Probability Without Replacement Flashcards: Draw-by-Draw Practice
Practice probability without replacement with 43 solved cards on changing counts, ordered and either-order draws, complements, and hidden versus observed results.
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Practice probability without replacement through 43 short cards about finite collections of counters, tickets and other simple objects. Track what remains after observed draws, calculate a specified colour order, include every permitted order, and use complements for at-least or not-both events. Each prompt states its own counts, selection rule and available information; answers start with the result and explain the key step.
The cards also distinguish an unseen removal from an observed result, calculate one probability after a truthful checker reports that at least one selected counter is red, and compare two reds with return and remixing versus without replacement. Error-repair prompts target unchanged denominators, incorrect remaining counts, missing orders and multiplying by an already known event. Most calculations use two draws, with several three-draw examples and one four-draw exact-count example.
The sequence begins with remaining counts, then builds ordered products, either-order and exact-count events, complements and information-sensitive decisions. Related variants are separated by at least three other cards. Every card uses the tag probability-without-replacement.
This focused deck assumes basic fraction arithmetic. It omits automatic reverse cards because a probability does not uniquely identify its original collection or event. It also excludes general definition drills, static two-way tables, hidden-source Bayes problems, distribution-formula memorization, large combination calculations, card-suit knowledge and examination preparation. It is a practice set for decisions within short problems, rather than a full probability course.
Questions, answers, organization and metadata were independently written with AI assistance. Mathematical facts were checked against Penn State conditional probability, its sampling-without-replacement model, and MIT conditional and total probability. No source exercises, wording or diagrams were copied. The cover is an original AI-generated illustration; the cards contain no images.
CC0 applies to the original expression and generated cover to the extent applicable rights exist. It does not claim ownership of mathematical facts or third-party material. This independent resource is not affiliated with or endorsed by Penn State, MIT or any examination provider.
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Kaart 1
Küsimus
A bag holds 5 red and 4 blue counters. One uniform draw without replacement is observed to be red. What remains?
Vastus
4 red and 4 blue counters: 8 in total.
Only the red count and the total decrease.
Kaart 2
Küsimus
A bag holds 3 green and 5 yellow counters. A yellow is removed. On the next uniform draw without replacement, what is the probability of green?
Vastus
3/7.
All 3 green counters remain among 7 counters.
Kaart 3
Küsimus
A box starts with 11 distinct tickets. Four are drawn uniformly without replacement. How many tickets are available for the fifth draw?
Vastus
7 tickets.
Four tickets have left the box, so the next denominator is 11 − 4.
Kaart 4
Küsimus
A bag holds 2 silver and 6 gold counters. Both silver counters have been removed. What is the probability of silver on the next uniform draw without replacement?
Vastus
0
No silver counters remain; all 6 remaining counters are gold.
Kaart 5
Küsimus
A bag holds 6 red and 5 blue counters. A red is removed. What is the probability of another red on the next uniform draw without replacement?
Vastus
5/10 = 1/2.
The same physical counter cannot return, but another counter of the same colour can be drawn.
Kaart 6
Küsimus
A bag starts with 6 green, 3 yellow and 2 white counters. Two uniform draws without replacement are observed to be green then yellow. What remains?
Vastus
5 green, 2 yellow and 2 white counters: 9 in total.
Update each colour actually removed.
Kaart 7
Küsimus
A bag has 8 counters. After one is removed, a learner uses 8 as the denominator for the next uniform draw without replacement. What denominator should they use?
Vastus
7
The next draw is from the 7 counters still in the bag.
Kaart 8
Küsimus
A bag has 7 red and 3 blue counters. Two red counters are removed. What is the probability of red on the next uniform draw without replacement?
Vastus
5/8.
There are 5 red counters left among 8 counters.
Kaart 9
Küsimus
A bag has 4 red and 5 blue counters. Draw two uniformly without replacement. What is the probability of red then red?
Vastus
1/6.
(4/9) × (3/8) = 12/72 = 1/6. The second red count is 3.
Kaart 10
Küsimus
A bag has 7 orange and 4 purple counters. Draw two uniformly without replacement. What is the probability of orange then purple?
Vastus
14/55.
(7/11) × (4/10) = 28/110 = 14/55. Removing orange leaves all 4 purple counters.
Kaart 11
Küsimus
A bag has 5 green and 4 yellow counters. Draw three uniformly without replacement. What is the probability of green, yellow, green in that order?
Vastus
10/63.
(5/9) × (4/8) × (4/7) = 80/504 = 10/63.
Kaart 12
Küsimus
Seven tickets, each labelled with a different letter, include A and B. Draw two uniformly without replacement. What is the probability of A first and B second?
Vastus
1/42.
(1/7) × (1/6) = 1/42. Only one ticket matches each specified letter.
Kaart 13
Küsimus
A bag has 4 red and 7 blue counters. For two uniform draws without replacement, a learner gives (4/11)² for two reds. Repair the calculation.
