The Central Idea
Suppose an event \(A\) can occur through several mutually exclusive cases
\[ B_1,B_2,\ldots,B_n. \]If these cases together cover the entire sample space, then every occurrence of \(A\) must occur through exactly one of them.
Therefore,
\[ \boxed{ P(A) = \sum_{i=1}^{n} P(B_i)P(A\mid B_i) }. \]This is the law of total probability.
A Partition of the Sample Space
The events
\[ B_1,B_2,\ldots,B_n \]form a partition of the sample space when:
- the events are mutually exclusive;
- their union is the entire sample space.
Symbolically,
\[ B_i\cap B_j=\varnothing \qquad\text{for }i\neq j \]and
\[ B_1\cup B_2\cup\cdots\cup B_n=S. \]Why Partitions Matter
A partition breaks one complicated probability problem into separate cases.
Because the cases do not overlap, the corresponding path probabilities can be added without double counting.
Because the cases cover the entire sample space, no possible route to the target event is omitted.
The Two-Case Version
The simplest partition uses an event \(B\) and its complement \(B^c\).
Since
\[ B\cap B^c=\varnothing \]and
\[ B\cup B^c=S, \]the event \(A\) can occur either with \(B\) or with \(B^c\).
Therefore,
\[ \boxed{ P(A) = P(B)P(A\mid B) + P(B^c)P(A\mid B^c) }. \]Deriving the Two-Case Formula
The event \(A\) can be split into two disjoint pieces:
\[ A=(A\cap B)\cup(A\cap B^c). \]Since these pieces are mutually exclusive,
\[ P(A) = P(A\cap B)+P(A\cap B^c). \]Apply the multiplication rule to each intersection:
\[ P(A\cap B)=P(B)P(A\mid B) \]and
\[ P(A\cap B^c)=P(B^c)P(A\mid B^c). \]Therefore,
\[ \boxed{ P(A) = P(B)P(A\mid B) + P(B^c)P(A\mid B^c) }. \]Multiply Along, Add Across
Probability trees provide a useful visual rule:
\[ \boxed{\text{multiply along each path}} \]and then
\[ \boxed{\text{add across alternative paths}}. \]The multiplication rule gives the probability of each complete path. The addition rule combines mutually exclusive paths leading to the same target event.
A First Example
Suppose \(60\%\) of customers are new customers and \(40\%\) are returning customers.
Let
\[ N=\{\text{new customer}\} \]and
\[ R=\{\text{returning customer}\}. \]Suppose the probability of purchasing is
\[ P(P\mid N)=0.20 \]for new customers and
\[ P(P\mid R)=0.50 \]for returning customers.
Calculate Each Purchase Path
The probability of being new and purchasing is
\[ P(N\cap P) = P(N)P(P\mid N) = 0.60(0.20) = 0.12. \]The probability of being returning and purchasing is
\[ P(R\cap P) = P(R)P(P\mid R) = 0.40(0.50) = 0.20. \]Add the Mutually Exclusive Purchase Paths
A customer cannot be both new and returning under this classification, so the two purchase paths are mutually exclusive.
Therefore,
\[ P(P) = 0.12+0.20 = \boxed{0.32}. \]Thus the overall purchase probability is \(32\%\).
The Overall Probability Is a Weighted Average
In the customer example,
\[ P(P) = 0.60(0.20)+0.40(0.50). \]The group-specific purchase probabilities,
\[ 0.20 \qquad\text{and}\qquad 0.50, \]are weighted by the population shares
\[ 0.60 \qquad\text{and}\qquad 0.40. \]Total probability can therefore be understood as a weighted average of conditional probabilities.
The Weights Must Sum to One
Because the partition events cover the sample space,
\[ P(B_1)+P(B_2)+\cdots+P(B_n)=1. \]These probabilities act as weights in
\[ P(A) = \sum_{i=1}^{n}P(B_i)P(A\mid B_i). \]The structure is exactly that of a weighted average.
The Overall Probability Lies Between the Conditional Probabilities
Because total probability is a weighted average, the overall probability cannot exceed the largest conditional probability or fall below the smallest one, provided the partition probabilities are nonnegative and sum to \(1\).
