SOLVED: Name Date Topic: Probability Mutually Exclusive Events ... - Free Printable

Here's the complete solution to Worksheet A3: Single Event Probability, with clear explanations for each problem.

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## Part 1: Names in a Hat

We are given 10 names:

> Mary, Jenny, Bob, Marilyn, Bill, Jack, Jerry, Tina, Connie, Joe

Total number of names = 10

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1. P(3-letter name) = 2/10 or 1/5 *(Given as example)*

✔ Explanation: The 3-letter names are Bob and Joe → 2 names.
So, probability = 2/10 = 1/5.

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2. P(4-letter name) = ________

Names with exactly 4 letters:
→ Mary, Bill, Jack, Tina, Joe? Wait — Joe is 3 letters.
Let’s list:

- Mary → 4 ✔
- Jenny → 5
- Bob → 3
- Marilyn → 7
- Bill → 4 ✔
- Jack → 4 ✔
- Jerry → 5
- Tina → 4 ✔
- Connie → 6
- Joe → 3

✔ So: Mary, Bill, Jack, Tina → 4 names

→ P(4-letter name) = 4/10 = 2/5

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3. P(name starting with B) = ________

Names starting with ‘B’:
→ Bob, Bill → 2 names

→ P(starting with B) = 2/10 = 1/5

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4. P(name starting with T) = ________

Names starting with ‘T’:
→ Tina → 1 name

→ P(starting with T) = 1/10

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5. P(7-letter name) = ________

Names with 7 letters:
→ Marilyn (7), Connie (6), others are shorter.

Only Marilyn → 1 name

→ P(7-letter name) = 1/10

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6. P(name starting with S) = ________

Names starting with ‘S’:
→ None! (Check: Mary, Jenny, Bob, Marilyn, Bill, Jack, Jerry, Tina, Connie, Joe) — no S.

→ P(starting with S) = 0/10 = 0

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7. P(name ending with Y) = ________

Names ending with ‘Y’:
→ Jenny, Marilyn, Jerry → 3 names

→ P(ending with Y) = 3/10

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## Part 2: Cards Drawn (12 cards total)

Cards shown:
> 10, 4, 7, J, S, 9, 10, 2, M, 5, 4, J

Total cards = 12

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8. P(2) = 1/12 *(Given as example)*

✔ Only one '2' → 1/12

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9. P(5) = ________

How many 5s? → One card is ‘5’

→ P(5) = 1/12

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10. P(J) = ________

How many ‘J’? → Two cards: positions 4 and 12 → 2 Js

→ P(J) = 2/12 = 1/6

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11. P(a number) = ________

Numbers are: 10, 4, 7, 9, 10, 2, 5, 4 → that’s 8 numbers

Letters: J, S, M, J → 4 letters

→ P(a number) = 8/12 = 2/3

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12. P(4) = ________

How many ‘4’s? → Two cards: position 2 and 11

→ P(4) = 2/12 = 1/6

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13. P(T) = ________

Is there any ‘T’? → No card has ‘T’

→ P(T) = 0/12 = 0

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14. P(a letter) = ________

Letters: J, S, M, J → 4 letters

→ P(a letter) = 4/12 = 1/3

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## Part 3: Standard 52-card Deck

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15. P(ace) = ________

There are 4 aces in a deck (one per suit)

→ P(ace) = 4/52 = 1/13

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16. P(face card - K, J, Q) = ________

Face cards: King, Queen, Jack — 3 per suit × 4 suits = 12 face cards

→ P(face card) = 12/52 = 3/13

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17. P(a red 10) = ________

Red suits: Hearts and Diamonds → each has one 10 → so 2 red 10s

→ P(red 10) = 2/52 = 1/26

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18. P(NOT a diamond) = ________

Total diamonds = 13 → NOT diamonds = 52 - 13 = 39

→ P(not a diamond) = 39/52 = 3/4

*(Alternatively: 1 - P(diamond) = 1 - 13/52 = 39/52 = 3/4)*

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## Part 4: Spinner (numbered 1–8)

Spinner has numbers: 1, 2, 3, 4, 5, 6, 7, 8 → total 8 outcomes

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19. P(an EVEN number) = ________

Even numbers: 2, 4, 6, 8 → 4 numbers

→ P(even) = 4/8 = 1/2

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20. P(a multiple of 3) = ________

Multiples of 3 between 1–8: 3, 6 → 2 numbers

→ P(multiple of 3) = 2/8 = 1/4

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21. P(a PRIME number) = ________

Prime numbers between 1–8:
→ 2, 3, 5, 7 → 4 primes
*(Note: 1 is not prime)*

→ P(prime) = 4/8 = 1/2

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22. P(9) = ________

Spinner only goes up to 8 → no 9

→ P(9) = 0/8 = 0

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## ✔ Final Answers Summary:

Part 1: Names

2. 2/5
3. 1/5
4. 1/10
5. 1/10
6. 0
7. 3/10

Part 2: Cards

9. 1/12
10. 1/6
11. 2/3
12. 1/6
13. 0
14. 1/3

Part 3: Deck

15. 1/13
16. 3/13
17. 1/26
18. 3/4

Part 4: Spinner

19. 1/2
20. 1/4
21. 1/2
22. 0

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