Independent events and the multiplication rule, several independent events together with complements for at-least questions, dependent events and conditional probability read from a two-way table, drawing with and without replacement, and probability tree diagrams.
Subject: Algebra 2 · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Algebra 2 · Chapter 10 — Counting Methods and Probability
Find Probabilities of Independent and Dependent Events
Objectives
Five outcomes. When does one event change the odds of another?
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.5 Find Probabilities of Independent and Dependent Events §10.5, pp. 717-721 — the lesson these objectives are drawn from
Warm-up
Lesson 10.4 combined events with OR. This lesson combines them with AND.
Discussion prompt
You toss a coin twice. Does the first toss change the chance of heads on the second? Now draw two cards without replacing the first. Does the first draw change the chance for the second?
Hint: What is left in each case?
Answer:
The coin has no memory: the second toss is still one half either way.
The deck does. After drawing a heart, only 12 hearts remain among 51 cards, so the second probability has changed.
\[ \text{coin: } \tfrac{1}{2}\cdot\tfrac{1}{2}; \qquad \text{cards: } \tfrac{13}{52}\cdot\tfrac{12}{51} \]
The first pair is independent and the second dependent, and telling them apart is the only decision this lesson asks for.
Concept
The probability that two events both occur is the probability of the first times the probability of the second. If the events are independent that second probability is unchanged; if they are dependent it must be conditioned on the first having happened.
conditional probability — The probability that B occurs given that A has occurred, written P of B given A. For dependent events it replaces P of B in the multiplication rule.
\[ P(A \text{ and } B) = P(A)\cdot P(B \mid A) \]
There is really one rule. For independent events the conditional probability equals the plain one, so the formula collapses to a simple product.
Figure (svg): Two columns comparing independent events with dependent ones
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.5 Find Probabilities of Independent and Dependent Events §10.5, pp. 717-718
Section
Section 1
Concept
Two events are independent when the occurrence of one has no effect on the occurrence of the other. Then the probability that both occur is the product of the two probabilities, and the rule extends to any number of events.
\[ P(A \text{ and } B) = P(A)\cdot P(B) \]
Independence is a claim about the situation, not about the arithmetic. Two separate raffles are independent; two draws from the same deck are not.
Figure (svg): Two independent events whose probabilities multiply, with the counts shown
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.5 Find Probabilities of Independent and Dependent Events §10.5, pp. 717-717 — Probability of Independent Events
Picture it
Example 1: five tickets in each of two draws.
Figure (svg): Two independent events whose probabilities multiply, with the counts shown
One fortieth times one fiftieth is one two-thousandth. Winning the first prize tells you nothing about the second, so the probabilities simply multiply.
Worked example
Example 1, a multiple-choice item.
\[ \text{With } 5 \text{ of } 200 \text{ tickets for one prize and } 5 \text{ of } 250 \text{ for another, find } P(\text{both}). \]
Find each probability
Why: Five over 200 and 5 over 250.
\[ \frac{1}{40}\text{ and } \frac{1}{50} \]
Check independence
Why: The two draws are separate and neither affects the other.
Multiply
Why: One fortieth times one fiftieth.
\[ \frac{1}{2000} \]
Interpret
Why: About one chance in two thousand.
\[ 0.0005 \]
Figure (svg): Two independent events whose probabilities multiply, with the counts shown
\[ \tfrac{1}{40}\cdot\tfrac{1}{50} = \tfrac{1}{2000} \]
Verify: sanity-check against each single chance
Why: The chance of the first prize alone is 1 in 40 and of the second alone 1 in 50, so winning both must be far rarer than either — and 1 in 2000 is. A probability of both events that came out larger than either single one would be impossible.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.5 Find Probabilities of Independent and Dependent Events §10.5, pp. 717-717
Sorting
Does the first change the second?
Sort into buckets
Sort each pair of events.
The word kept, or the phrase without replacement, is the usual signal for dependence. If whatever was drawn goes back, the events are independent.
