Probability Without Replacement Flashcards: Draw-by-Draw Practice

Question 1

A bag holds 5 red and 4 blue counters. One uniform draw without replacement is observed to be red. What remains?

Accepted Answer

4 red and 4 blue counters: 8 in total.

Only the red count and the total decrease.

Question 2

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?

Accepted Answer

3/7.

All 3 green counters remain among 7 counters.

Question 3

A box starts with 11 distinct tickets. Four are drawn uniformly without replacement. How many tickets are available for the fifth draw?

Accepted Answer

7 tickets.

Four tickets have left the box, so the next denominator is 11 − 4.

Question 4

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?

Accepted Answer

0

No silver counters remain; all 6 remaining counters are gold.

Question 5

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?

Accepted Answer

5/10 = 1/2.

The same physical counter cannot return, but another counter of the same colour can be drawn.

Question 6

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?

Accepted Answer

5 green, 2 yellow and 2 white counters: 9 in total.

Update each colour actually removed.

Question 7

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?

Accepted Answer

7

The next draw is from the 7 counters still in the bag.

Question 8

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?

Accepted Answer

5/8.

There are 5 red counters left among 8 counters.

Question 9

A bag has 4 red and 5 blue counters. Draw two uniformly without replacement. What is the probability of red then red?

Accepted Answer

1/6.

(4/9) × (3/8) = 12/72 = 1/6. The second red count is 3.

Question 10

A bag has 7 orange and 4 purple counters. Draw two uniformly without replacement. What is the probability of orange then purple?

Accepted Answer

14/55.

(7/11) × (4/10) = 28/110 = 14/55. Removing orange leaves all 4 purple counters.

Question 11

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?

Accepted Answer

10/63.

(5/9) × (4/8) × (4/7) = 80/504 = 10/63.

Question 12

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?

Accepted Answer

1/42.

(1/7) × (1/6) = 1/42. Only one ticket matches each specified letter.

Question 13

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.

Accepted Answer

(4/11) × (3/10) = 6/55.

After a red is drawn, only 3 reds remain among 10 counters.

Question 14

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?

Accepted Answer

5/28.

(6/8) × (2/7) × (5/6) = 60/336 = 5/28.

Question 15

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?

Accepted Answer

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.

Question 16

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?

Accepted Answer

15/28.

(5/8) × (3/7) + (3/8) × (5/7) = 30/56 = 15/28.

Question 17

A bag has 5 green and 2 yellow counters. Draw two uniformly without replacement. What is the probability that both have the same colour?

Accepted Answer

11/21.

Both green or both yellow: (5/7) × (4/6) + (2/7) × (1/6) = 22/42 = 11/21.

Question 18

A bag has 4 red and 5 blue counters. Draw three uniformly without replacement. What is the probability of exactly two reds?

Accepted Answer

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.

Question 19

A bag has 3 green and 6 yellow counters. Draw two uniformly without replacement. What is the probability of at least one green?

Accepted Answer

7/12.

Subtract the chance of no green: 1 − (6/9) × (5/8) = 1 − 5/12 = 7/12.

Question 20

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?

Accepted Answer

3/7.

Choosing together is without replacement. Count both sequential orders: 2 × (6/8) × (2/7) = 24/56 = 3/7.

Question 21

A box has 3 faulty and 7 working bulbs. Choose three uniformly without replacement. What is the probability that none is faulty?

Accepted Answer

7/24.

All three must work: (7/10) × (6/9) × (5/8) = 210/720 = 7/24.

Question 22

A bag has 2 red and 6 blue counters. Draw three uniformly without replacement. What is the probability of at most one red?

Accepted Answer

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.

Question 23

A bag has 3 red, 2 blue and 1 green counter. Draw two uniformly without replacement. What is the probability of different colours?

Accepted Answer

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.

Question 24

A box has 2 faulty and 6 working sensors. Choose four uniformly without replacement. What is the probability of exactly one faulty sensor?

Accepted Answer

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.

Question 25

Nine tickets include 2 prize tickets. Draw three uniformly without replacement. What is the probability of at least one prize?

Accepted Answer

7/12.

Subtract three non-prizes: 1 − (7/9) × (6/8) × (5/7) = 1 − 5/12 = 7/12.

Question 26

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?

Accepted Answer

The blue-then-red path.

Add (6/11) × (5/10). The total is 60/110 = 6/11; the learner counted only red then blue.

Question 27

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?

Accepted Answer

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.

Question 28

A bag has 3 green and 5 yellow counters. Draw three uniformly without replacement. What is the probability of at least two greens?

Accepted Answer

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.

Question 29

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?

Accepted Answer

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.

Question 30

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.

Accepted Answer

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.

Question 31

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?

Accepted Answer

2/3.

Either order leaves exactly 6 red and 3 blue counters. The next probability is 6/9 = 2/3.

Question 32

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.

Accepted Answer

6/8 = 3/4.

The total falls to 8, but the red count stays at 6 because the removed counter was blue.

Question 33

A box has 1 gold and 8 silver tokens. Draw three uniformly without replacement. What is the probability of at least two gold tokens?

Accepted Answer

0

Only one gold token exists, so no draw order can contain two gold tokens.

Question 34

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.

Accepted Answer

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.

Question 35

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.

Accepted Answer

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.

Question 36

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.

Accepted Answer

1/2.

All 4 reds remain among 8 counters: 4/8 = 1/2. The observed blue changes the information available.

Question 37

A bag has 5 red and 2 blue counters. Draw two uniformly without replacement. What is the probability that they are not both red?

Accepted Answer

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.

Question 38

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.

Accepted Answer

3/8.

(3/8) × (2/7) + (5/8) × (3/7) = 21/56 = 3/8. Weight each possible remaining state by its probability.

Question 39

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?

Accepted Answer

2/9.

There are 5 prizes among 10 tickets. Then (5/10) × (4/9) = 20/90 = 2/9.

Question 40

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?

Accepted Answer

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.

Question 41

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.

Accepted Answer

3/10.

There are 3 greens among 10 remaining counters. Unlike unseen removals, observed greens let you update the count exactly.

Question 42

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?

Accepted Answer

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.

Question 43

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?

Accepted Answer

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.

Về bộ thẻ này

Thẻ trong bộ này

Thẻ 1

Thẻ 2

Thẻ 3

Thẻ 4

Thẻ 5

Thẻ 6

Thẻ 7

Thẻ 8

Thẻ 9

Thẻ 10

Thẻ 11

Thẻ 12

Thẻ 13

Thẻ 14

Thẻ 15

Thẻ 16

Thẻ 17

Thẻ 18

Thẻ 19

Thẻ 20

Thẻ 21

Thẻ 22

Thẻ 23

Thẻ 24

Thẻ 25

Thẻ 26

Thẻ 27

Thẻ 28

Thẻ 29

Thẻ 30

Thẻ 31

Thẻ 32

Thẻ 33

Thẻ 34

Thẻ 35

Thẻ 36

Thẻ 37

Thẻ 38

Thẻ 39

Thẻ 40

Thẻ 41

Thẻ 42

Thẻ 43

Probability Without Replacement Flashcards: Draw-by-Draw Practice