Probability and conditional probability
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Choose the denominator before the numerator.
The central idea: identify the allowed pool
For a random choice among equally likely outcomes,
P(event)
The equally likely condition matters. Counting outcomes is not enough when those outcomes have unequal chances. For a uniform random selection of one record from a table, each record is equally likely, so counts give probabilities directly.
A probability lies between 0 and 1. Its percentage representation lies between 0% and 100%. An event with probability 0 cannot occur in the model; one with probability 1 is certain in the model.
Marginal, joint, and conditional probabilities
A two-way table classifies the same records by two categorical variables. Interior cells count joint categories; row and column totals count broader groups; the grand total counts all records once.
A marginal probability concerns a row or column category out of all records. A joint probability concerns an intersection such as “Grade 10 and bus rider,” again out of all records. A conditional probability restricts the allowed pool first: “among bus riders” uses only the bus-rider total as its denominator.
The denominator comes from the condition
In P(A ∣ B), read the vertical bar as “given B.” Keep only B outcomes, then ask what fraction also satisfy A:
P(A ∣ B)
Circle phrases such as “of those,” “among,” and “given that.” They tell you which outcomes remain eligible.
Reversing a condition usually changes the answer
P(A ∣ B) and P(B ∣ A) share a joint numerator but generally have different denominators. The fraction of bus riders who are tenth graders need not equal the fraction of tenth graders who ride the bus. The English sentences are not equivalent, even when the same two labels appear.
For missing counts, write the probability as a fraction before solving. If the table says there are x successes and 10 failures, a success probability of 3/5 gives x/(x + 10)
Combining events without double‐counting
Complement and inclusive “or”
The complement of A is “not A,” so P(not A)
“A or B” in mathematical probability is normally inclusive: A, B, or both. Use
P(A or B)
Subtract the overlap because it was included twice. If events are mutually exclusive, they cannot happen together and the overlap is zero. Do not treat “or” as “exactly one” unless the question says so.
Independence is different from mutual exclusivity
Events A and B are independent when conditioning on B does not change A’s probability, assuming P(B) > 0. Equivalently, P(A and B)
Mutually exclusive events with positive probabilities are not independent: knowing one occurred rules out the other. In addition, equal rates in one sample table do not prove universal independence in an entire future population.
Two‐stage choices: replacement changes the second pool
The general multiplication rule is P(A then B)
With replacement and fresh random selection, the original counts are restored. Without replacement, update both the relevant category count and the total count after the first choice. Do not apply an unchanged probability merely because the drawings happen at different times.
For “at least one,” the complement “none” may be simpler. For two independent trials with success probability p, the probability of at least one success is 1−(1−p)2. This is a useful consequence of counting and independence, not a separate combinatorics topic to memorize.
Area probabilities and uneven group sizes
Uniform location means area is the weight
When a point is selected uniformly from a region, an event’s probability is its favorable area divided by the eligible area. If the selection is conditioned on a smaller region, the smaller region becomes the denominator. Lengths alone do not determine an area probability. A shaded triangle contributes bh, not bh.
Uniform selection across area is an assumption. A point chosen by some other mechanism, such as a process concentrated near the center, need not have area-proportional probabilities.
Unequal groups need weighted probabilities
If 80% of items come from one source and 20% from another, the overall defect rate is the source-share-weighted average of the two defect rates. Use
P(D)
when A and B partition all items. Averaging the two defect rates without the source weights would describe equally sized sources, not this inventory.
Reverse a weighted situation with a friendly total
A difficult-looking conditional probability can become an ordinary two-way table. Choose a hypothetical total such as 10,000; calculate each source’s item count, then its defect count. Among all defective items, the share from a specified source is that source’s defect count divided by the total defect count. The convenient total cancels in the final ratio.
This method makes base rates visible. A source can have a higher defect rate yet contribute fewer defects because it supplies far fewer items. You do not need to memorize a named theorem to reason through the counts.
Four questions before calculating
What is selected? Are the elementary outcomes equally likely? Which condition restricts the eligible pool? Does the numerator count only outcomes within that pool? After calculating, confirm that the probability is between 0 and 1.
What to practice next
Examples 5.01–5.06 isolate changing denominators; 5.07–5.10 develop overlapping categories, independence, and area models; and 5.11–5.15 combine stages and unequal group sizes.
15 worked examples
Try the question before reading the solution. The examples progress from Foundation to Challenge.
5.01. Build a probability from equally likely selections
A bag contains 5 red, 3 blue, and 2 green counters. One counter is selected at random, with every counter equally likely to be selected. What is the probability of selecting a blue counter?
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Recognize the structure. Count favorable counters and all eligible counters. The categories themselves are not equally likely.
Work it through. The bag contains 5 + 3 + 2
P(blue) = = 0.3.
Random selection is uniform over counters, not over color names.
Answer: 3/10.
Check. The three color probabilities are 5/10, 3/10, and 2/10; they sum to 1.
Avoid the trap. The answer is not 1/3 merely because there are three colors. Equal likelihood must apply to the outcomes you count.
