Dependent events
You draw two cards from a pack without putting the first one back. A doctor interprets the result of a diagnostic test. In both cases, what has already happened changes the probabilities of what can happen next: these are dependent events, and we need a new tool to deal with them: conditional probability.
Conditional probability
In a class of $30$ students, $18$ are girls. It is known that $12$ girls and $5$ boys study French. If a student is selected at random from this class and turns out to…
In a class of $30$ students, $18$ are girls. It is known that $12$ girls and $5$ boys study French. If a student is selected at random from this class and turns out to be a girl, calculate the probability that she studies French.
In a survey of $200$ people about their preference between tea and coffee, the following data were obtained: $120$ prefer coffee, of whom $70$ are women. If there are…
In a survey of $200$ people about their preference between tea and coffee, the following data were obtained: $120$ prefer coffee, of whom $70$ are women. If there are $110$ women in total in the survey, calculate the probability that a person chosen at random prefers tea, given that they are a man.
Product rule for dependent events
When we draw without replacement (what has been drawn is not put back), the make-up of what is left changes for the next draw: the draws are dependent.
In a box there are $8$ LED light bulbs, $3$ of which are blown. If two bulbs are drawn at random, one after the other without replacement, calculate the probability that…
In a box there are $8$ LED light bulbs, $3$ of which are blown. If two bulbs are drawn at random, one after the other without replacement, calculate the probability that both work correctly.
In a drawer there are $5$ black socks and $3$ blue socks, all mixed together. If you take out two socks in the dark, one after another, calculate the probability that…
In a drawer there are $5$ black socks and $3$ blue socks, all mixed together. If you take out two socks in the dark, one after another, calculate the probability that both are blue.
Tree diagrams with dependent events
A tree diagram is the clearest way to organise several dependent stages: each branch is labelled with the probability conditional on what has already happened, and each complete path is calculated by multiplying its branches.
The following tree diagram shows the probabilities of a basketball player scoring ($E$) or missing ($F$) their first and second free throws. Observe the conditional…
The following tree diagram shows the probabilities of a basketball player scoring ($E$) or missing ($F$) their first and second free throws. Observe the conditional probabilities and calculate the probability that the player misses the first throw and scores the second.
In a board game, a chest contains $6$ gold coins and $4$ silver coins. A player draws two coins without looking, one after another without replacement. Based on the tree…
In a board game, a chest contains $6$ gold coins and $4$ silver coins. A player draws two coins without looking, one after another without replacement. Based on the tree diagram outline, calculate the probability that the two coins drawn are of different materials.
Total probability
When an outcome can be reached by several different paths that together cover all the possibilities (for example, going through one urn or another first, or through one population group or another), we add the probability of each path.
In a hospital, $40\%$ of the patients are admitted for respiratory diseases. Of these, $15\%$ are smokers. Among patients admitted for other reasons, $20\%$ are smokers.…
In a hospital, $40\%$ of the patients are admitted for respiratory diseases. Of these, $15\%$ are smokers. Among patients admitted for other reasons, $20\%$ are smokers. If a patient is chosen at random and turns out to be a smoker, what is the probability that they are admitted for a respiratory disease?
We have two urns and a fair six-sided die. We roll the die: if a number less than $3$ comes up, we draw a ball from Urn $A$; otherwise, we draw a ball from Urn $B$.…
We have two urns and a fair six-sided die. We roll the die: if a number less than $3$ comes up, we draw a ball from Urn $A$; otherwise, we draw a ball from Urn $B$. Observe the contents of the urns in the image and calculate the probability that the ball drawn is red.
Finding the cause from the effect
A common question in medicine is the reverse one: if a test result has been positive, what is the real probability of being ill? We have to "turn round" the conditional probability using the same formula for conditional probability, relying on total probability for the denominator.
A screening test for a disease has a reliability of $95\%$ for detecting sick people and $90\%$ for testing negative in healthy people. It is known that $2\%$ of the…
A screening test for a disease has a reliability of $95\%$ for detecting sick people and $90\%$ for testing negative in healthy people. It is known that $2\%$ of the population suffers from this disease. If a person takes the test and the result is positive, calculate the probability that they are actually sick.
In a technology company, $60\%$ of the employees are programmers, $30\%$ are designers and $10\%$ are managers. $80\%$ of the programmers, $40\%$ of the designers and…
In a technology company, $60\%$ of the employees are programmers, $30\%$ are designers and $10\%$ are managers. $80\%$ of the programmers, $40\%$ of the designers and $90\%$ of the managers speak fluent English. An employee is selected at random and found not to speak fluent English. Calculate the probability that this employee is a designer.
Back to the start: when drawing two cards without replacement from a pack of $40$, the probability of drawing two kings is $\dfrac{4}{40} \cdot \dfrac{3}{39} = \dfrac{1}{130}$: the first draw changes the remaining cards and conditions the second.