Conditional probability and independent events: notes and practice questions
- Conditional Probability: The probability of event given that has occurred:
- Independent Events: Events and are independent if:
or equivalently .
How it is examined
With or without replacement changes the second branch of a tree and is the standard discriminator. The guidance that problems can be solved without explicit formulae means a correct tree diagram plus a correct answer is worth full marks, so a mark scheme should not demand the formula. Paper 2, 5 to 8 marks. The difference from SL 4.6 is that here the formula is required rather than a diagram. Testing for independence means checking numerically and stating a conclusion, and the conclusion sentence is a mark. 3 to 5 marks.
- All four formulas are given.
- The conditional probability formula is given.
- Use of Venn diagrams, tree diagrams, sample space diagrams and tables of outcomes to calculate probabilities.
- Combined events: .
- Mutually exclusive events: .
- Conditional probability: .
Extended at AHL 4.13 (Bayes' theorem).
Linking questions
- Aim 8: use of probability in casinos.
- Other contexts: use of probability methods in medical studies to assess risk factors.
Practice questions
42 questions · 1 easy · 28 medium · 13 hardQuestion 1
EasyPaper 1 · no calculator5 marksIf there are two events A and B are its given that and . let
Find the value of if the events A and B are said to be mutually exclusive.
Find the value of if the events A and B are said to be independent.
When two events A and B are said to be mutually exclusive then .
When two events A and B are said to be independent then .
Question 2
MediumPaper 1 · no calculator6 marksIf two events A and B are said to be independent events where and . let
Find the value of .
Find
When two events A and B are said to be independent then .
Refer to the conditional probability formula,
Question 3
HardPaper 1 · no calculator16 marksA spinner with four sectors is spun. The sectors are numbered 1, 2, 3, and 4. Let be the score obtained when the spinner is spun. The probability distribution for is given in the following table.
| 1 | 2 | 3 | 4 | |
|---|---|---|---|---|
| **P()** | 0.1 | 0.3 |
(a) Find the value of .
(b) Find the value of .
A second spinner, B, is also spun. Let be the score obtained. The probability distribution for is given in the following table.
| 1 | 2 | 3 | 4 | |
|---|---|---|---|---|
| **P()** |
(c) (i) State the range of possible values of .
(ii) Hence, find the range of possible values of .
(d) Hence, find the range of possible values for .
Leo spins spinner A once and Mia spins spinner B once. The probability that Leo's score is greater than Mia's score is .
(e) Find the value of .
What is the sum of all probabilities in a probability distribution?
Recall the formula for the expected value of a discrete random variable, .
What is the fundamental range for any probability value?
Use the relationship between m and n from the fact that all probabilities sum to 1, combined with your answer from (c.i).
Express E(Y) in terms of a single variable (either m or n) and then use the range you found in part (c) to find the minimum and maximum possible values for E(Y).
First, list all the possible outcomes where Leo's score (X) is greater than Mia's score (Y). Then, write an expression for the total probability of this event in terms of m and n. Set this expression equal to the given probability and solve for m. Finally, use this value to calculate E(Y).
Question 4
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 5
HardPaper 1 · no calculator16 marksA biased four-sided spinner, A, is spun. Let be the score obtained. The probability distribution for is given in the following table.
| Score () | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
Find the value of .
Hence, find the value of .
A second biased four-sided spinner, B, is spun. Let be the score obtained. The probability distribution for is given in the following table.
| Score () | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
(i) State the range of possible values of .
(ii) Hence, find the range of possible values of .
Hence, find the range of possible values for .
Leo spins spinner A once and Mia spins spinner B once. The probability that Leo's score is greater than Mia's score is .
Find the value of .
The sum of all probabilities in a probability distribution must be equal to 1.
The expected value is the sum of each outcome multiplied by its probability. Use the value of you found in part (a).
What is the possible range for any probability value?
Use the fact that the probabilities for Spinner B must sum to 1. Express in terms of and then use the range of you found in the previous part.
First, write an expression for in terms of and . Then, use the relationship between and to express in terms of a single variable. Finally, use the range of that variable to find the range of the expected value.
