Probability (given no card cancellation): [ \frac{3000}{6840} = \frac{300}{684} = \frac{50}{114} = \frac{25}{57} \approx 0.4386 ]
Total possible ordered selections (without replacement from 20): (20 \times 19 \times 18 = 6840).
Just then, the bell rang. Three new customers entered: a nun, a clown, and a beekeeper.
"I bet," Chiara whispered, "the chance they all pick different toppings is 72%." Calcolo combinatorio e probabilita -Italian Edi...
The catch? The three chosen customers would pick , and the same topping could be chosen more than once. Enzo would then combine their choices into one bizarre, three-topping pizza. The First Mystery One rainy evening, a young data scientist named Chiara sat at the counter.
Every Saturday, Enzo offered a — a mystery pizza with random toppings chosen by a strange ritual. Customers would write their names on slips of paper, and Enzo would draw three names. Those three would each choose a topping from a list of ten: funghi, carciofi, salsiccia, peperoni, olive, cipolle, acciughe, rucola, gorgonzola, zucchine .
Enzo smiled, sliding her a free bruschetta . "Ah, combinatoria . Let’s reason." "I bet," Chiara whispered, "the chance they all
Enzo winked. " Probabilità doesn’t guarantee, but it guides. Now, who wants a slice?" If you'd like, I can rewrite this as a or turn each problem into a clean combinatorial formula for your Italian edition book. Just let me know.
Enzo’s eyes sparkled. "Now that is combinatorics with constraints ."
First person: 10 choices. Second: 9 choices (different from first). Third: 8 choices (different from first two). [ 10 \times 9 \times 8 = 720 ] The First Mystery One rainy evening, a young
Thus, overall probability that a pizza is made the customers are from three different towns: [ \frac{9}{10} \times \frac{25}{57} = \frac{225}{570} = \frac{45}{114} = \frac{15}{38} \approx 0.3947 ] The Revelation Chiara finished her wine. "Enzo, your pizza game is a lesson in combinatorics and probability."
Enzo laughed. "Life is random, cara mia . But understanding the combinations helps you not fear the uncertainty."
Total cards: 40. Cards with value 1: 4 (one per suit). [ P(\text{not drawing a '1'}) = \frac{36}{40} = \frac{9}{10} ]
Each of 3 people chooses 1 topping from 10: [ 10 \times 10 \times 10 = 1000 ]
Choose 1 from town A: 5 ways, 1 from B: 5, 1 from C: 5, 1 from D: 5, but we need exactly 3 towns — so first choose which 3 towns out of 4: (\binom{4}{3} = 4) ways. For each set of 3 towns: choose 1 person from each: (5 \times 5 \times 5 = 125) combinations. Then arrange them in order: (3! = 6) ways. Total favorable ordered selections: [ 4 \times 125 \times 6 = 3000 ]