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Use complements, overlap and conditional denominators while checking independence.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
Compute exact probabilities from explicit models, distinguishing conditional dependence from independent events.
An event is a set of possible outcomes. A union means at least one of two events occurs; an intersection means both occur; a complement means an event does not occur. In probability, or is normally inclusive unless stated otherwise. A Venn diagram can show overlap, but the regions must represent the stated counts. Fractions, decimals and percentages are alternative ways to express probabilities. An exact fraction often preserves information more cleanly than a rounded decimal.
| Term | What it means |
|---|---|
| Independent | Learning that one event occurred does not change the probability of the other. |
| Mutually exclusive | Two events cannot occur together. |
| Conditional probability | Probability within the outcomes consistent with specified information. |
| Complement | All outcomes in the sample space that are outside the event. |
A probability is a number from zero to one assigned to an event within a stated chance model. When the elementary outcomes are equally likely, it is the number of favorable outcomes divided by the total number of outcomes. Equal likelihood is a condition to justify, not a default for every list of possibilities. The possible sums of two dice are not equally likely, even though the thirty-six ordered pairs of fair die results are.
Before calculating, describe what is selected, how selection occurs and what information is already known. Drawing uniformly from all participants differs from drawing uniformly from the participants who meet a condition. Drawing a counter without replacement changes the population available on the next draw. These distinctions decide the denominator and whether probabilities can be multiplied directly.
Another way: steps
Specify the chance model; identify the event and any condition; choose direct counting, a complement or conditional multiplication; simplify and check bounds.
For two fair six-sided dice, distinguish the ordered results (1, 2) and (2, 1) because they arise from different die outcomes. The sum three has two such pairs, while the sum two has only (1, 1). Counting sums as if each were equally likely would therefore be wrong. When the model chooses a person uniformly from a group, each person is an elementary outcome, not each category label. A group with nine engineers and one designer does not make engineer and designer equally likely categories.
The complement of at least one success is no successes. For three independent attempts each with success probability one fifth, no successes has probability (4/5) cubed, so at least one success has probability 1 - (4/5) cubed. This avoids separately counting one, two and three successes. Check the wording: the complement of exactly one success is not no successes, because two or more also belong to that complement. Write the event in ordinary words before selecting a shortcut.
For events A and B, P(A or B) = P(A) + P(B) - P(A and B). The intersection was included once in each individual probability, so subtract one copy. If the events are mutually exclusive, the intersection is zero and simple addition is valid. If a question gives counts of members in two clubs and in both, apply the same formula to counts before dividing by the population. A resulting union count larger than the total group is a signal that overlap or interpretation was mishandled.
P(A given B) compares the outcomes in both A and B with the outcomes in B. The denominator is B's probability or count, not the full population. In a survey of one hundred people, suppose thirty use bicycles and twelve of those also use trains. Among cyclists the train-use probability is twelve thirtieths, not twelve hundredths. The latter is the joint probability of both habits in the whole group. Reversing the condition generally changes the answer: the proportion of cyclists among train users needs the total number of train users.
Independent events satisfy P(A and B) = P(A)P(B). Mutually exclusive events have intersection probability zero. If both events have positive probability and are mutually exclusive, observing one makes the other impossible, so they are not independent. Conversely, independent events can occur together. A pair of separate fair coin tosses is a standard independent model. Two membership categories are not automatically independent just because their labels differ; the question must state independence or supply information establishing it.
A bag with four red and three blue counters gives first-red probability four sevenths. After a red is removed, second-red probability is three sixths, so two reds has probability four sevenths times three sixths. With replacement and a fresh uniform draw, the second probability returns to four sevenths. The multiplication rule always uses the appropriate conditional probability for the second event. Independence is the special case in which that conditional value equals the unconditional one. Keep a small count table after each draw to avoid retaining an unavailable counter. A probability tree can organize more than one successful route: multiply conditional probabilities along each route and add probabilities of disjoint routes. For exactly one red in two draws, red then blue and blue then red are separate routes. Include both, update the counts on each, and avoid adding a route twice. The total probability of all complete, mutually exclusive routes should be exactly one; this checks whether the tree omitted or duplicated an outcome.
A probability cannot be negative or exceed one. An intersection cannot exceed either event's probability, and a union cannot be smaller than either event's probability. A conditional probability can exceed or fall below an unconditional probability because its reference group differs. These bounds do not prove an answer correct, but they quickly reject many arithmetic mistakes. For a fraction response, simplify when convenient, but mathematically equivalent fractions represent the same probability. Do not round an exact rational result unless the question explicitly asks for rounding.
A workshop selects one engineer to demonstrate equipment. If twenty attendees are engineers and eight of them completed a safety course, selection from engineers makes the relevant probability eight twentieths. The total number of attendees is irrelevant to that conditional denominator. A statement about all attendees would be a different probability and would require the number of course completers outside the engineering group as well.