5.02. Use a complement from a frequency table
A warehouse has 12 type A packages, 18 type B packages, and 30 type C packages. One package is selected at random from all 60 packages. What is the probability that it is not type C?
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Recognize the structure. “Not C” includes every other category. Use either the remaining count or a complement.
Work it through. The favorable count is 12 + 18
P(not C) = 1 − P(C) = 1 − = .
These categories exhaust all packages and do not overlap.
Answer: 1/2.
Check. Exactly half the packages are type C, leaving the other half outside C.
Avoid the trap. A complement is taken relative to the stated selection group. Changing the selection group would change the denominator.
5.03. Find a marginal probability in a two‐way table
A student is selected at random from the 120 students summarized in the table. What is the probability that the student is in grade 10?
Bus
Walk
Total
Grade 9
24
16
40
Grade 10
30
50
80
Total
54
66
120
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Recognize the structure. Only grade matters; combine both transportation categories within grade 10.
Work it through. The grade 10 row total is 80, and the full selection group contains 120 students. Thus
P(grade 10) = = .
A row or column total used this way gives a marginal probability.
Answer: 2/3.
Check. Grade 9 has probability 40/120
Avoid the trap. Using 30 counts only grade 10 students who ride the bus. The event includes both bus riders and walkers.
5.04. Find an intersection without changing the denominator
A student is selected at random from all 120 students in the table. What is the probability that the student is in grade 10 and rides the bus?
Bus
Walk
Total
Grade 9
24
16
40
Grade 10
30
50
80
Total
54
66
120
Show worked solutionHide worked solution for example 5.04
Recognize the structure. “And” asks for an overlap, but selection is still from all 120 students.
Work it through. The grade 10 and bus cell contains 30 students. Therefore
P(grade 10 and bus) = = .
The cell is the numerator; the grand total remains the denominator.
Answer: 1/4.
Check. The joint probability cannot exceed either P(grade 10)
Avoid the trap. Dividing by 80 would answer “bus, given grade 10.” Dividing by 54 would answer the reverse conditional question. Neither is the joint probability asked here.
5.05. Restrict the sample space after the word given
A student is selected at random from the students in the table. Given that the student is in grade 10, what is the probability that the student rides the bus?
Bus
Walk
Total
Grade 9
24
16
40
Grade 10
30
50
80
Total
54
66
120
Show worked solutionHide worked solution for example 5.05
Recognize the structure. The condition makes the grade 10 row the entire eligible group.
Work it through. Among the 80 grade 10 students, 30 ride the bus. Therefore
P(bus ∣ grade 10) = = .
The notation after the vertical bar names the condition.
Answer: 3/8.
Check. Within grade 10, the probability of walking is 50/80
Avoid the trap. The denominator is not 120 because grade 9 students have been ruled out by the condition. Write “eligible group: grade 10” before calculating.
5.06. Reverse the condition and rebuild the denominator
A student is selected at random from the students in the table. Given that the student rides the bus, what is the probability that the student is in grade 10?
Bus
Walk
Total
Grade 9
24
16
40
Grade 10
30
50
80
Total
54
66
120
Show worked solutionHide worked solution for example 5.06
Recognize the structure. Now the bus column, not the grade 10 row, is the eligible group.
Work it through. There are 54 bus riders, of whom 30 are in grade 10. Thus
P(grade 10 ∣ bus) = = .
The overlap count is the same as in the previous example, but the denominator changes.
Answer: 5/9.
Check. Within the bus column, grade 9 contributes 24/54
Avoid the trap. P(A ∣ B) generally differs from P(B ∣ A). Reversing the words without changing the denominator is a common conditional-probability error.
5.07. Count an inclusive or without double‐counting
Of 50 students, 28 play a sport, 20 play an instrument, and 8 do both. A student is selected at random from all 50. What is the probability that the student plays a sport or an instrument, or both?
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Recognize the structure. Adding the two category counts includes the overlap twice. Subtract it once.
Work it through. The number in at least one category is 28 + 20 − 8
P(sport or instrument) = = .
The remaining 10 students do neither.
Answer: 4/5.
Check. The disjoint groups are 20 sport only, 12 instrument only, 8 both, and 10 neither. Their counts sum to 50.
Avoid the trap. In this question, “or” explicitly includes both. Adding 28 and 20 without subtracting 8 counts those eight students twice.
5.08. Recover a missing count from a conditional probability
Among participants in an evening workshop, x complete a project and 12 do not. A participant is selected at random from the evening workshop. The probability that the participant completes the project is 3/4. What is x?
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Recognize the structure. The unknown count is part of both the favorable count and the eligible total.
Work it through. The completion probability is
Cross-multiplying gives 4x
Answer: 36.
Check. 36/(36 + 12) = 36/48 = 3/4.
Avoid the trap. Setting x/12
Another route. If 3/4 complete, 1/4 do not. The 12 noncompleters represent one quarter, so completers represent three equal quarters: 3(12)
5.09. Check independence by comparing conditional shares
The table records purchases by 150 visitors. One visitor is selected at random from this group. Are membership and purchasing independent for this selection?