Worked example
Guided Practice 1.
\[ \text{Find } P(\text{wins the certificate but NOT the passes}). \]
Probability of the first
Why: Five of 200 tickets.
\[ \frac{1}{40} \]
Probability of missing the second
Why: One minus 5 over 250.
\[ \frac{245}{250} = \frac{49}{50} \]
Check independence
Why: Still two separate draws.
Multiply
Why: One fortieth times 49 fiftieths.
\[ \frac{49}{2000},\text{ about } 0.0245 \]
Figure (svg): The solution to Worked example one prize but not the other shown as a ladder of expressions, one row per algebraic move
\[ \tfrac{1}{40}\cdot\tfrac{49}{50} = \tfrac{49}{2000} \]
Verify: check against the both-prizes answer
Why: Winning both is 1 over 2000 and winning only the first is 49 over 2000, so winning the first at all is 50 over 2000, which is 1 over 40 — matching. The two cases split the first event exactly, which confirms both computations at once.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.5 Find Probabilities of Independent and Dependent Events §10.5, pp. 717-717
Trap
\[ \text{two cards from one deck, no replacement} \]
Multiply the two unchanged probabilities
Why: The multiplication rule is applied without checking independence.
\[ \tfrac{13}{52}\cdot\tfrac{13}{52} \quad \text{(wrong)} \]
After the first heart is removed, only 12 hearts remain among 51 cards, so the second probability is not 13 over 52.
\[ \tfrac{13}{52}\cdot\tfrac{12}{51} \]
Condition the second probability on the first
Why: Removing a card changes both the favourable count and the total.
\[ P(A \text{ and } B) = P(A)\cdot P(B\mid A) \]
The book prints exactly this caution: decide first whether the events are independent or dependent, because the rule differs.
Fill the middle
Example 1.
Fill in the blanks
\frac2000___\cdot\frac______ = \frac______}
Why: Forty times 50 is 2000, so the probability of winning both is one in two thousand. Multiplying denominators is what multiplying the fractions amounts to when both numerators are 1.
Matching
Multiply the independent probabilities.
Match the pairs
Why: The last two are repeated identical events, so each is a single probability raised to a power: one eighth cubed and three tenths cubed. Repeated independent trials always give a power like that.
Prediction
Commit before reasoning.
Predict first
When is it wrong to multiply two probabilities to get the chance of both?
Correct: When the events are dependent, so the second probability changes.
\[ P(A\text{ and }B) = P(A)P(B\mid A), \text{ always} \]
Why: The rule for both events always multiplies, but the second factor must be the probability of B GIVEN that A happened. For independent events that equals the ordinary probability of B, so the simple product is correct. For dependent events it does not, and using the unconditioned probability gives an answer that is systematically wrong — too large or too small depending on which way the first event shifts things.
Section
Section 2
Concept
For several independent events, multiply all the probabilities. When the question asks for at least one success, computing the probability of no successes and subtracting from 1 is far shorter.
\[ P(\text{at least once}) = 1-[P(\text{not})]^n \]
At least one is the union of many overlapping cases, while none at all is a single condition — which is why the complement is almost always the shorter route.
Figure (svg): An at-least-once probability computed through the complement over five days
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.5 Find Probabilities of Independent and Dependent Events §10.5, pp. 718-718 — Use a complement to find a probability
Picture it
Example 3: hearing a favourite song at least once in a week.
Figure (svg): An at-least-once probability computed through the complement over five days
Three quarters chance of missing it on any one day, so three quarters to the fifth of missing it all week — leaving about 76 percent for hearing it.
