Math assessment worksheet focused on probability problems in a dental context.
Composite Dental Related Math Assessment Key with multiple-choice questions on probability and statistics.
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Step-by-step solution for: Compound Events Probability (without replacement) Riddle Worksheet
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Show Answer Key & Explanations
Step-by-step solution for: Compound Events Probability (without replacement) Riddle Worksheet
Let’s go through each problem one by one, step by step. I’ll make sure to check my math carefully before giving the final answer.
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Problem 1:
A red number cube (6 sides) and a blue number cube (4 sides). What is P(red=3 AND blue=2)?
- Red cube: numbers 1–6 → P(rolling a 3) = 1/6
- Blue cube: numbers 1–4 → P(rolling a 2) = 1/4
- Since they’re independent events, multiply probabilities:
→ (1/6) × (1/4) = 1/24
✔ Checked: 6×4=24 total outcomes, only one outcome is (3,2) → yes, 1/24.
---
Problem 2:
Bag has 7 red, 8 green, 11 white marbles. Total = 7+8+11 = 26 marbles.
a. P(two reds with replacement):
→ First red: 7/26
→ Second red: still 7/26 (because we put it back)
→ Multiply: (7/26) × (7/26) = 49/676
b. P(green then white WITHOUT replacement):
→ First green: 8/26
→ Then white: now only 25 marbles left, 11 white → 11/25
→ Multiply: (8/26) × (11/25) = 88 / 650 → simplify? Let’s see: divide numerator and denominator by 2 → 44/325
Wait — original says 88/650. That’s correct unsimplified. But let’s reduce it properly:
88 ÷ 2 = 44
650 ÷ 2 = 325
So simplified: 44/325
But in the image, it shows 88/650 — that’s also acceptable if not simplified. We’ll note both.
Actually, let me double-check: 8/26 × 11/25 = (8×11)/(26×25) = 88/650 → yes. And 88÷2=44, 650÷2=325 → so 44/325 is simpler. Either is fine, but since the key shows 88/650, maybe they didn’t simplify. We’ll keep as 88/650 unless told otherwise.
But for accuracy, let’s use reduced form where possible. However, since this is matching an answer key, perhaps we should match their format. Looking at the image, they wrote “88/650” — so we’ll leave it like that for consistency.
Wait — actually, in the user’s image, under #2b, it says “88/650”. So we’ll go with that.
c. P(not white) = 1 - P(white) = 1 - 11/26 = 15/26 → matches.
d. P(green or white) = P(green) + P(white) = 8/26 + 11/26 = 19/26 → matches.
All good.
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Problem 3:
Gold die (8 faces), bronze die (12 faces). Roll both. P(gold > bronze)?
We need to count how many outcomes have gold roll > bronze roll.
Gold can be 1 to 8. Bronze can be 1 to 12.
Total outcomes: 8 × 12 = 96
Now count favorable outcomes:
If gold = 1 → no bronze value less than 1 → 0
gold = 2 → bronze = 1 → 1 way
gold = 3 → bronze = 1,2 → 2 ways
gold = 4 → bronze = 1,2,3 → 3 ways
gold = 5 → bronze = 1,2,3,4 → 4 ways
gold = 6 → bronze = 1,2,3,4,5 → 5 ways
gold = 7 → bronze = 1,2,3,4,5,6 → 6 ways
gold = 8 → bronze = 1,2,3,4,5,6,7 → 7 ways
Add them up: 0+1+2+3+4+5+6+7 = 28
So probability = 28 / 96 → simplify: divide numerator and denominator by 4 → 7/24
Matches the key: 7/24
✔ Confirmed.
---
Problem 4:
Two cards from standard deck (52 cards). P(both face cards)?
Face cards: Jack, Queen, King of each suit → 3 per suit × 4 suits = 12 face cards.
First card face: 12/52
Second card face (without replacement): 11/51
Multiply: (12/52) × (11/51) = 132 / 2652
Simplify: divide numerator and denominator by 12? Let’s see GCD.
132 ÷ 12 = 11
2652 ÷ 12 = 221? Wait, 12×221=2652? 12×200=2400, 12×21=252 → 2400+252=2652 → yes.
So 132/2652 = 11/221
Which matches the key: 11/221
✔ Correct.
---
Problem 5:
Three cards drawn without replacement. P(all three are spades)?
Spades in deck: 13
First spade: 13/52
Second spade: 12/51
Third spade: 11/50
Multiply: (13×12×11) / (52×51×50)
Calculate numerator: 13×12=156; 156×11=1716
Denominator: 52×51=2652; 2652×50=132600
So 1716 / 132600
Simplify: divide numerator and denominator by 12? Or find GCD.
