Diagnostic Set 1 Study Guide

How to use this guide

Start with the chapters for the questions you missed or guessed on in the diagnostic. Read the core ideas, cover the walk-through and try the diagnostic question again, then do the three practice problems before revealing the answers.

1 · LearnCore ideas and toolkit formulas.

2 · ApplyRe-solve the diagnostic question.

3 · AvoidStudy the traps that catch most candidates.

4 · PractiseThree new problems per topic.

01Composite FunctionsAlgebra & Functions · Q1 02Probability Without ReplacementCounting & Probability · Q2 03Percents of Real CountsArithmetic & Percents · Q3 04Rate, Time & Distance: Time SavedWord Problems: Rates & Work · Q4 05Work Rates & “Percent Faster”Word Problems: Rates & Work · Q5 06Fractions of What RemainsArithmetic & Percents · Q6 07Percent Symmetry & Percent ArithmeticArithmetic & Percents · Q7 08Weighted AveragesStatistics · Q8 09Combinations: Counting PairsCounting & Probability · Q9 10Units Digits & CyclicityNumber Properties · Q10 11Two-Part Analysis: Logical ChainsData Insights: Two-Part · Q11 12Independent Events: Both, Neither, At Least OneData Insights: Two-Part · Q12

GMATGREDiagnostic Q1Algebra & Functions

Composite Functions

Functions nested inside other functions. The test checks whether you can work from the inside out, keep track of order, and respect domain rules along the way.

Where it appears. GMAT Problem Solving and GRE Problem Solving/QC. Also appears as “symbol functions” (x# = …, x◊y = …) that are just functions in disguise.

Core ideas

Inside out

In h(g(f(a))), evaluate f(a) first, feed that result into g, then feed that into h. The outermost function is always evaluated last.

Order matters

f(g(x)) and g(f(x)) are usually different. If f(x) = 2x and g(x) = x + 3, then f(g(1)) = 8 but g(f(1)) = 5.

Plug in expressions, not just numbers

If f(x) = x² − 1, then f(x + 2) = (x + 2)² − 1. Replace every x with the whole input, in brackets.

Domain rules

$\sqrt{x}$ needs x ≥ 0; a denominator can never be 0. If an intermediate value breaks a rule, the composite is undefined at that point.

Simplify radicals as you go

$\sqrt{12}$ = $\sqrt{4 \times 3}$ = 2 $\sqrt{3}$ . Clean intermediate values make the next function easy.

Toolkit

Composite

(f ∘ g)(x) = f(g(x))

Radical rule

\sqrt{a b}

\sqrt{a}

\sqrt{b}

Diagnostic Q1, step by step

$f (12) = \frac{\sqrt{12}}{2} = \frac{2 \sqrt{3}}{2} = \sqrt{3}$
$g (\sqrt{3}) =$ ( $\sqrt{3}$ )² + 9 = 3 + 9 = 12
h(12) = (12 + 1)² = 13² = 169
Answer: 169

Faster route. Write the chain as a pipeline before computing: 12 → f → g → h. Writing each output on a new line stops you from stopping early.

Traps the test sets

Stopping one function early. 13 is the value inside h before squaring.
Forgetting that ( $\sqrt{3}$ )² = 3, not 9.
Working outside-in: evaluating h first gives nonsense.

Practice

1f(x) = 2x − 3 and g(x) = x² + 1. Find g(f(2)) and f(g(2)).

f(2) = 1, so g(f(2)) = 2. g(2) = 5, so f(g(2)) = 7. Same inputs, different order, different answers.

2f(x) = 3x + k and f(f(1)) = 19. What is k?

f(1) = 3 + k. f(3 + k) = 3(3 + k) + k = 9 + 4k = 19, so k = 2.5.

3For x ≠ 0, x* is defined as

\frac{1}{x}

− 1. What is (2*)*?

2* =

\frac{1}{2}

− 1 = −

\frac{1}{2}

. (−

\frac{1}{2}

)* = −2 − 1 = −3.

TakeawayInside out, one line per function, and finish the outermost function before you pick an answer.

Lesson video

GMATGREDiagnostic Q2Counting & Probability

Probability Without Replacement

Drawing items one after another from a fixed group. Each draw changes what's left, so probabilities multiply with shrinking numbers.

Where it appears. GMAT and GRE Problem Solving; GRE QC; GMAT Data Insights probability tables.

