Extended. When you are told something has already happened, the group you choose from shrinks. Conditional probability is just ordinary probability inside that smaller group.
'Given that B' means look ONLY at the outcomes in B. Then \(\text{P(A given B)} = \dfrac{\text{number in both A and B}}{\text{number in B}}\). The denominator is the size of the given group, not the whole total.
Fig. 1: 40 students; M = studies music, D = studies drama. Given that a student studies drama, only the 18 in circle D count, and 8 of them study music
Section 01
Shrinking the group.
In the lab, pick a 'given that' condition and watch the table: everything outside the chosen row or column fades out. The probability is then worked out inside the row or column that is left.
Worked example From a Venn diagram
Use Fig. 1. A student is chosen at random. Find (a) the probability that the student studies music, given that they study drama (b) the probability that the student studies drama, given that they study music.
(a) Given drama: the group is circle D, \(8 + 10 = 18\) students.
Of these, 8 also study music: \(\dfrac{8}{18} = \dfrac{4}{9}\).
(b) Given music: the group is circle M, \(12 + 8 = 20\) students.
8 of them study drama: \(\dfrac{8}{20} = \dfrac{2}{5}\).
Answer: (a) \(\tfrac{4}{9}\) (b) \(\tfrac{2}{5}\)
Worked example From a two-way table
Of 120 people, 60 are male (9 left-handed, 51 right-handed) and 60 are female (7 left-handed, 53 right-handed). Find (a) P(left-handed, given female) (b) P(female, given left-handed).
(a) The group is the 60 females; 7 are left-handed: \(\dfrac{7}{60}\).
(b) The group is all left-handers: \(9 + 7 = 16\); 7 are female: \(\dfrac{7}{16}\).
'P(A given B)' and 'P(B given A)' are different. In Fig. 1 they are \(\tfrac{4}{9}\) and \(\tfrac{2}{5}\). Read which group is GIVEN: that group is the denominator.
Section 02
Conditions in sample spaces.
With two dice, a condition such as 'the total is 8' cuts the 36 cells down to the few that satisfy it. Count inside those cells only.
Worked example Given the total
Two fair dice are rolled. Given that the total is 8, find the probability that both dice show the same number.
Total 8: (2,6), (3,5), (4,4), (5,3), (6,2), which is 5 equally likely cells.
Only (4,4) is a double.
\(\dfrac{1}{5}\).
Answer: \(\tfrac{1}{5}\)
Worked example Given at least one six
Two fair dice are rolled. Given that at least one dice shows a 6, find the probability that the total is 10 or more.
Cells with at least one 6: the row and the column for 6, which is \(6 + 6 - 1 = 11\) cells ((6,6) is counted once).
Of these, total 10 or more: (4,6), (6,4), (5,6), (6,5), (6,6), which is 5 cells.
\(\dfrac{5}{11}\).
Answer: \(\tfrac{5}{11}\)
Your turn
Two fair dice are rolled. Given that the total is 10, find the probability that at least one dice shows a 5.
Total 10: (4,6), (5,5), (6,4), so 3 cells; only (5,5) contains a 5. Probability \(\tfrac{1}{3}\).
Examiner trap
Do not divide by 36 once a condition is given. 'Given that the total is 8' means the denominator is 5, not 36.
Stage 1 · Learn
Check your understanding
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What does 'given that the student is a girl' tell you to do?
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Use Fig. 1. What is the probability that a student studies drama, given that they study music?
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Of 60 males, 9 are left-handed. What is P(left-handed, given male)?
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On a tree diagram, the probabilities on the second set of branches are...
Section 03
Conditional probability on trees.
On a tree diagram, the second set of branches is already conditional: each one is a probability GIVEN what happened on the first branch. To reverse a tree (find the first event given the second), divide the path you want by the total for the second event.
Worked example Without replacement
A bag has 5 red and 3 green sweets. Two are taken without replacement. Given that the first is red, find the probability that the second is red.
Given the first is red, 7 sweets are left and 4 of them are red.
\(\dfrac{4}{7}\).
Answer: \(\tfrac{4}{7}\)
Worked example Reversing a tree
The probability that it rains is 0.3. If it rains, the probability that Sam is late is 0.4; if it does not rain, it is 0.1. Given that Sam is late, find the probability that it rained.
Path rain and late: \(0.3 \times 0.4 = 0.12\).
Path dry and late: \(0.7 \times 0.1 = 0.07\).
P(late) \(= 0.12 + 0.07 = 0.19\). This is the given group.
P(rain, given late) \(= \dfrac{0.12}{0.19} = 0.6315...\)
Answer: 0.632 (3 s.f.)
Examiner trap
The branch value 0.4 is P(late GIVEN rain), not P(rain GIVEN late). Reversing the condition needs the division by P(late).
Stage 2 · Exam
Exam-style questions
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Of 80 students, Year 10: 18 take the bus and 22 walk; Year 11: 26 take the bus and 14 walk. A student is chosen at random. Given that the student walks, find the probability that they are in Year 11.
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In a Venn diagram of 50 people, 14 are in A only, 6 in both, 18 in B only and 12 in neither. Given that a person is in B, find the probability that they are also in A.
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P(snow) = 0.25. If it snows, the probability that a train is delayed is 0.5; if not, it is 0.2. Given that the train is delayed, find the probability that it snowed.
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Two fair dice are rolled. Given that the total is 9, find the probability that at least one dice shows a 5.
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