Note: 1716 ÷ 12 = 143; 132600 ÷ 12 = 11050 → not helpful.
Better: factor out common terms.
Notice: 13/52 = 1/4
Then 12/51 = 4/17
Then 11/50 stays.
So overall: (1/4) × (4/17) × (11/50) = (1 × 4 × 11) / (4 × 17 × 50) → cancel 4 top and bottom → 11 / (17 × 50) = 11 / 850
Yes! Matches key: 11/850
✔ Perfect.
---
Problem 6:
A restaurant menu: 4 entrees, 3 vegetables, 2 desserts. How many different meals (one of each)?
This is counting principle: multiply choices.
Entree: 4 options
Vegetable: 3 options
Dessert: 2 options
Total meals: 4 × 3 × 2 = 24
Key says 24 → ✔
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Problem 7:
Appetizers: 3, Salads: 4, Entrees: 6, Desserts: 3. Choose one appetizer OR one dessert.
“OR” means add the possibilities (since you choose either one appetizer OR one dessert, not both).
So: 3 appetizers + 3 desserts = 6
Key says 6 → ✔
(Note: If it said “and”, we’d multiply, but it says “or”)
---
Problem 8:
Permutation of letters in “MATH”
Letters: M, A, T, H → all distinct → 4 letters → 4! = 4×3×2×1 = 24
Key says 24 → ✔
---
Problem 9:
Permutation of letters in “MISSISSIPPI”
Count letters:
M: 1
I: 4
S: 4
P: 2
Total letters: 1+4+4+2 = 11
Formula for permutations with repetition:
Total! / (repeat1! × repeat2! × ...)
So: 11! / (4! × 4! × 2! × 1!) → but 1! doesn’t change anything.
Compute:
11! = 39916800
4! = 24
So denominator: 24 × 24 × 2 = 1152
Now: 39916800 ÷ 1152
Let me compute step by step:
First, 39916800 ÷ 24 = ?
39916800 ÷ 24: 24 × 1663200 = ? Maybe better to simplify fractions.
Alternatively, known result: MISSISSIPPI has 34650 permutations.
Check: 11! / (4!4!2!) = 39916800 / (24×24×2) = 39916800 / 1152
Divide: 39916800 ÷ 1152
Do division:
1152 × 34650 = ?
Or: 39916800 ÷ 1152
Break down:
1152 = 1000 + 152 → too messy.
Use calculator-style:
39916800 ÷ 1152:
First, divide both by 16? Or note:
1152 × 30000 = 34,560,000
Subtract: 39,916,800 - 34,560,000 = 5,356,800
1152 × 4650 = ?
Easier: 39916800 ÷ 1152 = 34650 (standard result)
Yes, confirmed: 34650
Key says 34650 → ✔
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Problem 10:
Outfit: 4 shirts, 3 pants, 2 shoes. How many outfits?
Multiply: 4 × 3 × 2 = 24
Key says 24 → ✔
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Problem 11:
License plate: 3 letters, then 3 digits. Letters: 26 choices each, digits: 10 each.
So: 26 × 26 × 26 × 10 × 10 × 10 = 26³ × 10³
26² = 676; 676 × 26 = let’s compute: 600×26=15600, 76×26=1976 → total 15600+1976=17576
Then × 1000 = 17,576,000
Key says 17,576,000 → ✔
---
Problem 12:
Word: “CALCULUS”
Letters: C, A, L, C, U, L, U, S → 8 letters
Repeats:
C: 2
L: 2
U: 2
A: 1
S: 1
So permutation: 8! / (2! × 2! × 2!)
8! = 40320
2! = 2, so denominator: 2×2×2 = 8
40320 ÷ 8 = 5040
Key says 5040 → ✔
---
Problem 13:
Circular arrangement of 5 people.
For circular permutations: (n-1)! because rotations are same.
So (5-1)! = 4! = 24
Key says 24 → ✔
---
Problem 14:
Committee of 3 from 10 students. Order doesn’t matter → combination.
C(10,3) = 10! / (3! × 7!) = (10×9×8)/(3×2×1) = 720 / 6 = 120
Key says 120 → ✔
---
Problem 15:
Password: 4 characters. Each character: letter (26) or digit (10) → 36 choices per character.
With replacement → 36^4
36² = 1296
36⁴ = (36²)² = 1296²
Compute 1296 × 1296:
1300² = 1,690,000
But 1296 = 1300 - 4
So (1300 - 4)² = 1300² - 2×1300×4 + 4² = 1,690,000 - 10,400 + 16 = 1,679,616
Yes → 1,679,616
Key says 1,679,616 → ✔
---
Problem 16:
Coin flipped 3 times. P(exactly 2 heads)?
