14. A card is drawn from a deck of 52 cards and a die is tossed. What is the probability of: a. Drawing a heart and tossing a 1? b. Drawing a heart or tossing a 1? c. Drawing a face card and tossing an even number? d. Drawing a face card or tossing an even number? 15. Two cards are selected from a deck of 52 cards without replacement. What is the probability of: a. Drawing two diamonds? b. Drawing a club and then a heart? c. Drawing two black cards? 16. A box contains 3 oranges, 2 apples, and 5 bananas: You reach in the box and draw 3 pieces of fruit. What is the probability that: a. All three are bananas? b. You draw a blnana, apple, and orange in that order. c. You draw no apples? d. You draw two banana, and then an orange in that order?

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Let's solve each of the problems step by step, extracting the known conditions and calculating the probabilities.

### Problem 14: Card and Die

**Known Conditions:**
- A standard deck has 52 cards.
- There are 13 hearts in the deck.
- A die has 6 faces (1 to 6).

#### a. Drawing a heart and tossing a 1
The probability of drawing a heart (P(H)) is:
$$
P(H) = \frac{13}{52} = \frac{1}{4}
$$
The probability of tossing a 1 (P(1)) is:
$$
P(1) = \frac{1}{6}
$$
Since these two events are independent, the combined probability (P(H and 1)) is:
$$
P(H \text{ and } 1) = P(H) \times P(1) = \frac{1}{4} \times \frac{1}{6} = \frac{1}{24}
$$

#### b. Drawing a heart or tossing a 1
Using the formula for the probability of the union of two independent events:
$$
P(H \text{ or } 1) = P(H) + P(1) - P(H \text{ and } 1)
$$
Calculating:
$$
P(H \text{ or } 1) = \frac{1}{4} + \frac{1}{6} - \frac{1}{24}
$$
Finding a common denominator (24):
$$
P(H \text{ or } 1) = \frac{6}{24} + \frac{4}{24} - \frac{1}{24} = \frac{9}{24} = \frac{3}{8}
$$

#### c. Drawing a face card and tossing an even number
There are 12 face cards in a deck (3 face cards in each of the 4 suits). The probability of drawing a face card (P(F)) is:
$$
P(F) = \frac{12}{52} = \frac{3}{13}
$$
The even numbers on a die are 2, 4, and 6, so:
$$
P(\text{even}) = \frac{3}{6} = \frac{1}{2}
$$
Thus, the combined probability is:
$$
P(F \text{ and even}) = P(F) \times P(\text{even}) = \frac{3}{13} \times \frac{1}{2} = \frac{3}{26}
$$

#### d. Drawing a face card or tossing an even number
Using the union formula:
$$
P(F \text{ or even}) = P(F) + P(\text{even}) - P(F \text{ and even})
$$
Calculating:
$$
P(F \text{ or even}) = \frac{3}{13} + \frac{1}{2} - \frac{3}{26}
$$
Finding a common denominator (26):
$$
P(F \text{ or even}) = \frac{6}{26} + \frac{13}{26} - \frac{3}{26} = \frac{16}{26} = \frac{8}{13}
$$

### Problem 15: Two Cards Without Replacement

**Known Conditions:**
- A standard deck has 52 cards.

#### a. Drawing two diamonds
The probability of drawing the first diamond (P(D1)) is:
$$
P(D1) = \frac{13}{52}
$$
The probability of drawing a second diamond after the first (P(D2|D1)) is:
$$
P(D2|D1) = \frac{12}{51}
$$
Thus, the combined probability is:
$$
P(D1 \text{ and } D2) = P(D1) \times P(D2|D1) = \frac{13}{52} \times \frac{12}{51} = \frac{156}{2652} = \frac{1}{17}
$$

#### b. Drawing a club and then a heart
The probability of drawing a club first (P(C)) is:
$$
P(C) = \frac{13}{52}
$$
The probability of drawing a heart after that (P(H|C)) is:
$$
P(H|C) = \frac{13}{51}
$$
Thus, the combined probability is:
$$
P(C \text{ and } H) = P(C) \times P(H|C) = \frac{13}{52} \times \frac{13}{51} = \frac{169}{2652} = \frac{13}{204}
$$

#### c. Drawing two black cards
There are 26 black cards (13 clubs and 13 spades). The probability of drawing the first black card (P(B1)) is:
$$
P(B1) = \frac{26}{52}
$$
The probability of drawing a second black card (P(B2|B1)) is:
$$
P(B2|B1) = \frac{25}{51}
$$
Thus, the combined probability is:
$$
P(B1 \text{ and } B2) = P(B1) \times P(B2|B1) = \frac{26}{52} \times \frac{25}{51} = \frac{650}{2652} = \frac{25}{102}
$$

### Problem 16: Drawing Fruit

**Known Conditions:**
- 3 oranges, 2 apples, and 5 bananas (total 10 fruits).

#### a. All three are bananas
The probability of drawing three bananas (P(B1, B2, B3)) is:
$$
P(B1) = \frac{5}{10}, \quad P(B2|B1) = \frac{4}{9}, \quad P(B3|B1, B2) = \frac{3}{8}
$$
Thus, the combined probability is:
$$
P(B1 \text{ and } B2 \text{ and } B3) = \frac{5}{10} \times \frac{4}{9} \times \frac{3}{8} = \frac{60}{720} = \frac{1}{12}
$$

#### b. Drawing a banana, apple, and orange in that order
The probability is:
$$
P(B) = \frac{5}{10}, \quad P(A|B) = \frac{2}{9}, \quad P(O|B, A) = \frac{3}{8}
$$
Thus, the combined probability is:
$$
P(B \text{ and } A \text{ and } O) = \frac{5}{10} \times \frac{2}{9} \times \frac{3}{8} = \frac{30}{720} = \frac{1}{24}
$$

#### c. Drawing no apples
The total ways to draw 3 fruits from 10 is:
\[
\
**Problem 14: Card and Die** a. Probability of drawing a heart and tossing a 1: $$ \frac{1}{24} $$ b. Probability of drawing a heart or tossing a 1: $$ \frac{3}{8} $$ c. Probability of drawing a face card and tossing an even number: $$ \frac{3}{26} $$ d. Probability of drawing a face card or tossing an even number: $$ \frac{8}{13} $$ **Problem 15: Two Cards Without Replacement** a. Probability of drawing two diamonds: $$ \frac{1}{17} $$ b. Probability of drawing a club and then a heart: $$ \frac{13}{204} $$ c. Probability of drawing two black cards: $$ \frac{25}{102} $$ **Problem 16: Drawing Fruit** a. Probability of drawing three bananas: $$ \frac{1}{12} $$ b. Probability of drawing a banana, apple, and orange in that order: $$ \frac{1}{24} $$ c. Probability of drawing no apples: $$ \frac{1}{12} $$ d. Probability of drawing two bananas and then an orange in that order: $$ \frac{1}{20} $$

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