If
\[ 0.20\leq P(A\mid B_i)\leq0.70 \]for every partition group, then
\[ 0.20\leq P(A)\leq0.70. \]This provides a useful reasonableness check.
Example: Manufacturing Lines
A factory uses three production lines:
- Line 1 produces \(50\%\) of all units;
- Line 2 produces \(30\%\);
- Line 3 produces \(20\%\).
Their defect rates are:
\[ P(D\mid L_1)=0.01, \] \[ P(D\mid L_2)=0.03, \] \[ P(D\mid L_3)=0.05. \]Calculate the Defect Paths
Line 1 contributes
\[ P(L_1)P(D\mid L_1) = 0.50(0.01) = 0.005. \]Line 2 contributes
\[ 0.30(0.03)=0.009. \]Line 3 contributes
\[ 0.20(0.05)=0.010. \]Add the Defect Paths
The overall defect probability is
\[ P(D) = 0.005+0.009+0.010. \]Therefore,
\[ \boxed{P(D)=0.024}. \]The factory-wide defect rate is \(2.4\%\).
The Most Defective Line Need Not Produce Most Defects
Line 3 has the highest conditional defect rate:
\[ 5\%. \]But the number of defective units contributed by each line also depends on how much that line produces.
Total probability combines both:
\[ \text{group size}\times\text{within-group event rate}. \]This distinction is important whenever subgroups have different sizes.
Example: Medical Screening Population
Suppose \(2\%\) of a population has a condition:
\[ P(D)=0.02. \]Therefore,
\[ P(D^c)=0.98. \]A test returns positive with probability
\[ P(+\mid D)=0.95 \]among people with the condition and
\[ P(+\mid D^c)=0.04 \]among people without it.
Find the Overall Positive-Test Probability
A positive test can occur through two paths:
\[ D\cap+ \]or
\[ D^c\cap+. \]Therefore,
\[ P(+) = P(D)P(+\mid D) + P(D^c)P(+\mid D^c). \]Substitute:
\[ P(+) = 0.02(0.95)+0.98(0.04). \]Thus,
\[ P(+) = 0.019+0.0392 = \boxed{0.0582}. \]The overall positive-test rate is \(5.82\%\).
Why the Overall Positive Rate Exceeds the Disease Rate
The disease rate is only
\[ 2\%. \]However, positive results arise both from true positives and from false positives.
Total probability combines both sources into the overall positive-test probability.
A later lesson on diagnostic testing will examine this structure in more detail.
Example: Loan Default Across Risk Groups
Suppose a loan portfolio contains:
- \(50\%\) low-risk borrowers;
- \(35\%\) medium-risk borrowers;
- \(15\%\) high-risk borrowers.
Their default probabilities are
\[ 0.01,\qquad0.04,\qquad0.12. \]The overall default probability is
\[ 0.50(0.01)+0.35(0.04)+0.15(0.12). \]Therefore,
\[ P(D) = 0.005+0.014+0.018 = \boxed{0.037}. \]The portfolio-wide default rate is \(3.7\%\).
Example: Shipping Routes
A company sends packages through three routes:
\[ P(R_1)=0.50,\qquad P(R_2)=0.30,\qquad P(R_3)=0.20. \]The probabilities of a late delivery are
\[ P(L\mid R_1)=0.06, \] \[ P(L\mid R_2)=0.10, \] \[ P(L\mid R_3)=0.04. \]Therefore,
\[ P(L) = 0.50(0.06)+0.30(0.10)+0.20(0.04). \]Hence,
\[ \boxed{P(L)=0.068}. \]The overall late-delivery rate is \(6.8\%\).
Example: Customer Support Channels
Suppose customers contact support by:
- chat with probability \(0.50\);
- email with probability \(0.30\);
- phone with probability \(0.20\).
The probability that an issue is resolved on the first contact is
\[ 0.80 \]for chat,
\[ 0.65 \]for email, and
\[ 0.90 \]for phone.