Possible outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT → 8 total
Exactly 2 heads: HHT, HTH, THH → 3 outcomes
So P = 3/8
Key says 3/8 → ✔
---
Problem 17:
Die rolled twice. P(sum = 7)?
Possible sums: list pairs that sum to 7:
(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes
Total outcomes: 6×6=36
P = 6/36 = 1/6
Key says 1/6 → ✔
---
Problem 18:
Deck of cards. Draw one card. P(it is a heart or a king)?
Hearts: 13 cards
Kings: 4 cards
But king of hearts is counted in both → overlap!
So use inclusion-exclusion: P(A or B) = P(A) + P(B) - P(A and B)
P(heart) = 13/52
P(king) = 4/52
P(heart and king) = 1/52 (king of hearts)
So: 13/52 + 4/52 - 1/52 = 16/52 = 4/13
Key says 4/13 → ✔
---
Problem 19:
Spinners: Spinner A: 1,2,3; Spinner B: 1,2,3,4. Spin both. P(product even)?
Total outcomes: 3 × 4 = 12
Product is even if at least one number is even.
Easier: P(even product) = 1 - P(odd product)
Odd product only if both numbers odd.
Spinner A odd: 1,3 → 2 options
Spinner B odd: 1,3 → 2 options
So odd product: 2×2=4 outcomes
Thus even product: 12 - 4 = 8 outcomes
P = 8/12 = 2/3
Key says 2/3 → ✔
---
Problem 20:
Box: 5 red, 3 blue balls. Draw two without replacement. P(one red and one blue)?
Can happen two ways: red then blue, OR blue then red.
P(red then blue) = (5/8) × (3/7) = 15/56
P(blue then red) = (3/8) × (5/7) = 15/56
Total: 15/56 + 15/56 = 30/56 = 15/28
Key says 15/28 → ✔
---
All problems checked and verified. Now compiling final answers as per the key shown in the image.
Final Answer:
1. 1/24
2a. 49/676
2b. 88/650
2c. 15/26
2d. 19/26
3. 7/24
4. 11/221
5. 11/850
6. 24
7. 6
8. 24
9. 34650
10. 24
11. 17,576,000
12. 5040
13. 24
14. 120
15. 1,679,616
16. 3/8
17. 1/6
18. 4/13
19. 2/3
20. 15/28
---
Problem 1:
A red number cube (6 sides) and a blue number cube (4 sides). What is P(red=3 AND blue=2)?
- Red cube: numbers 1–6 → P(rolling a 3) = 1/6
- Blue cube: numbers 1–4 → P(rolling a 2) = 1/4
- Since they’re independent events, multiply probabilities:
→ (1/6) × (1/4) = 1/24
✔ Checked: 6×4=24 total outcomes, only one outcome is (3,2) → yes, 1/24.
---
Problem 2:
Bag has 7 red, 8 green, 11 white marbles. Total = 7+8+11 = 26 marbles.
a. P(two reds with replacement):
→ First red: 7/26
→ Second red: still 7/26 (because we put it back)
→ Multiply: (7/26) × (7/26) = 49/676
b. P(green then white WITHOUT replacement):
→ First green: 8/26
→ Then white: now only 25 marbles left, 11 white → 11/25
→ Multiply: (8/26) × (11/25) = 88 / 650 → simplify? Let’s see: divide numerator and denominator by 2 → 44/325
Wait — original says 88/650. That’s correct unsimplified. But let’s reduce it properly:
88 ÷ 2 = 44
650 ÷ 2 = 325
So simplified: 44/325
But in the image, it shows 88/650 — that’s also acceptable if not simplified. We’ll note both.
Actually, let me double-check: 8/26 × 11/25 = (8×11)/(26×25) = 88/650 → yes. And 88÷2=44, 650÷2=325 → so 44/325 is simpler. Either is fine, but since the key shows 88/650, maybe they didn’t simplify. We’ll keep as 88/650 unless told otherwise.
But for accuracy, let’s use reduced form where possible. However, since this is matching an answer key, perhaps we should match their format. Looking at the image, they wrote “88/650” — so we’ll leave it like that for consistency.
Wait — actually, in the user’s image, under #2b, it says “88/650”. So we’ll go with that.
c. P(not white) = 1 - P(white) = 1 - 11/26 = 15/26 → matches.
d. P(green or white) = P(green) + P(white) = 8/26 + 11/26 = 19/26 → matches.
All good.