First, identify all the pairs of scores where . Then, for each pair, calculate the probability of it occurring in terms of and/or . Sum these probabilities and set the total equal to the given value, . Solve for or , and then calculate the expected value.
Question 6
MediumPaper 2 · calculator6 marksEvents S and C are independent. The probability that a student passes a Statistics exam, P(S), is twice the probability that the student passes a Calculus exam, P(C).
Given that the probability a student passes at least one of these exams is 0.625, find the probability that the student passes the Calculus exam, P(C).
Recall the formula for the probability of the union of two events, , and the specific condition for independent events. Remember that probability values must be between 0 and 1.
Question 7
HardPaper 2 · calculator5 marksA factory produces two types of light bulbs: standard and long-life. The lifespan of the bulbs, in hours, can be modelled as normal distributions with the following parameters.
| Bulb type | Mean | Standard deviation |
|---|---|---|
| Standard | 800 h | 50 h |
| Long-life | 1000 h | 100 h |
(a) Find the percentage of standard bulbs that have a lifespan of less than 700 hours.
(b) The factory produces a large number of bulbs, of which 60% are standard bulbs. Both types of bulbs are produced and randomly mixed together for packaging. A quality control process identifies and removes all bulbs with a lifespan of less than 700 hours. An inspector randomly selects a bulb from this removed group. Find the probability that it is a standard bulb.
Use the normal distribution CDF with mean 800 and standard deviation 50 to find the probability of a lifespan less than 700 hours, then express it as a percentage.
This is a conditional probability problem. Let S be the event that a bulb is standard, and L be the event that its lifespan is less than 700 hours. You need to find P(S|L). Use the formula for conditional probability or Bayes' theorem. You will need to calculate the probability of a long-life bulb having a lifespan less than 700 hours first.
Question 8
MediumPaper 1 · no calculator5 marksA student, Chloe, travels to school by bus. She can take one of two routes, Route X or Route Y. She is equally likely to choose either route on any given day.
The probability that she arrives on time when taking Route X is .
The probability that she arrives on time when taking Route Y is .
(a) Find the probability that Chloe arrives on time for school on a randomly chosen day.
Let be the event that Chloe chooses Route X and let be the event that she arrives on time.
(b) Determine whether events and are independent.
Consider the two separate scenarios: choosing Route X and being on time, and choosing Route Y and being on time. The total probability is the sum of the probabilities of these two mutually exclusive events. Remember to account for the probability of choosing each route.
Two events A and B are independent if . Calculate both sides of this equation for the events X and T and compare them. Alternatively, you can check if .
Question 9
HardPaper 2 · calculator16 marks(a) The random variable follows a normal distribution with mean and standard deviation .
Find .
(b) The diameters of ball bearings produced by a factory, in mm, are normally distributed with mean and standard deviation . The ball bearings are categorized as defective, standard, large, or premium, according to their diameter. The following table shows the probability a ball bearing is classified into each category.
| Category | Probability |
|---|---|
| Defective | 0.03 |
| Standard | 0.65 |
| Large | 0.25 |
| Premium | 0.07 |
The maximum diameter of a defective ball bearing is 14.8 mm.
The minimum diameter of a premium ball bearing is 16.5 mm.
Find the value of and of .
(c) The factory rejects all defective ball bearings. The remaining ball bearings are sold.
Find the probability that a ball bearing chosen at random from those sold is categorized as
(i) standard;
(ii) large;
(iii) premium.
(d) The selling prices of the different categories of ball bearings at this factory are shown in the following table:
| Category | Selling Price ($) |
|---|---|
| Standard | 1.50 |
| Large | 1.80 |
| Premium | 2.50 |
The factory incurs a fixed cost of $300 for the production run and assumes it will sell the accepted ball bearings in exactly the same proportion as calculated in part (c).
According to this model, find the minimum number of accepted ball bearings that must be sold so that the net profit for the factory is at least $550.
Recall that for a normal distribution, you can standardize the variable to a standard normal variable using the formula . Then use your GDC to find the probability.