This distinction matters when reading percentages in reports. A high proportion of a small group can coexist with a low proportion in the full population, and reversing a conditional statement can change its meaning substantially. Translate given that into an explicit restriction on who or what remains eligible.
The practice uses an exact numeric input that accepts fractions written with a slash, such as 2/5. This is our shared learning interface; the official test may present separate fraction boxes. Equivalent fractions are mathematically interchangeable, but premature decimal rounding can lose exactness. Record whether an error came from the sample space, the condition, dependence or arithmetic before retrying a different question.
Two events can overlap without being independent. Mutually exclusive events with positive probabilities are dependent. Without replacement changes counts. The denominator of a conditional probability is the conditioning group. Exactly one and at least one are different events. A list of outcome categories is not necessarily an equally likely sample space.
Identify equally likely outcomes.
A box contains 5 striped and 7 plain cards; choose one uniformly.
Each of the twelve cards has the same chance.
Define the requested event.
The selected card is striped.
Five elementary outcomes qualify.
Form the favorable-to-total ratio.
Probability = 5/12.
The denominator includes all cards, not just the plain ones.
Check the result against the counts.
The probability is below one half.
Striped cards are fewer than plain cards.
State the desired event.
A fair die is rolled three times independently; find the chance of at least one 1.
One, two or three occurrences all qualify.
Describe its complement.
No roll shows 1.
This is one event covering all failures.
Compute the complement probability.
(5/6) cubed = 125/216.
Independence permits multiplying the three failure probabilities.
Subtract from the whole sample space.
1 - 125/216 = 91/216.
The event and its complement are disjoint and exhaustive.
Check the event wording.
This is not the probability of exactly one 1.
The answer includes multiple successes, as at least one requires.
Read the population counts.
Forty people use app A; twenty-five use app B; fifteen use both.
A person is selected uniformly from app B users.
Identify the conditioning event.
The selected person uses B.
Only the twenty-five B users remain eligible.
Identify qualifying eligible outcomes.
Fifteen B users also use A.
The intersection supplies the numerator.
Form the conditional probability.
15/25 = 3/5.
The denominator is the eligible group, not all A users.
Distinguish the reverse conditional.
Among A users, B use would be 15/40 = 3/8.
Changing the condition changes the reference population.
Check the claim's scope.
Three fifths describes A use among B users only.
It does not establish the probability of A use in the entire population.
Track the first selection.
A bag has 6 black and 4 white counters; first-black probability is 6/10.
The first selected counter is not replaced.
Update the second selection.
After a black draw, five black counters remain among nine.
Multiply the first probability by the conditional second probability.
Simplify and interpret the result.
A bag contains 3 red and 5 blue counters. One counter is drawn uniformly at random. What is the probability of red? Enter an exact fraction.
A fair six-sided die is rolled once. What is the probability that its result is greater than 4? Enter an exact fraction.
A fair coin is tossed three times independently. What is the probability of three heads? Enter an exact fraction.
A fair six-sided die is rolled twice independently. What is the probability of at least one 6? Enter an exact fraction.
A bag holds 4 green and 6 white counters. Two are drawn uniformly without replacement. What is the probability both are green? Enter an exact fraction.
Among 30 students, 18 study French, 14 study music and 8 study both. A student is selected uniformly. What is the probability the student studies French or music? Enter an exact fraction.
In a group, 12 members cycle to work, including 5 who also use a train. A member is selected uniformly from those who cycle. What is the probability that member also uses a train? Enter an exact fraction.
Events A and B are independent with probabilities 0.4 and 0.25. What is the probability that both occur? Enter an exact decimal.
Events C and D are mutually exclusive with probabilities 0.2 and 0.35. What is the probability that neither occurs? Enter an exact decimal.
A fair six-sided die is rolled. Given that the result is even, what is the probability it exceeds 3? Enter an exact fraction.
A workshop has 50 attendees. Of the 20 attendees who are engineers, 8 have completed a safety course. One engineer is chosen uniformly to demonstrate equipment. What is the probability the chosen person completed the course? Enter an exact fraction.
Answer three reserved questions without hints. These cases are separate from lesson practice.
Two independent events each have probability 0.3. What is the probability at least one occurs? Enter an exact decimal.
Of 40 participants, 25 read journal A, 20 read journal B and 10 read both. A participant is chosen uniformly. What is the probability the participant reads neither? Enter an exact fraction.
A bag holds 5 black and 3 orange counters. Two are drawn uniformly without replacement. What is the probability both are orange? Enter an exact fraction.
Tomorrow, explain why mutually exclusive positive-probability events cannot be independent. Then solve a without-replacement draw by multiplying conditional probabilities and an at-least-one question by its complement.
17. Complete a draw without replacement, step 3
(6/10)(5/9) = 1/3.
This is the probability of two black draws, not of at least one black draw.