Core ideas

Multiply along the sequence

P(first red and second red) = P(first red) × P(second red | first red).

Without replacement shrinks both parts

After one red is taken, there is one less red and one less marble: r/n, then (r − 1)/(n − 1).

Combinations give the same answer

P(2 red) = C(r,2) / C(n,2). Use whichever is faster; the order-based and combination-based methods always agree.

“One after the other” means without replacement

Unless the question says the item is returned, assume it is not.

Work backwards from a given probability

Set the product equal to the given value and solve, or test the answer choices.

Toolkit

Two draws, same colour

P =

\frac{r}{n}

\frac{r - 1}{n - 1}

Combination form

P =

\frac{C (r, 2)}{C (n, 2)}

Diagnostic Q2, step by step

Let r = red marbles out of 10. P(two red) = $\frac{r}{10}$ × $\frac{r - 1}{9}$ = $\frac{2}{15}$
r(r − 1) = 90 × 2/15 = 12
r = 4, because 4 × 3 = 12
Answer: 4

Faster route. Back-solve: try the choices in r(r − 1). 3 × 2 = 6, 4 × 3 = 12. Only 4 works.

Traps the test sets

Using $\frac{r}{10}$ × $\frac{r}{10}$ (with replacement) gives r² = 13.3, which isn't a choice.
Forgetting to subtract 1 from both the red count and the total on the second draw.

Practice

1A jar has 5 red and 3 blue marbles. Two are drawn without replacement. P(both blue)?

\frac{3}{8}

\frac{2}{7}

\frac{6}{56}

= $\frac{3}{28}$

2Same jar. P(one red and one blue)?

Two orders: 2 ×

\frac{5}{8}

\frac{3}{7}

\frac{30}{56}

= $\frac{15}{28}$ . Check: (5 × 3)/C(8,2) = 15/28.

3A group has exactly 3 women. The probability that two people chosen at random are both women is

\frac{1}{15}

. How many people are in the group?

3 × 2 / (n(n − 1)) = 1/15, so n(n − 1) = 90 and n = 10.

TakeawayTop and bottom both drop by one on each draw. Set up the product, then solve or back-solve.

Lesson video

GREDiagnostic Q3Arithmetic & Percents

Percents of Real Counts

When a percent is applied to people, animals or objects, the result must be a whole number. That hidden rule decides many Quantitative Comparison questions.

Where it appears. Mostly GRE Quantitative Comparison; also GMAT Data Sufficiency style reasoning.

Core ideas

Counts are integers

You can't have 2.5 brown dogs. If p% of N is a count, p% × N must be a whole number.

Turn the percent into a reduced fraction

10% = $\frac{1}{10}$ , 5% = $\frac{1}{20}$ , 35% = $\frac{7}{20}$ , 12.5% = $\frac{1}{8}$ . The denominator is the smallest group size that works.

Minimum totals

Add the smallest valid size of each group to get the least possible total.

QC strategy: test extremes

If Quantity A can only be larger than B at its minimum, A is greater. If it can be both larger and smaller, the answer is D.

Toolkit

Smallest valid N

p% =

\frac{a}{b}

in lowest terms ⇒ N is a multiple of b

Diagnostic Q3, step by step

10% = 1/10 ⇒ number of dogs is a multiple of 10 (minimum 10)
5% = 1/20 ⇒ number of cats is a multiple of 20 (minimum 20)
Smallest possible total = 10 + 20 = 30
30 > 25 in every case, so Quantity A is greater

Faster route. Picture it: one brown dog means at least 10 dogs; one orange cat means at least 20 cats.

Traps the test sets

Treating the number of animals as unknown and choosing D.
Using 5 cats (5% of 100?) instead of reducing 5% to 1/20.
Assuming the groups could be empty: the statement describes dogs and cats that are at the park.

Practice

135% of a company's employees are managers. What is the smallest possible number of employees?

35% = 7/20, so the total is a multiple of 20.

240% of the boys and 25% of the girls in a class wear glasses. What is the smallest possible class size (at least one boy and one girl)?

40% = 2/5 ⇒ boys multiple of 5. 25% = 1/4 ⇒ girls multiple of 4. Minimum 5 + 4 = 9.

312.5% of the books are hardcover and 30% of the magazines are glossy. Quantity A: books + magazines. Quantity B: 17.

Books multiple of 8, magazines multiple of 10 ⇒ at least 18 > 17. A is greater.

