Calcolo Combinatorio E Probabilita -italian Edi... Direct

This is always possible once we reach this stage. So the probability that a pizza gets made is just the probability of not drawing a '1' first:

"So," Chiara said, "a 1% chance. Rare, but possible."

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. Calcolo combinatorio e probabilita -Italian Edi...

Probability (given no card cancellation): [ \frac{3000}{6840} = \frac{300}{684} = \frac{50}{114} = \frac{25}{57} \approx 0.4386 ]

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.) This is always possible once we reach this stage

Enzo smiled, sliding her a free bruschetta . "Ah, combinatoria . Let’s reason."

Enzo laughed. "Life is random, cara mia . But understanding the combinations helps you not fear the uncertainty." The three chosen customers would pick , and

"So most of the time," Marco laughed, "the pizza is a mix of three distinct flavors!" That night, a boy named Luca asked the most curious question: "What if you drew the names without replacement from a total of 20 customers, but then the three chosen still pick toppings with repetition? And also, before picking toppings, you shuffle a deck of 40 Scoppia cards (Italian regional cards: four suits, numbered 1 to 10). If the first card is a '1' of any suit, you cancel the pizza game. If not, you proceed. What’s the chance we actually make a pizza?"

"But wait!" Luca interrupted. "What if you also require that the three chosen customers are all from different towns, and there are 4 towns with 5 customers each? And the selection without replacement must include one from each town — then what's the probability that a random ordered selection of 3 customers satisfies that?"

[ P(\text{pizza}) = \frac{9}{10} ]

[ \frac{720}{1000} = 0.72 \quad (72%) ]