Therefore,
\[ P(R) = 0.50(0.80)+0.30(0.65)+0.20(0.90). \]Thus,
\[ \boxed{P(R)=0.775}. \]The overall first-contact resolution rate is \(77.5\%\).
Total Probability from Counts
The law of total probability can also be applied with counts.
Suppose a school has:
- \(300\) first-year students with a \(70\%\) course-completion rate;
- \(200\) second-year students with an \(80\%\) completion rate.
The total student population is
\[ 500. \]The group weights are
\[ \frac{300}{500}=0.60 \]and
\[ \frac{200}{500}=0.40. \]Calculate the Overall Completion Rate
Apply total probability:
\[ P(C) = 0.60(0.70)+0.40(0.80). \]Therefore,
\[ P(C) = 0.42+0.32 = \boxed{0.74}. \]The overall completion rate is \(74\%\).
The Same Count Example Directly
The number of first-year students who complete the course is
\[ 300(0.70)=210. \]The number of second-year students who complete is
\[ 200(0.80)=160. \]Thus total completions equal
\[ 210+160=370. \]Therefore,
\[ \frac{370}{500}=0.74. \]The count method and total-probability method agree.
Total Probability from a Two-Way Table
Consider the following data:
| Purchased | Did Not Purchase | Total | |
|---|---|---|---|
| New Customer | 24 | 96 | 120 |
| Returning Customer | 40 | 40 | 80 |
| Total | 64 | 136 | 200 |
Directly from the table,
\[ P(P)=\frac{64}{200}=0.32. \]Verify the Table with Total Probability
The group probabilities are
\[ P(N)=\frac{120}{200}=0.60 \]and
\[ P(R)=\frac{80}{200}=0.40. \]The conditional purchase probabilities are
\[ P(P\mid N)=\frac{24}{120}=0.20 \]and
\[ P(P\mid R)=\frac{40}{80}=0.50. \]Therefore,
\[ P(P) = 0.60(0.20)+0.40(0.50) = \boxed{0.32}. \]Partitioning by More Than One Category
A partition may contain any finite number of mutually exclusive, collectively exhaustive categories.
For example:
\[ B_1=\text{small}, \qquad B_2=\text{medium}, \qquad B_3=\text{large}. \]Then
\[ P(A) = P(B_1)P(A\mid B_1) + P(B_2)P(A\mid B_2) + P(B_3)P(A\mid B_3). \]Partitions Must Not Overlap
Suppose we classify customers as:
- customers under age \(40\);
- customers who use mobile banking.
These groups can overlap, so they do not form a partition by themselves.
Adding path probabilities based on overlapping groups could double count customers belonging to both categories.
Partitions Must Cover All Possibilities
Suppose a population is divided into:
- ages \(18\) through \(29\);
- ages \(30\) through \(49\).
If people aged \(50\) or older are also part of the population, these two categories do not form a complete partition.
Any total-probability calculation based only on them would omit part of the sample space.
Check the Partition Before Calculating
A reliable partition should satisfy both:
\[ B_i\cap B_j=\varnothing \qquad(i\neq j) \]and
\[ \bigcup_{i=1}^{n}B_i=S. \]In plain language:
- no outcome belongs to more than one case;
- every outcome belongs to some case.
Example: Weather Categories
Suppose a day is classified as clear, cloudy, or rainy:
\[ P(C)=0.50,\qquad P(L)=0.30,\qquad P(R)=0.20. \]Suppose a delivery is late with probabilities
\[ 0.05,\qquad0.10,\qquad0.25 \]under these respective conditions.
Then
\[ P(\text{late}) = 0.50(0.05)+0.30(0.10)+0.20(0.25). \]Therefore,
\[ \boxed{P(\text{late})=0.105}. \]Example: Different Machines
A plant has two machines.
Machine \(A\) produces \(70\%\) of output and has a \(2\%\) defect rate.
Machine \(B\) produces \(30\%\) of output and has a \(6\%\) defect rate.
Then
\[ P(D) = 0.70(0.02)+0.30(0.06). \]Thus,
\[ \boxed{P(D)=0.032}. \]The overall defect rate is \(3.2\%\).
