Solved WORKSHEET 11: Two Rules of Probability Name: $ 9. | Chegg.com - Free Printable

Let's solve each part of the problem step by step using the rules of probability, especially focusing on:

- Addition Rule: $ P(A \cup B) = P(A) + P(B) - P(A \cap B) $
- Multiplication Rule (for independent events): $ P(A \text{ and } B) = P(A) \times P(B) $
- Complement Rule, etc.

We are working with a standard 52-card deck:
- 4 suits: Hearts (Red), Diamonds (Red), Spades (Black), Clubs (Black)
- Each suit has 13 cards: Ace, 2–10, Jack, Queen, King
- Face cards: Jack, Queen, King → 3 per suit → 12 total
- Even-numbered cards: 2, 4, 6, 8, 10 → 5 per suit → 20 total
- Red cards: Hearts and Diamonds → 26 total
- Black cards: Spades and Clubs → 26 total

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a. When one card is selected, P(Black and Ace)?

We want the probability that the card is both black and an ace.

- There are 2 black aces: Ace of Spades and Ace of Clubs.
- Total cards = 52

$$
P(\text{Black and Ace}) = \frac{2}{52} = \boxed{\frac{1}{26}}
$$

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b. When one card is selected, P(Red or even)?

Use the addition rule:

$$
P(\text{Red or Even}) = P(\text{Red}) + P(\text{Even}) - P(\text{Red and Even})
$$

- $ P(\text{Red}) = \frac{26}{52} = \frac{1}{2} $
- $ P(\text{Even}) = \frac{20}{52} $ (even cards: 2,4,6,8,10 in each suit → 5 × 4 = 20)
- $ P(\text{Red and Even}) $: Even cards that are red → even cards in Hearts and Diamonds → 5 per suit × 2 suits = 10

So:
$$
P(\text{Red or Even}) = \frac{26}{52} + \frac{20}{52} - \frac{10}{52} = \frac{36}{52} = \boxed{\frac{9}{13}}
$$

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c. When one card is selected, P(Face and Diamond)?

We want a card that is both a face card and a diamond.

- Face cards in Diamonds: Jack, Queen, King of Diamonds → 3 cards

$$
P(\text{Face and Diamond}) = \frac{3}{52} = \boxed{\frac{3}{52}}
$$

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d. When one card is selected, P(Face or Black)?

Again, use addition rule:

$$
P(\text{Face or Black}) = P(\text{Face}) + P(\text{Black}) - P(\text{Face and Black})
$$

- $ P(\text{Face}) = \frac{12}{52} $ (3 per suit × 4 suits)
- $ P(\text{Black}) = \frac{26}{52} $
- $ P(\text{Face and Black}) $: Face cards in Spades and Clubs → 3 per suit × 2 = 6

$$
P(\text{Face or Black}) = \frac{12}{52} + \frac{26}{52} - \frac{6}{52} = \frac{32}{52} = \boxed{\frac{8}{13}}
$$

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e. When one card is selected, P(Heart and Diamond)?

A card cannot be both a Heart and a Diamond — these are mutually exclusive.

$$
P(\text{Heart and Diamond}) = \boxed{0}
$$

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f. When one card is selected, P(Ace or Spade)?

Use addition rule:

$$
P(\text{Ace or Spade}) = P(\text{Ace}) + P(\text{Spade}) - P(\text{Ace and Spade})
$$

- $ P(\text{Ace}) = \frac{4}{52} $
- $ P(\text{Spade}) = \frac{13}{52} $
- $ P(\text{Ace and Spade}) = \frac{1}{52} $ (only Ace of Spades)

$$
P(\text{Ace or Spade}) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \boxed{\frac{4}{13}}
$$

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g. When one card is selected, P(Red and Heart)?

All hearts are red, so "Red and Heart" means just "Heart".

$$
P(\text{Red and Heart}) = P(\text{Heart}) = \frac{13}{52} = \boxed{\frac{1}{4}}
$$

Alternatively, since every heart is red, the intersection is just the set of hearts.

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h. When two cards are selected with replacement, what is the probability that both cards are Red cards?

Since selection is with replacement, the events are independent.

- $ P(\text{Red}) = \frac{26}{52} = \frac{1}{2} $
- So,
$$
P(\text{Both Red}) = P(\text{Red}) \times P(\text{Red}) = \frac{1}{2} \times \frac{1}{2} = \boxed{\frac{1}{4}}
$$

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✔ Final Answers:

a. $ \boxed{\frac{1}{26}} $
b. $ \boxed{\frac{9}{13}} $
c. $ \boxed{\frac{3}{52}} $
d. $ \boxed{\frac{8}{13}} $
e. $ \boxed{0} $
f. $ \boxed{\frac{4}{13}} $
g. $ \boxed{\frac{1}{4}} $
h. $ \boxed{\frac{1}{4}} $

Let me know if you'd like a visual explanation or diagram!