GRE · Data Analysis · TOPIC LESSON
Define the experiment and sample space before choosing addition, multiplication or a conditional denominator.
BY THE END, YOU CAN
BEFORE YOU START
Original educational lesson. Work through the examples before revealing the practice answers. This original lesson supports current GRE General Test study; it does not reproduce official questions, certify a writing score or predict test content.
LEARN THROUGH ORIGINAL PRACTICE
All teaching passages, writing prompts, examples and exercises in this lesson are original. No official ETS questions, prompts or explanations are reproduced. Current structure and observed public sample metadata are separate from this original teaching; no official score or question probability is implied.
01 / UNDERSTAND & APPLY
An outcome is one possible result of an experiment; an event is a collection of outcomes. For a finite sample space of equally likely outcomes, P(event)=favorable outcomes/total outcomes. Counting alone does not justify equal likelihood: two categories can contain different numbers of equally likely individual outcomes.
Probabilities range from 0 to 1. Probability 0 describes an impossible event in a finite equally likely experiment; probability 1 means certainty under its model. A model probability does not guarantee exact frequencies in a short run.
02 / UNDERSTAND & APPLY
The complement of A is every outcome not in A, so P(not A)=1−P(A). “At least one” is often easiest as 1−P(none). “At most one” includes zero and one; “exactly one” excludes both zero and two or more.
For repeated independent trials with success probability p, the probability of no success in n trials is (1−p)ⁿ. This product requires independence and the same p on every trial; without replacement, probabilities can change.
ORIGINAL WORKED EXAMPLE
Three independent components each fail with probability 0.1. What is the probability at least one fails?
Answer 0.271
03 / UNDERSTAND & APPLY
For any events A and B, P(A or B)=P(A)+P(B)−P(A and B). The union “or” normally includes outcomes in both events. Subtract the overlap because it was counted twice in the sum. If the events are mutually exclusive, their overlap probability is zero and simple addition applies.
Mutually exclusive means the events cannot occur together on the same trial. It does not mean independent. Two positive-probability mutually exclusive events are dependent: knowing one occurred makes the other impossible.
ORIGINAL WORKED EXAMPLE
A card is selected uniformly from cards numbered 1 through 20. Find the probability its number is divisible by 3 or 4.
Answer 1/2
04 / UNDERSTAND & APPLY
P(A given B), written P(A|B), uses only outcomes in B as its new sample space. For P(B)>0, P(A|B)=P(A and B)/P(B). In a count table, divide the overlap count by the count in the conditioning group.
P(A|B) and P(B|A) generally differ because they use different denominators. “Among people who…” identifies the denominator. If B has probability zero, this elementary ratio is undefined.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
ORIGINAL WORKED EXAMPLE
Among 50 club members, 30 cycle, 20 swim, and 12 do both. A member is selected uniformly from the swimmers. What is the probability that member cycles?
Answer 3/5
05 / UNDERSTAND & APPLY
For P(A)>0, P(A and B)=P(A)P(B|A). If A and B are independent, knowing A does not change B’s probability, and the product becomes P(A)P(B). Independence must be stated, justified by the experiment, or verified from probabilities; different labels alone do not establish it.
Independence can be checked using P(A and B)=P(A)P(B). Mutually exclusive positive-probability events cannot satisfy this equality, since their intersection is zero while the product is positive.
ORIGINAL WORKED EXAMPLE
Of 40 students, 20 study French, 16 study music, and 8 study both. For a uniformly chosen student, are French and music independent?
Answer Yes
06 / UNDERSTAND & APPLY
With replacement and a fresh uniform draw, the composition is restored, making successive draws independent under the usual model. Without replacement, the denominator drops and the relevant numerator may drop too. Track the type removed before computing the next probability.
For an ordered sequence, multiply along each path; for alternative disjoint paths, add their probabilities. A tree diagram labels conditional probabilities on branches, not repeated unconditional probabilities after the sample space changes.
ORIGINAL WORKED EXAMPLE
A bag holds 5 red and 3 blue beads. Two are drawn uniformly without replacement. Find the probability of exactly one red.
Answer 15/28
07 / UNDERSTAND & APPLY
Knowing P(A) and P(B) alone does not usually determine P(A and B). The intersection is at most the smaller marginal probability and at least max(0,P(A)+P(B)−1). These bounds come from set inclusion and the fact that a union cannot exceed probability 1.
For QC, never silently assume independence to fill a missing intersection. Construct compatible overlaps giving different comparisons. More data or an independence condition may be necessary.
08 / UNDERSTAND & APPLY
All examples and practice in this lesson are original. For quantitative comparison (QC), use A if Quantity A is greater, B if Quantity B is greater, C if they are equal, and D if the relationship cannot be determined from the information given. A, B or C must hold for every permitted case. To establish D, give two allowed cases producing different relationships; an inability to calculate is not itself evidence for D.
A single-answer question requires one choice. For multiple-answer questions, obey the stated selection count or select every correct option if instructed. Test each option separately; do not stop after finding one. Multiple-answer questions receive credit only when the complete correct set is selected, with no incorrect choices. Numeric entry requires the requested value, with units handled in the calculation; round only when the question directs you to. A single numeric-entry box accepts an integer or decimal; a pair of fraction boxes accepts an integer numerator and a nonzero integer denominator. Fractions need not be reduced, but reducing makes checks easier. Do not enter a percent sign when the requested number is a percentage. Preserve exact fractions until the last step.
A calculator can check arithmetic but cannot decide the domain, interpretation or denominator for you. Estimate sign and size first, enter parentheses carefully, and compare the result with the original conditions. Cancel factors or exploit structure when doing so is faster and more reliable than decimal entry.
09 / UNDERSTAND & APPLY
Unless restricted otherwise, each stated individual object is selected uniformly.
ORIGINAL PRACTICE · 1
2/5
There are 4 favorable individual tokens among 10 equally likely tokens. Two colors do not imply probability 1/2 for each color.
ORIGINAL PRACTICE · 2
B
Complement probabilities sum to 1, so 1−0.37=0.63. No independence assumption is needed.
ORIGINAL PRACTICE · 3
0.28
Independence permits multiplication: 0.4·0.7=0.28. Adding would relate to a union and still require overlap correction.
ORIGINAL PRACTICE · 4
A, C and D
No overlap gives A and the union formula C. Conditioning on B makes A impossible, giving D. Their positive marginal product means independence fails. Their disjoint probabilities cannot sum above 1.
ORIGINAL PRACTICE · 5
3/8
The condition restricts selection to 24 runners, and 9 qualify. The denominator is 24 regardless of how many nonrunners exist.
ORIGINAL PRACTICE · 6
1/15
P(first black)=2/6; after it, P(second black)=1/5. Multiply to get 2/30=1/15. Using 2/6 twice would assume replacement.
ORIGINAL PRACTICE · 7
D
If A=B, the intersection is 1/2, above 1/4. If A and B are complementary, it is 0, below 1/4. Both fit the marginals, so independence cannot be assumed.
ORIGINAL PRACTICE · 8
3/4
The complement is TT, with probability (1/2)(1/2)=1/4. Subtract from 1. Simply adding the two head probabilities would double-count HH.
KEEP THE BIG IDEAS
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