Triola Elementary Statistics · 11th Edition
Chapter 4: Probability
Every key formula from Triola Elementary Statistics Chapter 4, in one searchable page. Click a card to study it — worked examples included.
4.2 · PROBABILITY BASICS
Three approaches
1. Classical (equally likely outcomes): P(A) = (# favorable)/(# possible).
2. Relative frequency: past results — P(A) ≈ (times A occurred)/(trials).
3. Subjective: an educated estimate from judgment and experience.
2. Relative frequency: past results — P(A) ≈ (times A occurred)/(trials).
3. Subjective: an educated estimate from judgment and experience.
Rules of probability values
0 ≤ P(A) ≤ 1 · P(impossible) = 0 · P(certain) = 1
Express P(A) as a fraction, decimal, or percent — never a count like “3 in 5” alone.
KEY NOTES
- Classical approach requires equally likely outcomes — verify first.
- More trials ⇒ the relative-frequency estimate gets closer to the true probability.
4.4 · MULTIPLICATION RULE
Formal multiplication rule
P(A and B) = P(A) · P(B|A)
The second probability must account for the first event having happened.
Independent events
P(B|A) = P(B) ⇒ P(A and B) = P(A) · P(B)
One event’s outcome doesn’t affect the other’s.
WORKED EXAMPLE
Two independent radios, each fails with P = 0.001.
P(both fail) = 0.001 · 0.001 = 0.000001.
P(both fail) = 0.001 · 0.001 = 0.000001.
KEY NOTES
- 5% guideline: if the sample is ≤ 5% of the population, treat selections as independent even without replacement.
4.3 · ADDITION RULE
Formal addition rule
P(A or B) = P(A) + P(B) − P(A and B)
Subtract P(A and B) so overlapping outcomes are counted only once.
Disjoint events
P(A and B) = 0 ⇒ P(A or B) = P(A) + P(B)
Disjoint (mutually exclusive) = cannot occur at the same time.
WORKED EXAMPLE
Roll a die. P(even or 5):
P(even) = 3/6, P(5) = 1/6, disjoint ⇒
P = 3/6 + 1/6 = 4/6 = 2/3.
P(even) = 3/6, P(5) = 1/6, disjoint ⇒
P = 3/6 + 1/6 = 4/6 = 2/3.
Intuitive version: P(A or B) = (# ways A + # ways B, no double-count) / (total outcomes).
4.2 · COMPLEMENT
P(̄A) = 1 − P(A) · P(A) = 1 − P(̄A) · P(A) + P(̄A) = 1
The complement of A = all outcomes in which A does not occur.
WORKED EXAMPLE
62.4% of murders are cleared by arrest.
P(cleared) = 0.624 ⇒ P(not cleared) = 1 − 0.624 = 0.376.
P(cleared) = 0.624 ⇒ P(not cleared) = 1 − 0.624 = 0.376.
Complements are a shortcut: “not A” is often far easier to count than “A.”
4.6 · COUNTING
Factorial
n different items can be arranged n! ways
6 parks ⇒ 6! = 6 · 5 · 4 · 3 · 2 · 1 = 720 routes.
Permutations — order MATTERS
nPr = n!(n − r)!
Select r of n (without replacement); ABC ≠ CBA.
Combinations — order does NOT matter
nCr = n!(n − r)! r!
ABC = CBA — just the group. (Lottery jackpots: combinations!)
WORKED EXAMPLE
Choose 3 from 10:
10C3 = 10!/(7! 3!) = (10·9·8)/(3·2·1) = 120.
10C3 = 10!/(7! 3!) = (10·9·8)/(3·2·1) = 120.
4.5 · CONDITIONAL PROBABILITY
Conditional probability
P(B|A) = P(A and B)P(A)
Read “B given A” — the probability of B after A has occurred.
Positive test given lying: 42/98 ÷ 51/98 = 0.824.
Positive test given lying: 42/98 ÷ 51/98 = 0.824.
Complements: “at least one”
P(at least one) = 1 − P(none)
“At least one” = one or more; its complement is “none.”
WORKED EXAMPLE
Two babies: P(at least one girl) = 1 − P(no girls)
= 1 − (1/2 · 1/2) = 0.75.
= 1 − (1/2 · 1/2) = 0.75.
WATCH OUT!
Mistakes that cost points
- “Or” ⇒ add (subtract the overlap); “and” ⇒ multiply
- Addition: don’t double-count shared outcomes
- Multiplication: P(B|A) ≠ P(B) when events are dependent
- P(B|A) uses P(A) in the denominator — don’t flip them
- “At least one” ⇒ go straight to the complement
- Permutations vs combinations: does the order count? Decide first
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