Worked example
Example 2 and Guided Practice 2.
\[ \text{Find } P(\text{lane } 8 \text{ in all three heats of } 8) \text{ and } P(\text{three spins all perfect squares from } 1 \text{ to } 10). \]
Lanes: one heat
Why: One of 8 lanes.
\[ \frac{1}{8} \]
Lanes: three heats
Why: The heats are separate draws.
\[ (\frac{1}{8}) ^{3} = \frac{1}{512} \]
Spins: one spin
Why: The perfect squares are 1, 4 and 9.
\[ \frac{3}{10} \]
Spins: three spins
Why: Each spin is unaffected by the others.
\[ (\frac{3}{10}) ^{3} = \frac{27}{1000} \]
Figure (svg): Two independent events whose probabilities multiply, with the counts shown
\[ \tfrac{1}{512}; \qquad \tfrac{27}{1000} \]
Verify: check the second's perfect squares
Why: Between 1 and 10 the perfect squares are 1, 4 and 9 — three of the ten regions, so three tenths per spin. Sixteen is above 10 and does not count, which is the usual slip. Cubing gives 27 over 1000, about one chance in 37.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.5 Find Probabilities of Independent and Dependent Events §10.5, pp. 718-718
Fill the middle
Example 3.
Fill in the blanks
P(\text5 5 \text___) = \left(\frac______\right)^___} = \frac______
Why: Five days give five factors, so the probability is three quarters to the fifth. Repeated independent trials always produce a power.
Worked example
Example 3 and Guided Practice 3.
\[ \text{With } 4 \text{ of } 16 \text{ songs played daily, find } P(\text{favourite heard at least once in } 5 \text{ days}); \text{ then with } 12 \text{ songs.} \]
One day, 16 songs
Why: Choose 4 of the other 15, over 4 of all 16.
\[ \frac{1365}{1820} = \frac{3}{4} \]
Five days, then complement
Why: Three quarters to the fifth, subtracted from 1.
\[ 1 - \frac{243}{1024},\text{ about } 0.763 \]
One day, 12 songs
Why: Choose 4 of the other 11, over 4 of all 12.
\[ \frac{330}{495} = \frac{2}{3} \]
Five days, then complement
Why: Two thirds to the fifth, subtracted from 1.
\[ 1 - \frac{32}{243},\text{ about } 0.868 \]
Figure (svg): An at-least-once probability computed through the complement over five days
\[ 1-\left(\tfrac{3}{4}\right)^5 \approx 0.763; \quad 1-\left(\tfrac{2}{3}\right)^5 \approx 0.868 \]
Verify: check the daily probability differently
Why: With 16 songs, 4 are played, so the chance the favourite is among them is 4 over 16, or one quarter — leaving three quarters for missing it, matching the combination calculation. The simpler route works because every song is equally likely to be chosen.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.5 Find Probabilities of Independent and Dependent Events §10.5, pp. 718-718
Error analysis
A student finds the chance of hearing a favourite song at least once in five days.
Annotate
On: \( 5 \times \tfrac{1}{4} = \tfrac{5}{4} = 1.25 \)
Any probability above 1 flags the error at once. Adding works for disjoint events, and repeated trials are never disjoint.
Sorting
Compare the number of cases.
Sort into buckets
Sort each question.
The phrase at least one is close to a guarantee that the complement is the right move, since its opposite is always exactly one case.
Comparison
Fill the blanks. A shorter playlist raises every chance.
Comparison matrix
| Quantity | 16 songs | 12 songs |
|---|---|---|
| Chance of hearing it in a day | 4/16 = 1/4 | 4/12 = 1/3 |
| Chance of missing it in a day | 3/4 | 2/3 |
| Chance of missing it all week | (3/4)^5, about 0.237 | (2/3)^5, about 0.132 |
| Chance of hearing it at least once | about 0.763 | about 0.868 |
A daily chance rising from a quarter to a third pushes the weekly chance from 76 to 87 percent, because the daily advantage compounds across five days.
Prediction
Commit before reasoning.
Predict first
With a three quarters daily chance of missing the song, what happens as the number of days grows?
Correct: The chance of hearing it approaches 1 but never reaches it.
\[ \left(\tfrac{3}{4}\right)^n \to 0 \;\Longrightarrow\; 1-\left(\tfrac{3}{4}\right)^n \to 1 \]
Why: Three quarters to the n shrinks toward zero as n grows — at twenty days it is under 0.4 percent — so the complement approaches 1. But it never equals 1, since a positive number raised to any power stays positive. Given enough days you are almost certain to hear the song, and never quite certain, which is exactly the exponential decay of Lesson 7.2 doing probability work.