Example: Different Marketing Sources
Suppose website traffic arrives from:
- search: \(50\%\);
- referrals: \(30\%\);
- advertising: \(20\%\).
Conversion rates are:
\[ 4\%,\qquad6\%,\qquad10\%. \]The overall conversion probability is
\[ 0.50(0.04)+0.30(0.06)+0.20(0.10). \]Therefore,
\[ \boxed{P(C)=0.058}. \]The overall conversion rate is \(5.8\%\).
Changing the Group Mix Changes the Total
Suppose the conditional conversion rates remain unchanged, but traffic shifts toward the highest-converting source.
Even if no within-group rate changes, the overall conversion rate can change because the weights
\[ P(B_i) \]have changed.
Total probability therefore separates two drivers:
- the size of each subgroup;
- the event rate within each subgroup.
Composition Effects
An overall probability can rise or fall because the population composition changes, even when every conditional probability stays fixed.
This is sometimes called a composition effect.
Recognizing composition effects is important in business metrics, demography, risk analysis, education data, and many other applications.
Overall Rates Can Hide Subgroup Differences
A single unconditional probability such as
\[ P(A) \]summarizes the entire population.
But the conditional probabilities
\[ P(A\mid B_1), \quad P(A\mid B_2), \quad\ldots \]reveal how the event behaves within specific groups.
Total probability connects these two levels of description.
Example: Insurance Claims
Suppose an insurer divides policies into three classes:
\[ P(C_1)=0.50,\qquad P(C_2)=0.35,\qquad P(C_3)=0.15. \]Annual claim probabilities are
\[ 0.04,\qquad0.08,\qquad0.20. \]Then
\[ P(\text{claim}) = 0.50(0.04)+0.35(0.08)+0.15(0.20). \]Therefore,
\[ \boxed{P(\text{claim})=0.078}. \]The overall annual claim probability is \(7.8\%\).
Example: Production by Shift
Suppose \(65\%\) of units are produced during the day shift and \(35\%\) during the night shift.
The probability of passing inspection is
\[ 0.98 \]for day-shift units and
\[ 0.94 \]for night-shift units.
Therefore,
\[ P(\text{pass}) = 0.65(0.98)+0.35(0.94). \]Thus,
\[ \boxed{P(\text{pass})=0.966}. \]Using Complements Within Total Probability
Suppose a problem provides failure probabilities but asks for success.
For each group,
\[ P(S\mid B_i)=1-P(F\mid B_i). \]These complementary conditional probabilities can then be inserted into the total-probability formula:
\[ P(S) = \sum_i P(B_i)P(S\mid B_i). \]Example: Convert Failure Rates to Success Rates
Suppose two systems are used:
\[ P(S_1)=0.40,\qquad P(S_2)=0.60. \]Their failure probabilities are
\[ P(F\mid S_1)=0.10 \]and
\[ P(F\mid S_2)=0.04. \]Therefore,
\[ P(F) = 0.40(0.10)+0.60(0.04) = 0.064. \]Hence,
\[ P(F^c)=1-0.064=\boxed{0.936}. \]Total Probability and Probability Trees
A tree organizes total probability naturally.
At the first level, branch probabilities represent the partition:
\[ P(B_1),P(B_2),\ldots,P(B_n). \]At the second level, branches represent conditional probabilities such as
\[ P(A\mid B_i). \]Multiply each path:
\[ P(B_i)P(A\mid B_i). \]Then add all paths ending in \(A\).
A Tree Is an Organizational Tool
The law of total probability does not require drawing a tree.
A tree is useful because it makes three things visible:
- the partition at the first stage;
- the conditional probabilities within each branch;
- the separate paths leading to the target event.
The underlying calculation is still the same multiplication-and-addition structure.
