1. **Problem statement:** We deal 3 cards from a standard 52-card deck. Find the probabilities of: (a) 2 of a kind (exactly two cards of the same rank), (b) 3 of a kind (all three cards of the same rank), (c) 3 cards of the same suit (a flush), (d) 3 cards in consecutive order (a straight). 2. **Total number of 3-card hands:** $$\text{Total} = \binom{52}{3} = \frac{52 \times 51 \times 50}{3 \times 2 \times 1} = 22100$$ 3. **(a) Probability of exactly 2 of a kind:** - Choose the rank for the pair: 13 choices. - Choose 2 suits out of 4 for the pair: $\binom{4}{2} = 6$ ways. - Choose the third card rank different from the pair rank: 12 choices. - Choose the suit of the third card: 4 choices. Number of favorable hands: $$13 \times 6 \times 12 \times 4 = 3744$$ Probability: $$P(2\text{-of-a-kind}) = \frac{3744}{22100} \approx 0.1694$$ 4. **(b) Probability of 3 of a kind:** - Choose the rank: 13 choices. - Choose all 3 suits out of 4 for that rank: $\binom{4}{3} = 4$ ways. Number of favorable hands: $$13 \times 4 = 52$$ Probability: $$P(3\text{-of-a-kind}) = \frac{52}{22100} \approx 0.00235$$ 5. **(c) Probability of 3 cards of the same suit (flush):** - Choose the suit: 4 choices. - Choose any 3 cards from the 13 cards of that suit: $\binom{13}{3} = 286$ ways. Number of favorable hands: $$4 \times 286 = 1144$$ Probability: $$P(\text{flush}) = \frac{1144}{22100} \approx 0.0518$$ 6. **(d) Probability of 3 cards in consecutive order (straight):** - Number of possible 3-card straights: - Possible starting ranks: from Ace-2-3 up to Queen-King-Ace, counting Ace as low or high, total 12 sequences. - Each rank in the straight has 4 suits, so total ways per sequence: $$4^3 = 64$$ Number of favorable hands: $$12 \times 64 = 768$$ Probability: $$P(\text{straight}) = \frac{768}{22100} \approx 0.03475$$ **Final answers:** - (a) $\approx 0.1694$ - (b) $\approx 0.00235$ - (c) $\approx 0.0518$ - (d) $\approx 0.03475$