient step.
Ten assembly steps were conducted.
Find the probability that exactly three steps were efficient.
Write down the number of ways these three efficient steps could have occurred consecutively in a batch of 10.
Now consider a batch of assembly steps where it is given that exactly three efficient steps have occurred.
Write down an expression for the number of ways these three efficient steps could have occurred consecutively.
Find the greatest value of such that the probability of three consecutive efficient steps is more than 0.05, given that exactly three efficient steps have occurred in the batch.
Recall that the expected value for a continuous random variable is given by the integral of over its domain. Consider using a substitution method for integration.
Integrate the probability density function from the lower limit to 1.5. Recall the integral of .
Let be the probability of an efficient step from part (b). The probability of at least one efficient step in trials is . Set up an inequality and solve for .
This is a binomial probability problem. Identify , , and , then use the binomial probability formula .
Consider placing a block of 3 consecutive successes within the 10 trials. If the block starts at position 1, 2, etc., how many starting positions are there?
Generalize your approach from part (e) for trials instead of 10.
This is a conditional probability problem. The probability is the ratio of (number of ways for 3 consecutive successes) to (total number of ways for exactly 3 successes in trials). Set up an inequality and solve for .
Question 13
MediumPaper 2 · calculator8 marksA textile factory produces rolls of fabric. Due to manufacturing imperfections, of the fabric rolls produced are found to have minor defects. A quality control inspector randomly selects a batch of fabric rolls for inspection.
(a) Find the probability that exactly two of the selected fabric rolls have minor defects.
(b) Find the probability that no more than three of the selected fabric rolls have minor defects.
(c) Find the probability that at least two of the selected fabric rolls have minor defects.
(d) Find the variance of the number of fabric rolls with minor defects in a batch of .
This is a binomial probability problem. Identify the number of trials (), the probability of success (), and the number of successes () for the specific event.
For 'no more than three', you need to calculate the cumulative probability . This can be done using a GDC's binomial cumulative distribution function (binomCdf).
For 'at least two', consider the complement event. .
Recall the formula for the variance of a binomial distribution: .
Question 14
HardPaper 1 · no calculator8 marksA discrete random variable follows a binomial distribution, .
It is given that the probabilities , and for some integer form an arithmetic sequence, where .
(a) Show that .
(b) For a particular experiment, it is known that . Find the possible value(s) of .
Start by writing the condition for an arithmetic sequence in terms of probabilities, . Then use the formula for the binomial probability distribution and simplify the resulting equation by dividing by common factors.
Substitute the given value of into the equation you proved in part (a). This will give you a quadratic equation in terms of .
Question 15
MediumPaper 1 · no calculator5 marksA biased coin is flipped. The probability of getting a head is . Find the minimum number of times the coin must be flipped so that the probability of getting at least one tail is greater than .
Let be the number of flips. The event 'at least one tail' is the complement of which event? Set up an inequality using this information. You will need to solve an inequality involving exponents.
Question 16
HardPaper 2 · calculator8 marksA bakery is running a promotion where customers can win free pastries. A customer spins a special prize wheel 6 times. The wheel has three equally likely outcomes: 'Croissant', 'Muffin', or 'Danish'. A 'successful spin' is defined as landing on 'Croissant'.
Based on the number of 'Croissant' spins, the customer receives free pastries according to these rules:
- If the number of 'Croissant' spins is an even number, the customer receives 1 free pastry.
- If all 6 spins result in 'Croissant' OR all 6 spins result in a non-'Croissant' outcome (i.e., all 'Muffin' or 'Danish'), the customer receives 3 free pastries. This condition overrides the previous one if there is an overlap.
- In all other scenarios, the customer receives 0 free pastries.
Calculate the expected number of free pastries a customer wins and the expected number of times a customer wins nothing.
Identify the type of probability distribution for the number of 'Croissant' spins. Determine the probability of a 'successful spin'. Then, for each possible number of 'Croissant' spins, assign the corresponding number of free pastries. Finally, use the formula for expected value, . Remember that 'expected number of times an event occurs' in a single trial is equal to the probability of that event.
Question 17
MediumPaper 2 · calculator6 marksA telemarketing company is launching a new campaign. Based on previous data, the probability that a single call results in a successful sale is . A sales representative makes calls each week, and the outcome of each call is independent.