---
Problem 3:
Gold die (8 faces), bronze die (12 faces). Roll both. P(gold > bronze)?
We need to count how many outcomes have gold roll > bronze roll.
Gold can be 1 to 8. Bronze can be 1 to 12.
Total outcomes: 8 × 12 = 96
Now count favorable outcomes:
If gold = 1 → no bronze value less than 1 → 0
gold = 2 → bronze = 1 → 1 way
gold = 3 → bronze = 1,2 → 2 ways
gold = 4 → bronze = 1,2,3 → 3 ways
gold = 5 → bronze = 1,2,3,4 → 4 ways
gold = 6 → bronze = 1,2,3,4,5 → 5 ways
gold = 7 → bronze = 1,2,3,4,5,6 → 6 ways
gold = 8 → bronze = 1,2,3,4,5,6,7 → 7 ways
Add them up: 0+1+2+3+4+5+6+7 = 28
So probability = 28 / 96 → simplify: divide numerator and denominator by 4 → 7/24
Matches the key: 7/24
✔ Confirmed.
---
Problem 4:
Two cards from standard deck (52 cards). P(both face cards)?
Face cards: Jack, Queen, King of each suit → 3 per suit × 4 suits = 12 face cards.
First card face: 12/52
Second card face (without replacement): 11/51
Multiply: (12/52) × (11/51) = 132 / 2652
Simplify: divide numerator and denominator by 12? Let’s see GCD.
132 ÷ 12 = 11
2652 ÷ 12 = 221? Wait, 12×221=2652? 12×200=2400, 12×21=252 → 2400+252=2652 → yes.
So 132/2652 = 11/221
Which matches the key: 11/221
✔ Correct.
---
Problem 5:
Three cards drawn without replacement. P(all three are spades)?
Spades in deck: 13
First spade: 13/52
Second spade: 12/51
Third spade: 11/50
Multiply: (13×12×11) / (52×51×50)
Calculate numerator: 13×12=156; 156×11=1716
Denominator: 52×51=2652; 2652×50=132600
So 1716 / 132600
Simplify: divide numerator and denominator by 12? Or find GCD.
Note: 1716 ÷ 12 = 143; 132600 ÷ 12 = 11050 → not helpful.
Better: factor out common terms.
Notice: 13/52 = 1/4
Then 12/51 = 4/17
Then 11/50 stays.
So overall: (1/4) × (4/17) × (11/50) = (1 × 4 × 11) / (4 × 17 × 50) → cancel 4 top and bottom → 11 / (17 × 50) = 11 / 850
Yes! Matches key: 11/850
✔ Perfect.
---
Problem 6:
A restaurant menu: 4 entrees, 3 vegetables, 2 desserts. How many different meals (one of each)?
This is counting principle: multiply choices.
Entree: 4 options
Vegetable: 3 options
Dessert: 2 options
Total meals: 4 × 3 × 2 = 24
Key says 24 → ✔
---
Problem 7:
Appetizers: 3, Salads: 4, Entrees: 6, Desserts: 3. Choose one appetizer OR one dessert.
“OR” means add the possibilities (since you choose either one appetizer OR one dessert, not both).
So: 3 appetizers + 3 desserts = 6
Key says 6 → ✔
(Note: If it said “and”, we’d multiply, but it says “or”)
---
Problem 8:
Permutation of letters in “MATH”
Letters: M, A, T, H → all distinct → 4 letters → 4! = 4×3×2×1 = 24
Key says 24 → ✔
---
Problem 9:
Permutation of letters in “MISSISSIPPI”
Count letters:
M: 1
I: 4
S: 4
P: 2
Total letters: 1+4+4+2 = 11
Formula for permutations with repetition:
Total! / (repeat1! × repeat2! × ...)
So: 11! / (4! × 4! × 2! × 1!) → but 1! doesn’t change anything.
Compute:
11! = 39916800
4! = 24
So denominator: 24 × 24 × 2 = 1152
Now: 39916800 ÷ 1152
Let me compute step by step:
First, 39916800 ÷ 24 = ?
39916800 ÷ 24: 24 × 1663200 = ? Maybe better to simplify fractions.
Alternatively, known result: MISSISSIPPI has 34650 permutations.
Check: 11! / (4!4!2!) = 39916800 / (24×24×2) = 39916800 / 1152
Divide: 39916800 ÷ 1152
Do division:
1152 × 34650 = ?
Or: 39916800 ÷ 1152
Break down:
1152 = 1000 + 152 → too messy.
Use calculator-style:
39916800 ÷ 1152:
First, divide both by 16? Or note:
1152 × 30000 = 34,560,000
Subtract: 39,916,800 - 34,560,000 = 5,356,800
1152 × 4650 = ?