Vastus
(4/11) × (3/10) = 6/55.
After a red is drawn, only 3 reds remain among 10 counters.
Kaart 14
Küsimus
A bag has 6 red and 2 blue counters. Draw three uniformly without replacement. What is the probability of red, blue, red in that order?
Vastus
5/28.
(6/8) × (2/7) × (5/6) = 60/336 = 5/28.
Kaart 15
Küsimus
A bag starts with 6 red and 5 blue counters. The first draw is known to be red and is not returned. What is the probability that the next two uniform draws without replacement are blue then red?
Vastus
5/18.
The known red leaves 5 red and 5 blue. For the next two draws, (5/10) × (5/9) = 5/18. Do not multiply by the probability of the already known first red.
Kaart 16
Küsimus
A bag has 5 red and 3 blue counters. Draw two uniformly without replacement. What is the probability of one of each colour, in either order?
Vastus
15/28.
(5/8) × (3/7) + (3/8) × (5/7) = 30/56 = 15/28.
Kaart 17
Küsimus
A bag has 5 green and 2 yellow counters. Draw two uniformly without replacement. What is the probability that both have the same colour?
Vastus
11/21.
Both green or both yellow: (5/7) × (4/6) + (2/7) × (1/6) = 22/42 = 11/21.
Kaart 18
Küsimus
A bag has 4 red and 5 blue counters. Draw three uniformly without replacement. What is the probability of exactly two reds?
Vastus
5/14.
RRB, RBR and BRR each have probability (4 × 3 × 5)/(9 × 8 × 7) = 5/42. Add the three disjoint orders: 3 × 5/42 = 5/14.
Kaart 19
Küsimus
A bag has 3 green and 6 yellow counters. Draw two uniformly without replacement. What is the probability of at least one green?
Vastus
7/12.
Subtract the chance of no green: 1 − (6/9) × (5/8) = 1 − 5/12 = 7/12.
Kaart 20
Küsimus
A bag has 6 red and 2 blue counters. Two counters are selected together as a uniform pair without replacement. What is the probability of one red and one blue?
Vastus
3/7.
Choosing together is without replacement. Count both sequential orders: 2 × (6/8) × (2/7) = 24/56 = 3/7.
Kaart 21
Küsimus
A box has 3 faulty and 7 working bulbs. Choose three uniformly without replacement. What is the probability that none is faulty?
Vastus
7/24.
All three must work: (7/10) × (6/9) × (5/8) = 210/720 = 7/24.
Kaart 22
Küsimus
A bag has 2 red and 6 blue counters. Draw three uniformly without replacement. What is the probability of at most one red?
Vastus
25/28.
Zero reds: (6/8) × (5/7) × (4/6) = 5/14. Exactly one red: 3 × (2 × 6 × 5)/(8 × 7 × 6) = 15/28. Add them: 10/28 + 15/28 = 25/28.
Kaart 23
Küsimus
A bag has 3 red, 2 blue and 1 green counter. Draw two uniformly without replacement. What is the probability of different colours?
Vastus
11/15.
Subtract the same-colour cases: 1 − [(3/6) × (2/5) + (2/6) × (1/5)] = 1 − 8/30 = 11/15. Two greens are impossible.
Kaart 24
Küsimus
A box has 2 faulty and 6 working sensors. Choose four uniformly without replacement. What is the probability of exactly one faulty sensor?
Vastus
4/7.
There are 4 positions for the faulty sensor. Each order has probability (2 × 6 × 5 × 4)/(8 × 7 × 6 × 5) = 1/7. Add the 4 orders.
Kaart 25
Küsimus
Nine tickets include 2 prize tickets. Draw three uniformly without replacement. What is the probability of at least one prize?
Vastus
7/12.
Subtract three non-prizes: 1 − (7/9) × (6/8) × (5/7) = 1 − 5/12 = 7/12.
Kaart 26
Küsimus
A bag has 5 red and 6 blue counters. For one of each in two uniform draws without replacement, a learner uses only (5/11) × (6/10). What is missing?
Vastus
The blue-then-red path.
Add (6/11) × (5/10). The total is 60/110 = 6/11; the learner counted only red then blue.
Kaart 27
Küsimus
A bag has 3 red, 2 blue and 2 green counters. Draw three uniformly without replacement. What is the probability of one of each colour, in any order?
Vastus
12/35.
There are 6 colour orders. Each has probability (3 × 2 × 2)/(7 × 6 × 5) = 2/35, so the total is 6 × 2/35 = 12/35.
Kaart 28
Küsimus
A bag has 3 green and 5 yellow counters. Draw three uniformly without replacement. What is the probability of at least two greens?
Vastus
2/7.
Exactly two greens: 3 × (3 × 2 × 5)/(8 × 7 × 6) = 15/56. Three greens: (3 × 2 × 1)/(8 × 7 × 6) = 1/56. Total: 16/56 = 2/7.
Kaart 29
Küsimus
A bag starts with 5 red and 4 blue counters. One blue has been drawn and not returned. What is the probability that the next two uniform draws without replacement have the same colour?