(a) Find the probability that the sales representative makes exactly successful sales in a given week.
(b) Find the probability that the sales representative makes at least successful sales in a given week.
(c) Find the probability that the first successful sale occurs on the th call.
This scenario involves a fixed number of trials with two possible outcomes (success or failure) and a constant probability of success for each trial. This suggests a specific probability distribution. To find the probability of exactly successes, use the probability mass function for this distribution.
When calculating the probability of 'at least' a certain number of successes, it is often easier to calculate the complementary probability. Consider what events are included in 'at least 4' and how that relates to 'less than 4'.
This part asks for the probability of the first success occurring at a specific trial number. This implies a sequence of failures followed by a single success. Consider the probability of failure and success for each call.
Question 18
HardPaper 2 · calculator14 marksThe mass, grams, of mangoes grown in an orchard can be modelled by a normal distribution with mean and standard deviation . The masses of all mangoes are independent of each other.
of the mangoes have a mass of less than grams, while of the mangoes have a mass of more than grams.
Find the value of and the value of .
Hence, find the probability that a mango chosen at random has a mass of between and grams.
The farmer also grows avocados. Records show that of the avocados grown are classified as large.
The probability of an avocado being classified as large is independent of any other avocado.
On a particular day, avocados are randomly selected.
Find the probability that at least of these avocados are classified as large.
Given that at least of these avocados are classified as large, find the probability that more than are not classified as large.
Use the inverse normal function on your calculator to find the z-scores for the given probabilities, then set up a system of equations.
Use the normal cumulative distribution function (normal CDF) on your calculator with the mean and standard deviation you just found.
This is a binomial distribution problem. You need to find the probability of 15 or more successes out of 24 trials.
Use the conditional probability formula: . Think carefully about what 'more than 4 are not large' means in terms of the number of large avocados.
Question 19
MediumPaper 2 · calculator5 marksA manufacturing plant produces electronic components. Historically, of the components produced are found to be defective. A quality control inspector randomly selects a sample of components.
Determine the least value of such that the probability of finding at least one defective component in the sample is greater than .
Consider the complement event: the probability of finding no defective components. Use the properties of binomial distribution and logarithms to solve the inequality.
Question 20
HardPaper 2 · calculator16 marksLeo has a box of tokens which are coloured either silver, gold or bronze.
The box contains exactly 15 bronze tokens. The number of silver tokens is four times the number of gold tokens.
Leo plays a game where he takes 12 tokens out of the box, one at a time. He notes the colour of each token and returns it to the box before taking the next token out of the box.
The probability that the first token is bronze is 0.3.
Show that there are 28 silver tokens in the box.
Find the probability that at least 7 of the 12 tokens that Leo takes are bronze. Give your answer correct to five significant figures.
Leo has to pay to take part in the game. If he takes at least 7 tokens of the same colour, he wins a prize. If he does not take at least 7 tokens of the same colour, then he does not win a prize. There is a different prize for each colour, as shown in the following table, where .
| Outcome | At least 7 gold tokens | At least 7 bronze tokens | At least 7 silver tokens |
|---|---|---|---|
| Prize |
Let the random variable represent Leo's net gain in dollars when he plays the game once. For example, if he takes at least 7 gold tokens, his net gain is since he gains .
The probability distribution of is shown in the following table, with probabilities given correct to four decimal places, where .
| 90 | ||||
|---|---|---|---|---|
| 0.0004 | 0.0386 | 0.5552 |
Write down the value of .
Use the probabilities in the table to find the value of .
Determine the smallest integer value of for which Leo could expect to make a positive net gain.
Leo wants to play the game until he wins a prize.
Find the minimum number of times Leo needs to play the game in order that the probability of winning at least one prize is greater than 0.999.
Use the probability of drawing a bronze token to find the total number of tokens in the box first.
Model the number of bronze tokens drawn as a binomial distribution.
The value represents the net gain if Leo wins the prize for silver tokens. Remember to subtract the cost of playing the game.
The sum of all probabilities in a probability distribution table must equal 1.
Set up an inequality where the expected value is greater than 0, and solve for .
The probability of winning at least one prize is .
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