Section
Section 3
Concept
The probability of B given that A has occurred is written P of B given A. In a two-way table it is computed by restricting attention to A's row or column, so the denominator becomes that group's total rather than the grand total.
\[ P(B \mid A) \]
A conditional probability may be larger than, smaller than, or equal to the unconditional one. Equal means the events are independent.
Figure (svg): A two-way table of storm types by hemisphere, with two probabilities computed from it
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.5 Find Probabilities of Independent and Dependent Events §10.5, pp. 718-719 — Find a conditional probability
Picture it
Example 4: cyclone types by hemisphere.
Figure (svg): A two-way table of storm types by hemisphere, with two probabilities computed from it
Overall 760 of 1575 cyclones are hurricanes, about 0.483; among northern ones, 545 of 1142, about 0.477. Restricting to the north barely changes it.
Worked example
Example 4.
\[ \text{From the cyclone table, find } P(\text{hurricane}) \text{ and } P(\text{hurricane given northern hemisphere}). \]
Total every cell
Why: One hundred ninety-nine plus 18 plus 398 plus 200 plus 545 plus 215.
\[ 1575 \]
Count all hurricanes
Why: Five hundred forty-five plus 215.
\[ 760 \]
Divide for the first
Why: Seven hundred sixty over 1575.
\[ \text{about } 0.483 \]
Restrict to the north and divide
Why: Five hundred forty-five over the northern total of 1142.
\[ \text{about } 0.477 \]
Figure (svg): A two-way table of storm types by hemisphere, with two probabilities computed from it
\[ \tfrac{760}{1575} \approx 0.483; \quad \tfrac{545}{1142} \approx 0.477 \]
Verify: check the column total
Why: The northern column is 199 plus 398 plus 545, which is 1142, and the southern is 18 plus 200 plus 215, which is 433 — and 1142 plus 433 is 1575, the grand total. Verifying that the margins add up catches a misread cell before it propagates.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.5 Find Probabilities of Independent and Dependent Events §10.5, pp. 719-719
Fill the middle
Example 4b.
Fill in the blanks
P(\text1142\mid\text___) = \frac______}
Why: The northern column totals 1142, and conditioning on northern makes that the denominator. Using 1575 would answer a different question entirely.
Worked example
Guided Practice 4.
\[ \text{Find } P(\text{tropical storm}) \text{ and } P(\text{tropical storm given southern hemisphere}). \]
Count all tropical storms
Why: Three hundred ninety-eight plus 200.
\[ 598 \]
Divide by the grand total
Why: Five hundred ninety-eight over 1575.
\[ \text{about } 0.380 \]
Restrict to the south
Why: Two hundred storms out of the southern total.
\[ \frac{200}{433} \]
Divide
Why: Two hundred over 433.
\[ \text{about } 0.462 \]
Figure (svg): The solution to Worked example two more from the table shown as a ladder of expressions, one row per algebraic move
\[ \tfrac{598}{1575} \approx 0.380; \quad \tfrac{200}{433} \approx 0.462 \]
Verify: notice how much this one shifts
Why: The overall storm share is 38 percent but the southern share is 46 percent — a real difference, unlike the hurricane figures which barely moved. So storm type and hemisphere are NOT independent: knowing the hemisphere genuinely changes the estimate.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.5 Find Probabilities of Independent and Dependent Events §10.5, pp. 719-719
Trap
\[ P(\text{hurricane}\mid\text{north}) = \frac{545}{1575} \]
Divide the northern hurricanes by everything
Why: The grand total is used out of habit.
\[ \approx 0.346 \quad \text{(wrong)} \]
That is the probability of being a northern hurricane, not the probability of being a hurricane once you already know it is northern.
\[ P(\text{hurricane}\mid\text{north}) = \frac{545}{1142} \approx 0.477 \]
Restrict the denominator to the given group
Why: Conditioning throws away everything outside that group.
\[ \text{northern cyclones only: } 1142 \]
The given condition always names the new denominator. Underlining the word given, and then the group it names, makes the right total obvious.
Comparison
Fill the blanks. Only the denominator changes.
