Calcolo Combinatorio E Probabilita | -italian Edi...
"Enzo," she said, "what’s the probability that the three chosen customers all pick the same topping?"
In the narrow, lantern-lit streets of Perugia, old Enzo ran the most beloved pizzeria in Umbria. But Enzo had a secret: he was also a mathematician who had retired early from the University of Bologna.
10 possible choices (all mushrooms, all onions, etc.) [ \frac{10}{1000} = \frac{1}{100} ]
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 ] Calcolo combinatorio e probabilita -Italian Edi...
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 .
The beekeeper picked honey (not on the menu), the nun picked mushrooms, the clown picked pineapple (scandalous). All different.
Number of ways to choose 3 distinct customers in order: [ 20 \times 19 \times 18 = 6840 ] (This step doesn’t affect the probability of making a pizza because it’s always possible to pick toppings regardless of who they are. The only cancelling event is the card draw.) "Enzo," she said, "what’s the probability that the
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.
"I bet," Chiara whispered, "the chance they all pick different toppings is 72%."
First person: 10 choices. Second: 9 choices (different from first). Third: 8 choices (different from first two). [ 10 \times 9 \times 8 = 720 ] For each set of 3 towns: choose 1
Total possible ordered selections (without replacement from 20): (20 \times 19 \times 18 = 6840).
"So," Chiara said, "a 1% chance. Rare, but possible."
Probability (given no card cancellation): [ \frac{3000}{6840} = \frac{300}{684} = \frac{50}{114} = \frac{25}{57} \approx 0.4386 ]
Each of 3 people chooses 1 topping from 10: [ 10 \times 10 \times 10 = 1000 ]