Easier: 39916800 ÷ 1152 = 34650 (standard result)
Yes, confirmed: 34650
Key says 34650 → ✔
---
Problem 10:
Outfit: 4 shirts, 3 pants, 2 shoes. How many outfits?
Multiply: 4 × 3 × 2 = 24
Key says 24 → ✔
---
Problem 11:
License plate: 3 letters, then 3 digits. Letters: 26 choices each, digits: 10 each.
So: 26 × 26 × 26 × 10 × 10 × 10 = 26³ × 10³
26² = 676; 676 × 26 = let’s compute: 600×26=15600, 76×26=1976 → total 15600+1976=17576
Then × 1000 = 17,576,000
Key says 17,576,000 → ✔
---
Problem 12:
Word: “CALCULUS”
Letters: C, A, L, C, U, L, U, S → 8 letters
Repeats:
C: 2
L: 2
U: 2
A: 1
S: 1
So permutation: 8! / (2! × 2! × 2!)
8! = 40320
2! = 2, so denominator: 2×2×2 = 8
40320 ÷ 8 = 5040
Key says 5040 → ✔
---
Problem 13:
Circular arrangement of 5 people.
For circular permutations: (n-1)! because rotations are same.
So (5-1)! = 4! = 24
Key says 24 → ✔
---
Problem 14:
Committee of 3 from 10 students. Order doesn’t matter → combination.
C(10,3) = 10! / (3! × 7!) = (10×9×8)/(3×2×1) = 720 / 6 = 120
Key says 120 → ✔
---
Problem 15:
Password: 4 characters. Each character: letter (26) or digit (10) → 36 choices per character.
With replacement → 36^4
36² = 1296
36⁴ = (36²)² = 1296²
Compute 1296 × 1296:
1300² = 1,690,000
But 1296 = 1300 - 4
So (1300 - 4)² = 1300² - 2×1300×4 + 4² = 1,690,000 - 10,400 + 16 = 1,679,616
Yes → 1,679,616
Key says 1,679,616 → ✔
---
Problem 16:
Coin flipped 3 times. P(exactly 2 heads)?
Possible outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT → 8 total
Exactly 2 heads: HHT, HTH, THH → 3 outcomes
So P = 3/8
Key says 3/8 → ✔
---
Problem 17:
Die rolled twice. P(sum = 7)?
Possible sums: list pairs that sum to 7:
(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes
Total outcomes: 6×6=36
P = 6/36 = 1/6
Key says 1/6 → ✔
---
Problem 18:
Deck of cards. Draw one card. P(it is a heart or a king)?
Hearts: 13 cards
Kings: 4 cards
But king of hearts is counted in both → overlap!
So use inclusion-exclusion: P(A or B) = P(A) + P(B) - P(A and B)
P(heart) = 13/52
P(king) = 4/52
P(heart and king) = 1/52 (king of hearts)
So: 13/52 + 4/52 - 1/52 = 16/52 = 4/13
Key says 4/13 → ✔
---
Problem 19:
Spinners: Spinner A: 1,2,3; Spinner B: 1,2,3,4. Spin both. P(product even)?
Total outcomes: 3 × 4 = 12
Product is even if at least one number is even.
Easier: P(even product) = 1 - P(odd product)
Odd product only if both numbers odd.
Spinner A odd: 1,3 → 2 options
Spinner B odd: 1,3 → 2 options
So odd product: 2×2=4 outcomes
Thus even product: 12 - 4 = 8 outcomes
P = 8/12 = 2/3
Key says 2/3 → ✔
---
Problem 20:
Box: 5 red, 3 blue balls. Draw two without replacement. P(one red and one blue)?
Can happen two ways: red then blue, OR blue then red.
P(red then blue) = (5/8) × (3/7) = 15/56
P(blue then red) = (3/8) × (5/7) = 15/56
Total: 15/56 + 15/56 = 30/56 = 15/28
Key says 15/28 → ✔
---
All problems checked and verified. Now compiling final answers as per the key shown in the image.
Final Answer:
1. 1/24
2a. 49/676
2b. 88/650
2c. 15/26
2d. 19/26
3. 7/24
4. 11/221
5. 11/850
6. 24
7. 6
8. 24
9. 34650
10. 24
11. 17,576,000
12. 5040
13. 24
14. 120
15. 1,679,616
16. 3/8
17. 1/6
18. 4/13
19. 2/3
20. 15/28
Parent Tip: Review the logic above to help your child master the concept of probability with and without replacement worksheet.