TakeawayReduce the percent to a fraction; its denominator is the smallest group that can exist.

Lesson video

GREGMATDiagnostic Q4Word Problems: Rates & Work

Rate, Time & Distance: Time Saved

Two trips over the same distance at different speeds. The difference in their times gives you an equation.

Where it appears. Both exams, Problem Solving. The same setup appears with “one hour late”, “arrive earlier” or “take longer”.

Core ideas

D = R × T

Time = Distance ÷ Rate. Write one time for the actual trip and one for the hypothetical trip.

The difference is the equation

Slower time − faster time = hours saved.

Product shortcut

For distance D, speed change Δr and time saved Δt: r(r + Δr) = D·Δr/Δt. Here 60 × 2 / 1 = 120 = r(r + 2).

Back-solve

Choices are speeds. Divide the distance by a choice and by that choice + 2, and check the difference.

Toolkit

Time

T =

\frac{D}{R}

Time saved

\frac{D}{r}

−

\frac{D}{r + Δ r}

= Δt

Diagnostic Q4, step by step

$\frac{60}{r}$ − $\frac{60}{r + 2}$ = 1
60(r + 2) − 60r = r(r + 2) ⇒ 120 = r² + 2r
r² + 2r − 120 = 0 ⇒ (r + 12)(r − 10) = 0
r = 10 mph. Check: 60/10 = 6 h, 60/12 = 5 h. 10 mph

Faster route. Back-solve with 10: 6 hours vs 5 hours, exactly one hour saved.

Traps the test sets

Subtracting the times in the wrong order and getting a negative speed.
Choosing 12 mph, which is the faster hypothetical speed.

Practice

1A car travels 180 miles. If it had gone 15 mph faster, the trip would have taken 1 hour less. What was its speed?

r(r + 15) = 180 × 15 = 2700 ⇒ r = 45 mph (4 h vs 3 h).

2Walking 12 km at 1 km/h slower than usual takes 1 hour longer. What is the usual speed?

r(r − 1) = 12 ⇒ r = 4 km/h (3 h vs 4 h).

3You drive to a town at 60 mph and return at 40 mph. What is your average speed?

Equal distances: 2 × 60 × 40 / (60 + 40) = 48 mph, not 50.

TakeawayWrite both times, subtract, then back-solve or factor.

Lesson video

GMATGREDiagnostic Q5Word Problems: Rates & Work

Work Rates & “Percent Faster”

Machines, pipes or people working together. Rates add; times do not.

Where it appears. GMAT Problem Solving (very common), GRE Problem Solving.

Core ideas

Rate = 1 / time

A machine that finishes in 5 hours does 1/5 of the job per hour.

Rates add

Working together: combined rate = rate A + rate B. Combined time = 1 ÷ combined rate.

“50% faster” describes rate

A's rate = 1.5 × B's rate, so A's time = B's time ÷ 1.5.

Two-machine shortcut

Together time = $\frac{a b}{a + b}$ when the solo times are a and b.

Toolkit

Combined

\frac{1}{a}

\frac{1}{b}

\frac{1}{T}

Two workers

T =

\frac{a b}{a + b}

Diagnostic Q5, step by step

Let B's rate = b. Then A's rate = 1.5b
Together: 1.5b + b = 2.5b = 1/4 job per hour ⇒ b = 1/10
A's rate = 1.5/10 = 3/20 ⇒ A's time = 20/3 ≈ 6.67 hours
Answer: approximately 7 hours

Faster route. Check with the shortcut: A = 20/3, B = 10 ⇒ (20/3 × 10)/(20/3 + 10) = 4 hours ✓.

Traps the test sets

Choosing 10 hours, which is B's time.
Treating “50% faster” as “takes 50% less time”.

Practice

1Pipe A fills a tank in 6 hours and pipe B in 3 hours. How long together?

6 × 3 / 9 = 2 hours.

2A is twice as fast as B. Together they finish in 6 hours. How long does A take alone?

2b + b = 1/6 ⇒ b = 1/18, A = 1/9 ⇒ 9 hours.

3A alone takes 10 hours; A and B together take 6 hours. How long does B take alone?

1/6 − 1/10 = 1/15 ⇒ 15 hours.

TakeawayConvert everything to rates, add them, and flip back to time at the end.

Lesson video

GREGMATDiagnostic Q6Arithmetic & Percents

Fractions of What Remains

A quantity is reduced in stages, each stage taking a fraction of whatever is left. Multiply what remains, never add what was used.