Use the given probabilities and boundary values to find the corresponding z-scores. Then, set up two simultaneous equations involving and and solve them.
This is a conditional probability problem. The new sample space consists only of non-defective ball bearings.
Remember to use the new sample space (non-defective ball bearings) for this conditional probability.
The denominator for the conditional probability remains the probability of a non-defective ball bearing.
First, calculate the expected revenue per accepted ball bearing using the probabilities from part (c) and the selling prices. Then, set up an inequality for the total profit.
Question 10
MediumPaper 1 · no calculator5 marksAt a high school, students can study Chemistry and Physics. Let C be the event that a randomly selected student studies Chemistry and P be the event that they study Physics.
It is known that , and .
Find the probability that a randomly selected student studies Physics, .
You are given three pieces of information and need to find one unknown probability. Write down the standard probability formulas that connect these terms: the formula for the union of two events and the formula for conditional probability. You will end up with a system of two equations that you can solve.
Question 11
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 12
MediumPaper 1 · no calculator5 marksTwo archers, Clara and David, are competing. To decide who shoots an arrow, a fair six-sided die is rolled. If the die shows a 1, Clara is chosen. If the die shows any other number, David is chosen.
The probability that Clara hits the target is . The probability that David hits the target is .
(a) Find the probability that the target is hit.
(b) Let be the event that Clara is chosen and let be the event that the target is hit. Determine, with a reason, whether events and are independent.
First, determine the probability of choosing each archer based on the die roll. Then, use a tree diagram or the law of total probability to find the overall probability of hitting the target.
To check for independence between two events A and B, you can verify if or if .
Question 13
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 14
MediumPaper 1 · no calculator5 marksEvents and are such that , and .
Find .
You need to use two key probability formulas. One relates to the union of two events, , and the other relates to conditional probability, . Write down these two formulas using the given information. You will end up with a system of two equations with two unknowns, which you can then solve.
Question 15
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 16
MediumPaper 1 · no calculator6 marksConsider events and such that , and .
(a) Find .
(b) Determine if events and are independent. Justify your answer.
First, find the probability of event C occurring using the given probability of its complement, C'. Then, use the formula for conditional probability, , to find the probability of the intersection.
To determine if two events are independent, you need to check if or if . You will first need to find using the formula for the union of two events: .
Question 17
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 18
MediumPaper 2 · calculator6 marks(a) The lifespan of a certain type of rechargeable battery, in hours, can be modelled by a normal distribution with a mean of 1500 hours and a standard deviation of 50 hours. A battery is deemed faulty and rejected if its lifespan is less than 1425 hours.
Find the probability that a randomly selected battery is rejected.
(b) Estimate the number of batteries that will be rejected from a random sample of 200 batteries.
(c) Given that a battery is not rejected, find the probability that it has a lifespan greater than 1575 hours.
Use the normal cumulative distribution function (CDF) to find the probability that the lifespan is less than the rejection threshold. Remember to use the given mean and standard deviation.
Multiply the probability of a single battery being rejected by the total number of batteries in the sample.
This is a conditional probability problem. You need to find . Remember the formula for conditional probability: .
Question 19
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 20
MediumPaper 2 · calculator8 marksAt a tech company, employees can participate in various initiatives. 70% of employees attend a professional development workshop (), and 20% of employees work on a special innovation project (). 20% of employees do neither activity.
An employee is selected at random.
Find the probability that the employee attends a professional development workshop and works on a special innovation project.
Find the probability that the employee works on a special innovation project, but does not attend a professional development workshop.
At the company, 40% of the employees are in the Marketing department. Of those in the Marketing department, 30% work on a special innovation project.
An employee is selected at random. Let be the event "the employee is in the Marketing department" and let be the event "the employee works on a special innovation project".
Find .
Determine if the events and are independent. Justify your answer.
Consider using the formula for the union of two events or a Venn diagram. Remember that .
Think about how to express 'event B but not event A' in terms of probabilities you already know or can calculate.
Recall the definition of conditional probability: . Rearrange this to find .
For two events and to be independent, . Alternatively, check if or .
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