Comparison matrix
| Probability | Numerator | Denominator |
|---|---|---|
| P(hurricane) | 760 | 1575, the grand total |
| P(hurricane given north) | 545 | 1142, the northern total |
| P(tropical storm) | 598 | 1575 |
| P(tropical storm given south) | 200 | 433, the southern total |
Every conditional probability trades the grand total for a group total, and the numerator narrows to match.
Sorting
Read what comes after the word given.
Sort into buckets
Sort each probability by its denominator.
The last item conditions on hurricane rather than on hemisphere, so its denominator is 760 — the hurricane ROW total rather than a column one.
Prediction
Commit before reasoning.
Predict first
P(hurricane) is 0.483 and P(hurricane given north) is 0.477. What does that near-equality suggest?
Correct: Hemisphere barely affects whether a cyclone is a hurricane, so the two events are close to independent.
\[ P(B\mid A) = P(B) \;\Longleftrightarrow\; A \text{ and } B \text{ independent} \]
Why: Independence means exactly that conditioning changes nothing: P of B given A equals P of B. Here the two differ by less than a percentage point, so knowing the hemisphere tells you almost nothing about whether the cyclone reached hurricane strength. Contrast the tropical storm figures, 0.380 against 0.462, which differ enough to show real dependence.
Section
Section 4
Concept
Drawing with replacement leaves the pool unchanged, so the events are independent. Drawing without replacement removes an object, so the second probability must be conditioned on the first.
\[ \tfrac{39}{52}\cdot\tfrac{13}{52} \text{ against } \tfrac{39}{52}\cdot\tfrac{13}{51} \]
The rule extends to more draws: each factor uses the counts remaining at that point, so three draws without replacement have three different denominators.
Figure (svg): The same two draws computed with replacement and without
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.5 Find Probabilities of Independent and Dependent Events §10.5, pp. 719-720 — Comparing independent and dependent events
Picture it
Example 5: a non-heart then a heart.
Figure (svg): The same two draws computed with replacement and without
With replacement gives about 0.188 and without gives about 0.191 — close, because removing one card from 52 changes little.
Worked example
Example 5, both parts.
\[ \text{Find } P(\text{first not a heart, second a heart}) \text{ with and without replacement.} \]
First probability
Why: Thirty-nine non-hearts of 52.
\[ \frac{39}{52} = \frac{3}{4} \]
With replacement
Why: The deck is whole again, so 13 hearts of 52.
\[ (\frac{3}{4}) (\frac{1}{4}) = \frac{3}{16} \]
Without replacement
Why: Fifty-one cards remain, still with 13 hearts.
\[ (\frac{3}{4}) (\frac{13}{51}) \]
Simplify
Why: Thirty-nine over 204.
\[ \frac{13}{68},\text{ about } 0.191 \]
Figure (svg): The same two draws computed with replacement and without
\[ \tfrac{3}{16}; \qquad \tfrac{13}{68} \]
Verify: explain why the second is slightly larger
Why: Removing a non-heart leaves all 13 hearts among only 51 cards, so the second draw is slightly MORE likely to be a heart. The first event helped the second, which is why the dependent answer edges above the independent one — a small effect here, but real.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.5 Find Probabilities of Independent and Dependent Events §10.5, pp. 719-719
Fill the middle
Example 5b.
Fill in the blanks
\frac51___\cdot\frac______} = \frac______
Why: After one card is drawn and kept, 51 remain. The hearts count is unchanged at 13 because the card removed was not a heart.