Total Probability with Four Groups
Suppose
\[ P(B_1)=0.10,\quad P(B_2)=0.20,\quad P(B_3)=0.30,\quad P(B_4)=0.40, \]and
\[ P(A\mid B_1)=0.20, \] \[ P(A\mid B_2)=0.30, \] \[ P(A\mid B_3)=0.40, \] \[ P(A\mid B_4)=0.50. \]Then
\[ P(A) = 0.10(0.20) +0.20(0.30) +0.30(0.40) +0.40(0.50). \]Therefore,
\[ \boxed{P(A)=0.40}. \]Recovering an Unknown Conditional Probability
Total probability can also be used as an equation.
Suppose
\[ P(B)=0.60, \qquad P(B^c)=0.40, \] \[ P(A\mid B)=0.30, \]and
\[ P(A)=0.42. \]Let
\[ x=P(A\mid B^c). \]Then
\[ 0.42=0.60(0.30)+0.40x. \]Solve for the Missing Conditional Probability
Continue:
\[ 0.42=0.18+0.40x. \]Subtract \(0.18\):
\[ 0.24=0.40x. \]Therefore,
\[ \boxed{x=0.60}. \]Thus,
\[ P(A\mid B^c)=0.60. \]Recovering an Unknown Group Weight
Suppose a population consists of groups \(B\) and \(B^c\).
Let
\[ P(B)=x, \qquad P(B^c)=1-x. \]If
\[ P(A\mid B)=0.80, \] \[ P(A\mid B^c)=0.30, \]and
\[ P(A)=0.50, \]then
\[ 0.50=0.80x+0.30(1-x). \]Solve for the Group Weight
Expand:
\[ 0.50=0.80x+0.30-0.30x. \]Combine terms:
\[ 0.20=0.50x. \]Therefore,
\[ \boxed{x=0.40}. \]Thus \(40\%\) of the population belongs to \(B\).
Total Probability as Decomposition
The formula
\[ P(A) = \sum_i P(B_i)P(A\mid B_i) \]decomposes an overall probability into subgroup contributions.
Each term
\[ P(B_i)P(A\mid B_i) \]is the joint probability
\[ P(A\cap B_i). \]Therefore,
\[ P(A) = \sum_i P(A\cap B_i). \]The Joint Contributions Sum to the Total
Because the partition events are mutually exclusive,
\[ (A\cap B_i)\cap(A\cap B_j)=\varnothing \qquad(i\neq j). \]Because the partition covers the sample space,
\[ A = (A\cap B_1) \cup (A\cap B_2) \cup \cdots \cup (A\cap B_n). \]The total probability formula follows by adding these disjoint pieces.
Why This Matters for Bayesian Reasoning
Bayes' theorem often requires a denominator such as
\[ P(A). \]When \(A\) can occur through several underlying groups, total probability supplies that denominator:
\[ P(A) = \sum_i P(B_i)P(A\mid B_i). \]The next lesson will use this structure to reverse conditional probabilities.
Common Error: Forgetting a Path
Suppose an event \(A\) can occur through three partition groups:
\[ B_1,\quad B_2,\quad B_3. \]A calculation using only
\[ P(B_1)P(A\mid B_1) + P(B_2)P(A\mid B_2) \]is incomplete if the \(B_3\) path can also produce \(A\).
Every relevant partition path must be included.
Common Error: Using Overlapping Groups
Total probability requires a partition.
If two groups overlap, an outcome in the overlap can be counted twice when path probabilities are added.
Before using the formula, verify that the cases are mutually exclusive.
Common Error: Using Groups That Do Not Cover the Sample Space
If the selected cases omit possible outcomes, then the sum
\[ \sum_i P(B_i) \]will be less than \(1\).
Any resulting total probability may therefore omit part of the target event.
Common Error: Adding Conditional Probabilities Directly
Suppose
\[ P(A\mid B_1)=0.20 \]and
\[ P(A\mid B_2)=0.60. \]We cannot simply write
\[ P(A)=0.20+0.60. \]Each conditional probability must first be weighted by the probability of its group:
\[ P(A) = P(B_1)P(A\mid B_1) + P(B_2)P(A\mid B_2). \]Common Error: Multiplying All Paths Together
Alternative partition paths represent mutually exclusive routes to the same event.
Multiply within each route:
\[ P(B_i)P(A\mid B_i). \]Then add the route probabilities:
\[ \sum_i P(B_i)P(A\mid B_i). \]Do not multiply probabilities from separate alternative paths together.