Vastus
13/28.
The remaining bag has 5 red and 3 blue. Add (5/8) × (4/7) and (3/8) × (2/7): 26/56 = 13/28.
Kaart 30
Küsimus
A bag has 4 red and 5 blue counters. The first uniform draw is removed unseen. What is the probability that the second uniform draw is red? No selected item is returned.
Vastus
4/9.
Before any colour is observed, each original counter is equally likely to occupy the second position. Removing an unseen counter does not tell you which colour count fell.
Kaart 31
Küsimus
A bag starts with 7 red and 4 blue counters. Draw three counters uniformly without replacement; no counter is returned. You know the first two draws contained one of each colour, but not their order. What is the probability of red on the third draw?
Vastus
2/3.
Either order leaves exactly 6 red and 3 blue counters. The next probability is 6/9 = 2/3.
Kaart 32
Küsimus
A bag has 6 red and 3 blue counters. A blue is removed. A learner gives 5/8 for red on the next uniform draw. Repair the probability. No selected item is returned.
Vastus
6/8 = 3/4.
The total falls to 8, but the red count stays at 6 because the removed counter was blue.
Kaart 33
Küsimus
A box has 1 gold and 8 silver tokens. Draw three uniformly without replacement. What is the probability of at least two gold tokens?
Vastus
0
Only one gold token exists, so no draw order can contain two gold tokens.
Kaart 34
Küsimus
A bag has 5 green and 7 yellow counters. Two uniform draws are removed unseen. What is the probability that the third uniform draw is green? No selected item is returned.
Vastus
5/12.
With no observed colours, every draw position has the original green proportion. The remaining green count is unknown, so 5/10 would be unjustified.
Kaart 35
Küsimus
A bag starts with 5 red and 6 blue counters. A red has already been observed and removed. A learner uses (5/11) × (4/10) for red on the next uniform draw without replacement. Repair the answer.
Vastus
4/10 = 2/5.
The question is conditional on the known first red. Multiplying by 5/11 instead finds the probability of two reds before either draw is observed.
Kaart 36
Küsimus
A bag has 4 red and 5 blue counters. The first uniform draw is observed to be blue and is removed. What is the probability that the second uniform draw is red? No selected item is returned.
Vastus
1/2.
All 4 reds remain among 8 counters: 4/8 = 1/2. The observed blue changes the information available.
Kaart 37
Küsimus
A bag has 5 red and 2 blue counters. Draw two uniformly without replacement. What is the probability that they are not both red?
Vastus
11/21.
1 − (5/7) × (4/6) = 1 − 10/21 = 11/21. This includes mixed colours and two blues; it does not mean neither is red.
Kaart 38
Küsimus
A bag has 3 red and 5 blue counters. One uniformly selected counter is removed unseen. Use the two possible first colours to find the probability that the next uniformly selected counter is red. No selected item is returned.
Vastus
3/8.
(3/8) × (2/7) + (5/8) × (3/7) = 21/56 = 3/8. Weight each possible remaining state by its probability.
Kaart 39
Küsimus
Twelve tickets include 5 prize tickets. Two non-prizes are observed and removed. What is the probability that the next two uniform draws without replacement are both prizes?
Vastus
2/9.
There are 5 prizes among 10 tickets. Then (5/10) × (4/9) = 20/90 = 2/9.
Kaart 40
Küsimus
A bag has 5 red and 3 blue counters. For two reds in uniform draws without replacement, a learner doubles (5/8) × (4/7) to allow either order. Is that valid?
Vastus
No. The probability is 5/14.
(5/8) × (4/7) = 20/56 = 5/14 already counts red then red. Swapping the two colour positions creates no new colour path.
Kaart 41
Küsimus
A bag starts with 5 green and 7 yellow counters. Two greens are observed and removed. What is the probability of green on the next uniform draw? No selected item is returned.
Vastus
3/10.
There are 3 greens among 10 remaining counters. Unlike unseen removals, observed greens let you update the count exactly.
Kaart 42
Küsimus
A bag has 2 red and 5 blue counters. Two are drawn uniformly without replacement. A truthful checker always reports whether at least one is red; the report is yes. What is the probability that both are red?
Vastus
1/11.
Of the 21 equally likely unordered pairs, 11 contain at least one red: 21 − 10 all-blue pairs. Only 1 pair contains both reds. Among the 11 eligible pairs, the probability is 1/11.
Kaart 43
Küsimus
A bag has 3 red and 5 blue counters. What is the probability of two reds if each uniform draw is followed by return and remixing, compared with two uniform draws without replacement?
Vastus
With return and remixing: 9/64. Without replacement: 3/28.
Returning the first counter restores 3 reds among 8, so (3/8) × (3/8) = 9/64. Without replacement, use (3/8) × (2/7) = 3/28.
43 kaarti
Probability Without Replacement Flashcards: Draw-by-Draw Practice
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