Worked example
Example 6 and Guided Practice 6.
\[ \text{Find } P(\text{three friends choose different costumes from } 15) \text{ and } P(\text{a jack then another jack, no replacement}). \]
Costumes: the first friend
Why: Any of the 15 works.
\[ \frac{15}{15} \]
Costumes: the second and third
Why: Fourteen then 13 of the 15 remain unused.
\[ (\frac{14}{15}) (\frac{13}{15}) \]
Costumes: multiply
Why: One hundred eighty-two over 225.
\[ \text{about } 0.809 \]
Jacks: no replacement
Why: Four of 52, then 3 of 51.
\[ \frac{12}{2652} = \frac{1}{221} \]
Figure (svg): The solution to Worked example three dependent draws shown as a ladder of expressions, one row per algebraic move
\[ \tfrac{182}{225}; \qquad \tfrac{1}{221} \]
Verify: check the costume denominators
Why: The store has duplicates, so the POOL of 15 costumes never shrinks — only the set of unused ones does. That is why every denominator stays 15 while the numerators fall. Had there been one of each costume, the denominators would have fallen too.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.5 Find Probabilities of Independent and Dependent Events §10.5, pp. 720-720
Error analysis
A student computes the chance of two jacks drawn without replacement.
Annotate
On: \( \frac{4}{52}\cdot\frac{3}{52} = \frac{12}{2704} \)
Both parts of the second fraction change when a card is removed. Writing out what remains after the first draw, in words, prevents the half-adjustment.
Matching
With replacement, or without.
Match the pairs
Why: In the spade-then-club pair the removal helps slightly, since no club was taken; in the jack pair it hurts, since a jack was. Which way the shift goes depends on whether the first draw removed one of the second event's favourable outcomes.
Sorting
Ask what the first draw removed.
Sort into buckets
Sort each without-replacement pair.
Both effects come from the same removal, and which dominates is decided entirely by whether the first card belonged to the second event.
Prediction
Commit before reasoning.
Predict first
Removing one card from 52 changed the answer by under two percent. When would replacement matter more?
Correct: With a small pool, where removing one object changes the proportions sharply.
\[ \tfrac{3}{5}\cdot\tfrac{2}{4} = 0.30 \text{ against } \tfrac{3}{5}\cdot\tfrac{3}{5} = 0.36 \]
Why: Drawing two red marbles from a bag of three red and two blue gives three fifths times two quarters, which is 0.3, against three fifths squared, or 0.36, with replacement — a 17 percent difference. From a deck of 52 the shift was under 2 percent. The smaller the pool, the more one removal matters, which is the same reason the no-repetition penalty in Lesson 10.1 was heavier for digits than for letters.
Section
Section 5
Concept
A tree diagram records conditional probabilities on its branches. Multiplying along a path gives that path's probability, and adding the paths that reach an outcome gives the outcome's probability.
\[ P(C) = P(A)P(C\mid A)+P(B)P(C\mid B) \]
The branches leaving any single point must add to 1, since they list every possibility at that stage. Checking that is the fastest way to spot a missing branch.
Figure (svg): A probability tree with two stages and the paths to one outcome added together
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.5 Find Probabilities of Independent and Dependent Events §10.5, pp. 720-720 — Multi-step problem
Picture it
Example 7: adults and students wearing seat belts.
Figure (svg): A probability tree with two stages and the paths to one outcome added together
Two of the four paths end with a belted student, and adding their probabilities gives 0.536.
Worked example
Example 7.
\[ \text{Adults belt } 69\% \text{ of the time; students belt } 66\% \text{ when the adult does and } 26\% \text{ when not. Find } P(\text{student belts}). \]
Draw the first stage
Why: Adult belted, 0.69, or not, 0.31.
\[ \text{two branches summing to } 1 \]
Draw the second stage
Why: Student belted with probability 0.66 or 0.26 depending on the adult.
Multiply along each wanted path
Why: Point six nine times 0.66, and 0.31 times 0.26.
\[ 0.4554\text{ and } 0.0806 \]
Add the two paths
Why: Both end with a belted student.
\[ 0.536 \]
Figure (svg): A probability tree with two stages and the paths to one outcome added together
\[ (0.69)(0.66)+(0.31)(0.26) = 0.536 \]
Verify: check that all four paths total 1
Why: The other two paths give 0.69 times 0.34, which is 0.2346, and 0.31 times 0.74, which is 0.2294. Adding all four gives 0.4554 plus 0.0806 plus 0.2346 plus 0.2294, exactly 1 — so every possibility has been accounted for and none double-counted.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.5 Find Probabilities of Independent and Dependent Events §10.5, pp. 720-720
Fill the middle
Example 7.