Common Error: Forgetting That the Weights Are Probabilities
The partition weights must satisfy
\[ P(B_i)\geq0 \]and
\[ \sum_iP(B_i)=1. \]If given percentages sum to \(100\%\), they form valid population weights. If they do not, the classification may be incomplete or inconsistent.
Common Error: Confusing \(P(A)\) with \(P(A\mid B_i)\)
The quantity
\[ P(A\mid B_i) \]is the event rate within group \(B_i\).
The quantity
\[ P(A) \]is the overall event rate across the full population.
Total probability is the bridge between them.
Common Error: Expecting the Overall Rate to Equal an Unweighted Average
Suppose two groups have event rates
\[ 0.20 \]and
\[ 0.80. \]Their simple average is
\[ 0.50. \]But the overall probability equals \(0.50\) only if the two groups have equal probability weights.
In general, use the weighted average supplied by total probability.
A Reliable Procedure
- Identify the target event \(A\).
- Identify a partition \(B_1,\ldots,B_n\) of the sample space.
- Verify that the partition groups are mutually exclusive.
- Verify that their probabilities sum to \(1\).
- Find \(P(A\mid B_i)\) for each group.
- Multiply \(P(B_i)P(A\mid B_i)\) for each path.
- Add all path probabilities.
- Check that the result is between the smallest and largest relevant conditional rates.
- Interpret the result as the overall probability of \(A\).
Example: Full Two-Group Workflow
Suppose \(75\%\) of orders are domestic and \(25\%\) are international.
The probability of an order arriving late is
\[ 0.04 \]for domestic orders and
\[ 0.12 \]for international orders.
Let \(L\) mean late.
Then
\[ P(L) = 0.75(0.04)+0.25(0.12). \]Therefore,
\[ P(L) = 0.03+0.03 = \boxed{0.06}. \]The overall late-delivery rate is \(6\%\).
Example: Full Three-Group Workflow
A university course enrolls students from three programs:
\[ P(P_1)=0.50,\qquad P(P_2)=0.30,\qquad P(P_3)=0.20. \]Completion probabilities are
\[ P(C\mid P_1)=0.80, \] \[ P(C\mid P_2)=0.70, \] \[ P(C\mid P_3)=0.90. \]Therefore,
\[ P(C) = 0.50(0.80)+0.30(0.70)+0.20(0.90). \]Thus,
\[ P(C)=0.40+0.21+0.18=\boxed{0.79}. \]The overall completion probability is \(79\%\).
Practice 1: Two-Group Total Probability
Suppose
\[ P(B)=0.70, \qquad P(B^c)=0.30, \] \[ P(A\mid B)=0.40, \]and
\[ P(A\mid B^c)=0.20. \]Find \(P(A)\).
Practice 2: Three Groups
Suppose
\[ P(B_1)=0.50,\quad P(B_2)=0.30,\quad P(B_3)=0.20, \]and
\[ P(A\mid B_1)=0.10,\quad P(A\mid B_2)=0.20,\quad P(A\mid B_3)=0.40. \]Find \(P(A)\).
Practice 3: Manufacturing
Machine 1 produces \(60\%\) of a factory's output with a \(2\%\) defect rate. Machine 2 produces \(40\%\) with a \(5\%\) defect rate.
Find the overall defect probability.
Practice 4: Customer Segments
A company has \(30\%\) premium customers and \(70\%\) standard customers. Premium customers renew with probability \(0.90\); standard customers renew with probability \(0.65\).
Find the overall renewal probability.
Practice 5: Test Positivity
Suppose
\[ P(D)=0.05, \] \[ P(+\mid D)=0.90, \]and
\[ P(+\mid D^c)=0.08. \]Find \(P(+)\).
Practice 6: Missing Conditional Probability
Suppose
\[ P(B)=0.40, \] \[ P(A\mid B)=0.70, \] \[ P(A)=0.52. \]Find \(P(A\mid B^c)\).
