Use totals, intersections, complements and conditional information.
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.
Check the conditions
Each count describes the same population.
Use the conditioning-event total as the denominator.
Three worked examples
Example 1 · Events and conditional probability
A group contains 72 people. 21 belong to A, 30 belong to B, and 9 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)
\(P(A\mid B)=9/30=\frac{3}{10}\)
Answer: \(\frac{3}{10}\).
(b)
\(P(A\cup B)=(21+30-9)/72=\frac{7}{12}\)
Answer: \(\frac{7}{12}\).
(c)
\(P(B\mid A^c)=(30-9)/(72-21)=\frac{7}{17}\)
Answer: \(\frac{7}{17}\).
(d)
\(1-P(A\cup B)=\frac{5}{12}\)
Answer: \(\frac{5}{12}\).
Example 2 · Events and conditional probability
A group contains 76 people. 22 belong to A, 32 belong to B, and 10 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)
\(P(A\mid B)=10/32=\frac{5}{16}\)
Answer: \(\frac{5}{16}\).
(b)
\(P(A\cup B)=(22+32-10)/76=\frac{11}{19}\)
Answer: \(\frac{11}{19}\).
(c)
\(P(B\mid A^c)=(32-10)/(76-22)=\frac{11}{27}\)
Answer: \(\frac{11}{27}\).
(d)
\(1-P(A\cup B)=\frac{8}{19}\)
Answer: \(\frac{8}{19}\).
Example 3 · Events and conditional probability
A group contains 80 people. 23 belong to A, 34 belong to B, and 11 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)
\(P(A\mid B)=11/34=\frac{11}{34}\)
Answer: \(\frac{11}{34}\).
(b)
\(P(A\cup B)=(23+34-11)/80=\frac{23}{40}\)
Answer: \(\frac{23}{40}\).
(c)
\(P(B\mid A^c)=(34-11)/(80-23)=\frac{23}{57}\)
Answer: \(\frac{23}{57}\).
(d)
\(1-P(A\cup B)=\frac{17}{40}\)
Answer: \(\frac{17}{40}\).
Common mistakes
Do not interchange P(A|B) and P(B|A).
Being disjoint is different from being independent.