Purchased
Did not purchase
Total
Member
60
40
100
Nonmember
30
20
50
Total
90
60
150
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Recognize the structure. For independence, knowing membership must not change the purchase probability.
Work it through. The overall purchase probability is 90/150
Answer: Yes, for random selection from this table.
Check. The joint probability is 60/150
Avoid the trap. Equal conditional percentages do not require equal counts. This table-based conclusion also does not prove exact independence for all future visitors.
5.10. Use area only when location is uniform
A point is selected uniformly at random from a rectangle of width 10 and height 6. The shaded right triangle has base 6 and height 4 and lies entirely in the lower strip of height 4. Given that the point lies in that lower strip, what is the probability that it lies in the triangle?
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Recognize the structure. The condition restricts the eligible area to the lower strip, not the whole rectangle.
Work it through. The triangle’s area is (6)(4)
P(triangle ∣ lower strip) = = .
Answer: 3/10.
Check. Without the condition, the triangle’s probability would be 12/60
Avoid the trap. Dividing by 60 ignores the condition. Also, an area ratio is a probability only when the selection mechanism is uniform over the relevant region.
5.11. Multiply independent probabilities with replacement
A bag contains 4 red and 6 blue counters. A counter is chosen at random, replaced, and the bag is mixed. A second counter is then chosen independently at random. What is the probability of red first and blue second?
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Recognize the structure. Replacement restores the original counts, and independence is specified.
Work it through. The first-red probability is 4/10. The second-blue probability remains 6/10. Thus
P(red then blue) = × = = 0.24.
The order is part of the requested event.
Answer: 6/25.
Check. Among 10 × 10
Avoid the trap. Adding the two probabilities would not describe both events occurring. Do not double the result unless the question also allows blue first and red second.
5.12. Update the second probability without replacement
A bag contains 4 red and 6 blue counters. Two counters are chosen at random in sequence without replacement. What is the probability of red first and blue second?
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Recognize the structure. After a red counter is removed, the eligible total changes to 9 while the blue count stays 6.
Work it through. Use the probability of the first event times the conditional probability of the second:
P(red then blue) = × = .
The draws are not independent; the second fraction describes the bag after the specified first draw.
Answer: 4/15.
Check. A red removal raises the second-blue chance from 6/10 to 6/9. The result is therefore slightly larger than the with-replacement answer 0.24.
Avoid the trap. Using 6/10 for the second draw ignores the changed total. Using 5/9 would incorrectly remove a blue counter after the first draw was red.
5.13. Use the complement of at least one
Each of two independent shipments has probability 0.10 of arriving late. What is the probability that at least one of the shipments arrives late?
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Recognize the structure. The complement of “at least one late” is the event that neither shipment is late.
Work it through. Each shipment has on-time probability 1 − 0.10
P(at least one late) = 1 − 0.81 = 0.19.
Answer: 0.19.
Check. Exactly one late has probability 2(0.10)(0.90)
Avoid the trap. Adding 0.10 + 0.10
5.14. Weight conditional rates by their base groups
Supplier A provides 70% of a company’s parts, and 2% of A’s parts are defective. Supplier B provides the other 30%, and 5% of B’s parts are defective. One part is selected at random from the company’s parts. What is the probability that it is defective?
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Recognize the structure. The overall defect rate is a weighted combination of supplier-specific rates.
Work it through. Defective parts can come from either supplier, and those supplier events are disjoint:
P(defective) = (0.70)(0.02) + (0.30)(0.05) = 0.014 + 0.015 = 0.029.
Equivalently, in a hypothetical 10,000-part collection matching these shares, A supplies 140 defective parts and B supplies 150.
Answer: 0.029, or 2.9%.
Check. The overall rate lies between 2% and 5% and is closer to 2%, the rate of the larger source.
Avoid the trap. The simple average 3.5% would weight suppliers equally. The stated supply shares are 70% and 30%, not 50% each.
5.15. Reverse a conditional statement with a base‐rate table
Production line A makes 20% of all components, and 4% of A’s components are defective. Line B makes the remaining 80%, and 1% of B’s components are defective. A component is selected at random and is found to be defective. What is the probability that it came from line A?
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Recognize the structure. The eligible group is all defective components. Build counts for both sources before conditioning.
Work it through. Use a convenient total of 10,000 components to represent the given shares:
Line | All components | Defective components |
|---|---|---|
A | 2,000 | 80 |
B | 8,000 | 80 |
Among the 80 + 80
Answer: 1/2.
Check. P(A and defective) = 0.20(0.04) = 0.008. The overall defect probability is 0.008 + 0.80(0.01)
Avoid the trap. The requested probability is not 4%. Four percent is defective given A; the question asks for A given defective. The much larger output of B matters.
Section 5: exit questions
Try the five section-exit questions before checking the explanations. Use scratch paper for your reasoning and enter the requested numerical value for a question without choices.