Practice 7: Missing Group Weight
Suppose
\[ P(A\mid B)=0.80, \] \[ P(A\mid B^c)=0.40, \]and
\[ P(A)=0.60. \]Find \(P(B)\).
Practice 8: Check the Partition
A business divides customers into three groups with probabilities
\[ 0.45,\qquad0.35,\qquad0.15. \]Can these three probabilities alone represent a complete partition? Explain.
Practice 9: Weighted Average Check
Three group-specific event probabilities are
\[ 0.20,\qquad0.50,\qquad0.70. \]Could the overall probability under a valid partition equal \(0.80\)? Explain.
Practice 10: Identify the Error
A population is split into two equally sized groups. The event probabilities are
\[ P(A\mid B_1)=0.30 \]and
\[ P(A\mid B_2)=0.50. \]A student writes
\[ P(A)=0.30+0.50=0.80. \]Explain the error and give the correct result.
Practice Answers
Practice 1
\[ P(A) = 0.70(0.40)+0.30(0.20). \]Therefore,
\[ \boxed{P(A)=0.34}. \]Practice 2
\[ P(A) = 0.50(0.10)+0.30(0.20)+0.20(0.40). \]Thus,
\[ P(A)=0.05+0.06+0.08=\boxed{0.19}. \]Practice 3
\[ P(D) = 0.60(0.02)+0.40(0.05). \]Therefore,
\[ \boxed{P(D)=0.032}. \]Practice 4
\[ P(R) = 0.30(0.90)+0.70(0.65). \]Hence,
\[ P(R)=0.27+0.455=\boxed{0.725}. \]Practice 5
First,
\[ P(D^c)=0.95. \]Then
\[ P(+) = 0.05(0.90)+0.95(0.08). \]Therefore,
\[ P(+)=0.045+0.076=\boxed{0.121}. \]Practice 6
Since
\[ P(B^c)=0.60, \]let
\[ x=P(A\mid B^c). \]Then
\[ 0.52=0.40(0.70)+0.60x. \]Thus,
\[ 0.52=0.28+0.60x, \] \[ 0.24=0.60x, \]so
\[ \boxed{x=0.40}. \]Practice 7
Let
\[ P(B)=x. \]Then
\[ P(B^c)=1-x. \]Use total probability:
\[ 0.60=0.80x+0.40(1-x). \]Therefore,
\[ 0.60=0.40+0.40x, \] \[ 0.20=0.40x, \]so
\[ \boxed{x=0.50}. \]Practice 8
No. The probabilities sum to
\[ 0.45+0.35+0.15=0.95. \]A complete partition must have total probability \(1\). Another group or missing \(0.05\) probability must be accounted for.
Practice 9
No. A weighted average using valid nonnegative weights that sum to \(1\) must lie between the smallest and largest group-specific probabilities.
Since the largest conditional probability is \(0.70\), an overall probability of \(0.80\) is impossible under those conditions.
Practice 10
The conditional probabilities must be weighted by the group probabilities. Since the groups are equally sized,
\[ P(B_1)=P(B_2)=0.50. \]Therefore,
\[ P(A) = 0.50(0.30)+0.50(0.50) = 0.15+0.25. \]Thus,
\[ \boxed{P(A)=0.40}. \]What to Remember
If
\[ B_1,B_2,\ldots,B_n \]form a partition of the sample space, then
\[ \boxed{ P(A) = \sum_{i=1}^{n}P(B_i)P(A\mid B_i) }. \]- The partition cases must be mutually exclusive.
- The partition cases must cover the full sample space.
- Multiply within each path.
- Add across alternative paths leading to the target event.
- The group probabilities act as weights.
- The overall probability is a weighted average of conditional probabilities.
- Changing subgroup proportions can change the overall rate even when within-group rates stay fixed.
The central habit is: decompose the event into complete, nonoverlapping paths before adding their probabilities.
Where We Go Next
Total probability moves from subgroup-specific probabilities to an overall probability. The next step reverses the direction of the question.
The next lesson, Bayes' Theorem, uses joint probabilities and total probability to calculate the probability of an underlying cause or group after an outcome has been observed.