You may well have tried the Monty Hall problem, a question based on the game show 'Let's Make a Deal'. In the game show, extra information is given about the whereabouts of a prize. If you use this extra information, you are more likely to choose the correct strategy and win the car! This is the basis of conditional probability: extra information changes probabilities. You should be able to recognise these sort of questions, as they often contain the words "given that".
On this page, you should learn about
- conditional probability
- using Venn diagrams to visualise conditional probability, the probability of A given B
\(P(A|B)=\frac { P(A\cap B) }{ P(B) } \)
\(P(A|B)=\frac { c }{ c+b } \)
- Bayes theorem
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You can practise some of the skills in this page with the following quiz
START QUIZ!
Probability that the outcome of rolling a die is
odd and a prime number odd or a prime number odd if it is a prime number
Put the statement next to the correct mathematical notation
\(P(A\cap B)\) =
\(P(A|B)\) =
\(P(A\cup B)\) =
The following two-way table shows the number of females and males in grade 11 and 12 of a school
Complete the table and answer the probability questions below. The answers to the probabilities have been simplified.
| Grade 11 | Grade 12 | Total | |
| Females | 15 | | 23 |
| Males | | | |
| Total | 24 | 55 |
A student is selected at random to represent the student group on the school council. The probability that the student is
a) female and from grade 12 = \(\frac{a}{b}\)
a =
b =
b) male given that he/she is from grade 11 = \(\frac{c}{d}\)
c =
d =
c) a grade 11 student given that he is male = \(\frac{e}{f}\)
e =
f =
The possibility space below shows the outcomes when two dice are rolled and their scores are added together
The answers to the probabilities below have been simplified.
Probability that the sum equals 5 = \(\frac{a}{b}\)
a =
b =
Probability that the sum is less than 6 =\(\frac{c}{d}\)
c =
d =
Probability that the sum equals 5 given that the sum is less than 6 =\(\frac{e}{f}\)
e =
f =
The possibility space below shows the outcomes when two dice are rolled and their scores are multiplied together
The answers to the probabilities below have been simplified.
Probability that the product is a multiple of 4 = \(\frac{a}{b}\)
a =
b =
Probability that the product is a multiple of 3 =\(\frac{c}{d}\)
c =
d =
Probability that the product is a multiple of 3 and the product is a multiple of 4 =\(\frac{e}{f}\)
e =
f =
Probability that the product is a multiple of 3 given that the product is a multiple of 4 =\(\frac{g}{h}\)
g =
h =
a)
b)
c)
d)
In a class of 50 students, 21 study Visual Arts (A), 27 study Business Studies (B) and 12 study neither.
a) Work out the values of a , b , c and d in the Venn diagram
a =
b =
c =
d =
b) Find the answers below if the fractions have been simplified.
If a student is selected at random, the probability that she/he...
- studies Visual Arts AND Business Studies = \(\frac{e}{f}\)
e = f =
- studies Visual Arts given that she/he studies Business Studies =\(\frac{g}{h}\)
g = h =
- studies Business Studies given that she/he studies Visual Arts =\(\frac{i}{j}\)
i = j =
a)
There are 50 students altogether and 12 that take neither Visual Arts nor Business Studies. Hence there are 38 (50 - 12) who take either of the two subjects. Within the 21 students that study Visual Arts there are some that study Business Studies. We find the number in the intersection, c
21 + 27 - c = 38
b)
studies Visual Arts AND Business Studies = \(\frac{10}{50}=\frac{1}{5}\)
studies Visual Arts given that she/he studies Business Studies =\(P(A|B)=\frac { P(A\cap B) }{ P(B) } =\frac{\frac{10}{50}}{\frac{27}{50}}=\frac{10}{27}\)
studies Business Studies given that she/he studies Visual Arts =\(P(B|A)=\frac { P(B\cap A) }{ P(A) } =\frac{\frac{10}{50}}{\frac{21}{50}}=\frac{10}{21}\)
In a class of 24 students 16 students own a games console and 14 students own a scooter, 3 own neither.
The probability that a students owns both a games console and a scooter = \(\frac{a}{b}\)
a =
b =
The probability that a students owns a games console given she/he owns a scooter = \(\frac{c}{d}\)
c =
d =
a) \(\frac{9}{24}\)
For the events A and B, it is known that P(A) = 0.6, P(B) = 0.5 and P(A') = 0.4
Give the following answers as decimals
P(A\(\cap \)B) =
P(A|B) =
\(P(A|B)=\frac { P(A\cap B) }{ P(B) } =\frac{0.2}{0.5}=\frac{2}{5}=0.4\)
A number is selected at random from the list below
1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10
Giving your answer as a decimal, find the probability that the number selected is
a) less than 3 given that it is less than 6 =
b) less than 6 given that it is more than 2 =
c) less than 3 given that it is even =
d) prime given that it is odd =
For the events A and B, it is known that P(A) = 0.45, P(B) = 0.5 and P(A\(\cap\)B)' = 0.8
The following answers as fractions in their simplest form
a) P(A\(\cap\)B) = \(\frac{a}{b}\)
a =
b =
b) P(B|A) = \(\frac{c}{d}\)
c =
d =
c) P(A'|B') = \(\frac{e}{f}\)
e =
f =
c) P(A'|B') = \(\frac { P(A'\cap B') }{ P(B') } =\frac{0.25}{0.5}=\frac{1}{2}=0.5\)
You know that your mathematics teacher Mr O'Cop has 2 children. He tells you that he has a son called Rob.
Find the probability that Rob has a brother
Think about this carefully!
The probability that Rob has a brother = \(\frac{a}{b}\)
a =
b =
You might have thought that the probability of having a brother is 0.5. Don't worry, most people give this wrong answer when asked this question!
Here are the possibilities for his 2 children.
If he has a son, then he cannot have 2 girls so there are 3 possiblities.
Only one would have Rob having a brother!
Henri Tarr travels to school by bike \(\frac{3}{5}\) of the week and by car the rest of the time.
If he travels by bike, the probability that he is late is \(\frac{1}{8}\).
If he travels by car, the probability that he is late is \(\frac{3}{8}\).
a) Copy and complete the following tree diagram
b) Find the probability that Henri goes to school by car and is late for school.
c) Find the probability that he is late for school.
d) Given that he is late, find the probability that he travels to school by car.
Hint
Full Solution
A and B are independent events. P(A) = 0.3 and P(B) = 0.4
a) Find \(P(A'\cap B')\)
b) Hence find P(A'|B')
Hint
Full Solution
In the Olympic Games, it is believed that 5% of sprinters are taking anabolic steroids. A test is developed that gives a positive result for an athlete who has taken anabolic steroids in the last month in 90% of cases. The test gives a positive result for an athlete who has not taken anabolic steroids in the last month in 15% of cases. A sprinter in chosen at random.
a) Find the probability that she/he tests positive anabolic steroids
b) Given that the athlete tests positive, find the probability that the athlete has taken anabolic steroids in the last month.
Hint
Full Solution
How much of Conditional Probability have you understood?
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