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?

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: \[ \