Probability Worksheets - Free Printable

Let's solve each of these probability problems step by step. We are working with a standard 52-card deck (no Jokers), and in all cases, the card is replaced after the first draw, meaning the two draws are independent events.

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Basic Facts about a Standard Deck:

- Total cards: 52
- Suits: Clubs, Spades (black), Hearts, Diamonds (red)
- Each suit has 13 cards: A, 2–10, J, Q, K
- Face cards: Jack, Queen, King → 3 per suit, so 12 total
- Number cards: 2 through 10 → 9 ranks per suit
- Red cards: Hearts & Diamonds → 26
- Black cards: Clubs & Spades → 26

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Now let’s go through each question:

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1) Probability of drawing a Club on the first draw, replacing it, then drawing a Heart on the second draw.

- P(Club) = 13/52 = 1/4
- P(Heart) = 13/52 = 1/4
- Since independent:
$$
P = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}
$$

✔ Answer: 1/16

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2) Probability of drawing a 2 through 6 on the first draw, replacing it, then drawing a Club on the second draw.

- Cards 2 through 6: 5 ranks × 4 suits = 20 cards
- P(2–6) = 20/52 = 5/13
- P(Club) = 13/52 = 1/4
- Independent:
$$
P = \frac{5}{13} \times \frac{1}{4} = \frac{5}{52}
$$

✔ Answer: 5/52

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3) Probability of drawing a red face card on the first draw, replacing it, then drawing a Spade on the second draw.

- Red face cards: Hearts and Diamonds → 2 suits × 3 face cards = 6 red face cards
- P(Red face card) = 6/52 = 3/26
- P(Spade) = 13/52 = 1/4
- Independent:
$$
P = \frac{3}{26} \times \frac{1}{4} = \frac{3}{104}
$$

✔ Answer: 3/104

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4) Probability of drawing a red card on the first draw, replacing it, then drawing a 2 on the second draw.

- P(Red card) = 26/52 = 1/2
- There are 4 twos (one per suit)
- P(2) = 4/52 = 1/13
- Independent:
$$
P = \frac{1}{2} \times \frac{1}{13} = \frac{1}{26}
$$

✔ Answer: 1/26

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5) Probability of drawing a Club 6 through 9 on the first draw, replacing it, then drawing a Spade on the second draw.

- Club 6 through 9: 4 cards (6,7,8,9 of Clubs)
- P(Club 6–9) = 4/52 = 1/13
- P(Spade) = 13/52 = 1/4
- Independent:
$$
P = \frac{1}{13} \times \frac{1}{4} = \frac{1}{52}
$$

✔ Answer: 1/52

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6) Probability of drawing a black card on the first draw, replacing it, then drawing a red card on the second draw.

- P(Black card) = 26/52 = 1/2
- P(Red card) = 26/52 = 1/2
- Independent:
$$
P = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}
$$

✔ Answer: 1/4

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7) Probability of drawing a black 6 through 9 on the first draw, replacing it, then drawing a 6 on the second draw.

- Black suits: Clubs and Spades → 2 suits
- 6 through 9: 4 ranks → 4 cards per suit → 2×4 = 8 black cards
- P(Black 6–9) = 8/52 = 2/13
- P(6) = 4/52 = 1/13
- Independent:
$$
P = \frac{2}{13} \times \frac{1}{13} = \frac{2}{169}
$$

✔ Answer: 2/169

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8) Probability of drawing a face card that is a Diamond on the first draw, replacing it, then drawing a black card on the second draw.

- Diamond face cards: Jack, Queen, King of Diamonds → 3 cards
- P(Diamond face card) = 3/52
- P(Black card) = 26/52 = 1/2
- Independent:
$$
P = \frac{3}{52} \times \frac{1}{2} = \frac{3}{104}
$$

✔ Answer: 3/104

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9) Find the probability of drawing a 8.

This is a single draw — no replacement or second draw.

- There are 4 eights (one per suit)
- P(8) = 4/52 = 1/13

✔ Answer: 1/13

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10) Probability of drawing a 4 on the first draw, replacing it, then drawing a face card on the second draw.

- P(4) = 4/52 = 1/13
- Face cards: 12 total (3 per suit × 4 suits)
- P(Face card) = 12/52 = 3/13
- Independent:
$$
P = \frac{1}{13} \times \frac{3}{13} = \frac{3}{169}
$$

✔ Answer: 3/169

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✔ Final Answers Summary:

| Question | Answer |
|--------|--------------|
| 1 | 1/16 |
| 2 | 5/52 |
| 3 | 3/104 |
| 4 | 1/26 |
| 5 | 1/52 |
| 6 | 1/4 |
| 7 | 2/169 |
| 8 | 3/104 |
| 9 | 1/13 |
| 10 | 3/169 |

Let me know if you'd like this formatted as a printable answer sheet!