Where it appears. Both exams; also written as percents (“spends 20% of the rest”).

Core ideas

Track what's left

If 1/4 is used, 3/4 remains. Multiply the remaining fractions stage by stage.

“Of the remaining” vs “of the original”

Fractions of the remainder multiply; fractions of the original add.

Smart numbers

Start with the LCM of the denominators (here 60) to keep everything whole.

Toolkit

Remaining after stages

(1 − f₁)(1 − f₂)(1 − f₃)…

Diagnostic Q6, step by step

After hour 1: 1 × $\frac{3}{4}$ = $\frac{3}{4}$ remains
After hour 2: $\frac{3}{4}$ × $\frac{1}{3}$ = $\frac{1}{4}$ remains
After hour 3: $\frac{1}{4}$ × $\frac{2}{5}$ = $\frac{1}{10}$ remains
Answer: $\frac{1}{10}$

Faster route. Smart number: 60 mL. Drinks 15 ⇒ 45 left. Drinks 30 ⇒ 15 left. Drinks 9 ⇒ 6 left. 6/60 = 1/10.

Traps the test sets

Adding 1/4 + 2/3 + 3/5, which is more than 1.
Multiplying the fractions drunk (1/4 × 2/3 × 3/5) instead of the fractions left.

Practice

1Ada spends 1/3 of her salary on rent, 1/4 of the remainder on food, and saves 1/2 of what's left. What fraction is neither spent nor saved?

\frac{2}{3}

\frac{3}{4}

\frac{1}{2}

= $\frac{1}{4}$

2A tank loses 1/5 of its water, then 1/4 of what remains. 60 L are left. What is its capacity?

Remaining =

\frac{4}{5}

\frac{3}{4}

\frac{3}{5}

⇒ capacity = 100 L

3Same tank, but it loses 1/5 and then 1/4 of the original amount. What fraction remains?

1 −

\frac{1}{5}

−

\frac{1}{4}

= $\frac{11}{20}$

TakeawayWrite the remaining fraction for each stage and multiply.

Lesson video

GREDiagnostic Q7Arithmetic & Percents

Percent Symmetry & Percent Arithmetic

x% of y always equals y% of x. Knowing a few percent identities saves long calculations.

Where it appears. GRE Quantitative Comparison favourite; GMAT mental-math shortcut.

Core ideas

Symmetry

x% of y = $\frac{x}{100}$ × y = $\frac{x y}{100}$ = y% of x.

Mental-math use

8% of 25 is hard; 25% of 8 = 2 is easy.

Successive changes multiply

+20% then −20% ⇒ 1.2 × 0.8 = 0.96, a 4% net decrease.

Percent vs percentage points

A rate moving from 10% to 12% is a 2-point rise but a 20% increase.

Toolkit

Symmetry

x% of y = y% of x

Successive change

(1 + p)(1 + q) − 1

Diagnostic Q7, step by step

Quantity A = 14/100 × 9,778,443,000
Quantity B = 9,778,443,000/100 × 14
Both equal (14 × 9,778,443,000)/100
The two quantities are equal

Faster route. Spot the swap in the two quantities and answer in seconds without calculating.

Traps the test sets

Assuming the huge percent in B must make B bigger.
Spending two minutes on the arithmetic.

Practice

1What is 36% of 25?

25% of 36 = 9.

2Quantity A: 250% of 8. Quantity B: 8% of 250.

Both are 20. Equal.

3A price rises 20% and then falls 20%. Net change?

1.2 × 0.8 = 0.96 ⇒ 4% decrease.

TakeawayWhen two percents look like mirror images, they are equal.

Lesson video

GREDiagnostic Q8Statistics

Weighted Averages

Two groups with different averages combine into one. The combined average sits closer to the bigger group, and the distances tell you the ratio.

Where it appears. GRE Multiple Answer and QC; GMAT mixtures and Data Insights tables.

Core ideas

The equation

Average A × count A + average B × count B = combined average × total count.

The balance (lever) method

Distances from the combined average are inversely proportional to the group sizes.

Equivalent ratios

On multi-select questions every equivalent form (2:1, 4:2, 8:4) is correct.

Mixtures are weighted averages

Acid concentrations, prices per kg and test scores all use the same method.