Fill in the blanks
(0.69)(0.66) = 0.4554
Why: Multiplying gives 0.4554, the probability that the adult belts AND the student belts. Each path's probability is the product of its branch probabilities.
Worked example
Using the same tree for a different question.
\[ \text{Given that a student is belted, find the probability the adult was belted too.} \]
Identify the wanted paths
Why: Only the path with both belted.
\[ 0.4554 \]
Identify the condition's total
Why: Every path ending with a belted student.
\[ 0.536 \]
Divide
Why: Point four five five four over 0.536.
\[ \text{about } 0.850 \]
Interpret
Why: Most belted students were riding with a belted adult.
\[ 85 \% \]
Figure (svg): A probability tree with two stages and the paths to one outcome added together
\[ \frac{0.4554}{0.536} \approx 0.850 \]
Verify: compare with the unconditional figure
Why: Overall 69 percent of adults belt, but among the cars where the student is belted the figure rises to 85 percent. Knowing the student's behaviour genuinely changes the estimate for the adult, which is what dependence means — and the tree supports questions in either direction.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.5 Find Probabilities of Independent and Dependent Events §10.5, pp. 720-720
Trap
\[ \text{adult belted } 0.69, \text{ then student belted } 0.66 \]
Add the two branch probabilities
Why: The path is treated as a union of two events.
\[ 0.69+0.66 = 1.35 \quad \text{(wrong)} \]
A path means both stages happen, so the probabilities multiply. Adding them exceeds 1, which is impossible.
\[ (0.69)(0.66) = 0.4554 \]
Multiply along a path and add across paths
Why: Along a path is AND; across paths is OR.
\[ 0.4554+0.0806 = 0.536 \]
The two operations do different jobs in the same diagram, and confusing them is the commonest tree-diagram error.
Sorting
Along a path, or across paths.
Sort into buckets
Sort each operation on a tree diagram.
The two operations correspond exactly to AND and OR, which is the same distinction that has run through the whole chapter.
Comparison
Fill the blanks. All four together must total 1.
Comparison matrix
| Path | Computation | Probability |
|---|---|---|
| Both belted | (0.69)(0.66) | 0.4554 |
| Adult belted, student not | (0.69)(0.34) | 0.2346 |
| Adult not, student belted | (0.31)(0.26) | 0.0806 |
| Neither belted | (0.31)(0.74) | 0.2294 |
The four probabilities total exactly 1, which confirms that the tree covers every possibility once and only once.
Prediction
Commit before reasoning.
Predict first
The branches leaving any single point of a tree always add to 1. Why?
Correct: Because they list every possibility at that stage, and something must happen.
\[ 0.66+0.34 = 1; \quad 0.26+0.74 = 1 \]
Why: The branches from a point are exhaustive and mutually exclusive, so they are a complete partition of what can happen next — and the probabilities of a complete partition always total 1. That makes it a free check on every stage: if a set of branches does not sum to 1, a possibility has been left out or a probability mistyped.
Comparison
Fill the blanks. Only the second factor differs.
Comparison matrix
| Situation | Rule | Example |
|---|---|---|
| Independent events | P(A) * P(B) | two separate raffles |
| Dependent events | P(A) * P(B given A) | two cards, kept |
| Repeated independent trials | a single probability raised to a power | (3/4)^5 for five days |
| At least one success | 1 minus the probability of none | 1 - (3/4)^5 |
The general rule always multiplies by a conditional probability; independence is simply the case where conditioning changes nothing.
Pattern
Decide independence, then multiply.
A conditional probability restricts the denominator to the group named after the word given.
OpenStax Algebra and Trigonometry 2e, §13.7 Probability §13.7
Check
Independent events multiply.
Check your understanding
With 5 of 200 tickets for one prize and 5 of 250 for another, what is the probability of winning both?
Answer: A
Why: One fortieth times one fiftieth is one two-thousandth.
Check
Conditioning changes the denominator.
Check your understanding
Of 1142 northern cyclones, 545 were hurricanes. What is P(hurricane given northern)?
