Use totals, intersections, complements and conditional information. Track changing totals and use conditional probability after an observation.
Notes and key methods
For events A and B, \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). Conditional probability changes the sample space: \(P(A\mid B)=P(A\cap B)/P(B)\), with P(B)>0.
Independence means \(P(A\cap B)=P(A)P(B)\). Mutually exclusive nonzero-probability events cannot be independent, because their intersection has probability zero.
Multiply probabilities along a branch of a tree and add mutually exclusive branches that describe the required event. Without replacement, both the number of available objects and the total may change.
To reverse a condition, first calculate the probability of the observed event across all possible branches, then divide the desired joint probability by that total.
Check the conditions
Each count describes the same population.
Use the conditioning-event total as the denominator.
Objects are selected uniformly from those still available.
The denominator for the second selection is one less than the original total.
Three worked examples
Example 1 · Events and conditional probability
A group contains 68 people. 20 belong to A, 28 belong to B, and 8 belong to both.
Find P(A|B).
Find P(A union B).
Find P(B|not A).
Find the probability of belonging to neither event.
Show the reasoning
(a)
\(P(A\mid B)=8/28=\frac{2}{7}\)
Answer: \(\frac{2}{7}\).
(b)
\(P(A\cup B)=(20+28-8)/68=\frac{10}{17}\)
Answer: \(\frac{10}{17}\).
(c)
\(P(B\mid A^c)=(28-8)/(68-20)=\frac{5}{12}\)
Answer: \(\frac{5}{12}\).
(d)
\(1-P(A\cup B)=\frac{7}{17}\)
Answer: \(\frac{7}{17}\).
Example 2 · Trees and sampling without replacement
A container has 7 red and 6 blue counters. Two are selected uniformly, one after the other, without replacement.
Find the probability that both are red.
Find the probability of different colours.
Given that the second counter is red, find the probability that the first was red.