Toolkit

Balance

\frac{n < s u b > A < / s u b >}{n < s u b > B < / s u b >}

\frac{B - a v g}{a v g - A}

Diagnostic Q8, step by step

100a + 160b = 120(a + b)
100a + 160b = 120a + 120b ⇒ 40b = 20a
a : b = 2 : 1
Equivalent choices: 2:1 and 8:4

Faster route. Lever: A is 20 below 120, B is 40 above. The ratio of counts is 40 : 20 = 2 : 1.

Traps the test sets

Picking 3:2 by reversing the ratio.
Selecting only 2:1 and missing 8:4 on a select-all question.

Practice

1Men average 180 lb and women 130 lb; the group averages 160 lb. Ratio of men to women?

Distances 20 and 30 ⇒ men : women = 30 : 20 = 3 : 2.

230 students average 70 and 20 students average 80. Overall average?

(2100 + 1600)/50 = 74.

3How much 10% acid and 40% acid make 60 L of 20% acid?

Distances 10 and 20 ⇒ 2 : 1 ⇒ 40 L of 10%, 20 L of 40%.

TakeawayMeasure each group's distance from the combined average; the counts are in the inverse ratio.

Lesson video

GREGMATDiagnostic Q9Counting & Probability

Combinations: Counting Pairs

Choosing groups where order doesn't matter. Round-robin games, handshakes and committees all count the same way.

Where it appears. Both exams; GRE QC often compares a combination with a permutation.

Core ideas

Does order matter?

Teams A vs B and B vs A are the same game ⇒ combination. President then VP ⇒ permutation.

Pairs formula

C(n, 2) = $\frac{n (n - 1)}{2}$

General formula

C(n, r) = $\frac{n!}{r! (n - r)!}$ ; P(n, r) = $\frac{n!}{(n - r)!}$

Home and away doubles it

If each pair plays twice, the count is n(n − 1).

Toolkit

Pairs

C(n,2) =

\frac{n (n - 1)}{2}

Diagonals of an n-gon

\frac{n (n - 3)}{2}

Diagnostic Q9, step by step

Each game is an unordered pair of teams
C(7, 2) = (7 × 6)/2
= 42/2 = 21
Answer: 21 games

Faster route. Team 1 plays 6 new opponents, team 2 plays 5 new, and so on: 6 + 5 + 4 + 3 + 2 + 1 = 21.

Traps the test sets

Choosing 42, which counts every game twice.
Choosing 14 (7 × 2).

Practice

110 people each shake hands with every other person once. How many handshakes?

C(10, 2) = 45.

28 teams each play every other team twice (home and away). How many games?

8 × 7 = 56.

3From 7 people, how many 3-person committees? How many ways to choose a president, VP and treasurer?

Committees: C(7, 3) = 35. Officers: 7 × 6 × 5 = 210.

TakeawayAsk whether swapping two members creates something new. If not, divide by the repeats.

Lesson video

GREGMATDiagnostic Q10Number Properties

Units Digits & Cyclicity

Huge powers have predictable last digits. Only the base's units digit and the exponent's remainder when divided by 4 matter.

Where it appears. GMAT Problem Solving and Data Sufficiency; GRE Problem Solving and QC.

Core ideas

Only the last digit of the base matters

33⁴³ ends the same way as 3⁴³.

Cycles

2: 2,4,8,6 · 3: 3,9,7,1 · 7: 7,9,3,1 · 8: 8,4,2,6 · 4: 4,6 · 9: 9,1 · 0, 1, 5, 6 never change.

Use the remainder mod 4

Divide the exponent by 4. Remainder 1, 2, 3 ⇒ 1st, 2nd, 3rd term; remainder 0 ⇒ 4th term.

Combine last digits only

For sums and products, add or multiply the units digits and keep the last digit.

Toolkit

Units digit of aⁿ

cycle of a's last digit, position = n mod 4 (0 ⇒ 4th)

Diagnostic Q10, step by step

Both bases end in 3; the cycle is 3, 9, 7, 1
33⁴³: 43 ÷ 4 remainder 3 ⇒ units digit 7
43³³: 33 ÷ 4 remainder 1 ⇒ units digit 3
7 + 3 = 10 ⇒ units digit 0

Faster route. List the first four powers of 3 (3, 9, 27, 81) to rebuild the cycle if you forget it.

Traps the test sets

Adding the base digits (3 + 3 = 6) and ignoring the exponents.
Mis-handling remainder 0: it points to the last term of the cycle, not the first.

Practice

1Units digit of 7¹⁰³?