Answer: A
Why: Conditioning on northern makes 1142 the denominator, giving 545 over 1142.
Check
Without replacement, both parts change.
Check your understanding
Drawing two cards without replacement, what is the probability of a jack then another jack?
Answer: A
Why: One jack is gone, so the second draw is 3 jacks among 51: 4/52 times 3/51.
Real world
A system has three independent backup components, each of which fails with probability 0.1 in a given year. The system fails only if all three fail.
Discussion prompt
Find the probability of system failure, then find how many components would be needed to push it below one in a million.
Hint: Failures are independent, so multiply.
Answer:
\[ P(\text{all three fail}) = (0.1)^3 = 0.001 \]
\[ (0.1)^n < 10^{-6} \;\Longrightarrow\; n > 6 \]
Three components give a failure probability of one in a thousand, and seven components would push it below one in a million.
Each extra component divides the failure probability by ten, which is why redundancy is so effective and why it is the standard approach in aircraft, data centres and spacecraft. The catch is the independence assumption: if all three components share a power supply, or were built in the same faulty batch, their failures are dependent and the product rule overstates the reliability badly. Engineers call this common-cause failure, and it is the reason redundant systems are deliberately built to different designs. The mathematics is one line; making the assumption true is the hard part.
Commit first
Answer, then rate your confidence honestly.
Predict first
Drawing two cards without replacement, is the chance of two hearts equal to 13 over 52, squared?
Correct: No — after one heart is kept, only 12 hearts remain among 51 cards.
\[ \tfrac{13}{52}\cdot\tfrac{12}{51} \approx 0.0588 \text{ against } \left(\tfrac{13}{52}\right)^2 = 0.0625 \]
Why: The correct product is 13 over 52 times 12 over 51, which is about 0.0588, against 0.0625 for the squared version. Both the favourable count and the total changed, and using the unconditioned second probability ignores both changes. With a deck of 52 the gap is small, but with a bag of five marbles the same error would be badly wrong. The general rule always multiplies by the conditional probability; independence is the special case where conditioning changes nothing.
Explain it
They multiply probabilities without thinking about it.
Discussion prompt
In four sentences or fewer, explain when multiplying two probabilities is not enough.
Hint: Think about a bag of marbles.
Answer:
If you toss a coin twice, the first toss tells you nothing about the second, so you just multiply the two halves.
But if you take a marble from a bag and keep it, the bag has changed for the second draw. So the second probability has to be worked out from what is actually left, not from the original bag.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For independence, ask whether anything was removed or changed between the events. For removals, write out in words what remains before writing the second fraction. For tables, underline the word given and use that group's total. For trees, multiply along paths and add across them, and check that branches from each point sum to 1.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a multiplication page. Top: write both rules side by side and one situation of each kind, with a sentence on what makes events independent. Middle left: work Example 5 both ways in parallel columns and explain in one line why the dependent answer came out slightly larger. Middle right: copy the cyclone table with all its margins, compute one unconditional and two conditional probabilities from it, and note which pair suggests independence. Bottom left: work the CD problem through the complement, and write the answer for 12 songs beside it. Bottom right: draw the seat belt tree in full with all four paths, compute each path, check they total 1, and use it to answer the reverse question about the adult.
If your four tree paths do not total exactly 1, recheck each pair of branches: the probabilities leaving any single point must add to 1.
Recap
Five things, and one question behind all of them.
| If you see | Then |
|---|---|
| With replacement, or separate trials | Independent; multiply the plain probabilities |
| Without replacement, or objects kept | Dependent; condition the second probability |
| The word given | Restrict the denominator to that group |
| At least one | Compute the probability of none and subtract from 1 |
| Different conditional probabilities per branch | Draw a tree |
| A path on a tree | Multiply along it; add across paths |
Lesson 10.6 puts repeated independent trials together into a binomial distribution, counting the ways each number of successes can occur.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.5 Find Probabilities of Independent and Dependent Events §10.5, pp. 717-721 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Algebra 2 — $55/session, free consultation.