103 ÷ 4 remainder 3 ⇒ third term of 7, 9, 3, 1 ⇒ 3.

2Units digit of 2⁵⁰ + 5⁵⁰?

50 ÷ 4 remainder 2 ⇒ 4; 5 stays 5 ⇒ 4 + 5 = 9.

3Units digit of 12¹² × 13¹³?

2¹²: remainder 0 ⇒ 6. 3¹³: remainder 1 ⇒ 3. 6 × 3 = 18 ⇒ 8.

TakeawayLast digit of the base, exponent mod 4, then combine last digits.

Lesson video

GMATDiagnostic Q11Data Insights: Two-Part

Two-Part Analysis: Logical Chains

GMAT Two-Part Analysis asks for one selection per column that together satisfy a condition. Ordering questions reward building a chain through a shared link.

Where it appears. GMAT Focus Data Insights (Two-Part Analysis). The chain logic also helps with GRE logic-based reading.

Core ideas

Transitivity

If C > B and B > A, then C > A. You need a middle term that both statements share.

Column constraints

Each column restricts which statements are allowed. The A-column needs a statement that mentions A.

Direction matters

“A is older than B” and “B is older than A” are opposites. Write each as an arrow.

Eliminate dead ends

A statement whose other artifact never links to the target can't complete the chain.

Toolkit

Chain

C > X and X > A ⇒ C > A

Diagnostic Q11, step by step

Target: C > A (C is older than A)
C-column options: C > D or C > B
A-column options with a link: B > A (D never appears older than A)
Chain C > B > A ⇒ select B older than A and C older than B

Faster route. Draw arrows from older to younger for every option. The only path from C to A goes through B.

Traps the test sets

Picking “A is older than B”, which points the wrong way.
Choosing C > D, since no option places D older than A.

Practice

1You want “P is taller than R”. Options: Q taller than R · R taller than Q · P taller than Q · Q taller than P. Pick one P-statement and one R-statement.

P taller than Q and Q taller than R ⇒ P > Q > R.

2You want “X arrived before Z”. Options: X before Y · Y before X · Y before Z · Z before Y.

X before Y and Y before Z.

3Do “A is older than B” and “C is older than B” tell you whether A is older than C?

No. Both point at B; there is no chain between A and C.

TakeawayDraw arrows, find the shared middle term, and check the direction.

Lesson video

GMATGREDiagnostic Q12Data Insights: Two-Part

Independent Events: Both, Neither, At Least One

When two events don't influence each other, their probabilities multiply. Everything else follows from that and from complements.

Where it appears. GMAT Data Insights and Problem Solving; GRE Problem Solving and QC.

Core ideas

Independent ⇒ multiply

P(A and B) = P(A) × P(B).

Neither

P(neither) = (1 − P(A)) × (1 − P(B)).

At least one

P(at least one) = 1 − P(neither) = P(A) + P(B) − P(A)P(B).

Exactly one

P(exactly one) = P(A) + P(B) − 2P(A)P(B).

Independent ≠ mutually exclusive

Mutually exclusive events can't happen together; independent events happen together with probability P(A)P(B).

Toolkit

Both

P(A)·P(B)

Neither

(1 − P(A))(1 − P(B))

At least one

1 − neither

Diagnostic Q12, step by step

Both = 0.35 × 0.2 = 0.07
No swan = 0.65, no heron = 0.8
Neither = 0.65 × 0.8 = 0.52
Check: at least one = 0.35 + 0.2 − 0.07 = 0.48 = 1 − 0.52 ✓. Select 0.07 and 0.52

Faster route. Draw a 1 × 1 square split 0.35 / 0.65 one way and 0.2 / 0.8 the other. Each rectangle's area is a probability.

Traps the test sets

Choosing 0.48 for “Neither”; that's the probability of at least one.
Adding 0.35 + 0.2 for “Both”.

Practice

1P(rain) = 0.3 and P(traffic) = 0.4, independent. P(exactly one)?

0.3 + 0.4 − 2(0.12) = 0.46.

2Two archers hit the target with probabilities 0.8 and 0.6. P(the target is hit at least once)?

1 − 0.2 × 0.4 = 0.92.

3Can two events with non-zero probabilities be both mutually exclusive and independent?

No. Mutually exclusive means P(A and B) = 0, but independence needs P(A)P(B) > 0.

TakeawayMultiply for both, multiply complements for neither, subtract from 1 for at least one.

Lesson video