Probability of Winning Tickets in a Box with 80 Tickets and 10 People

AI Thread Summary
In a scenario where 10 people each buy 10 tickets from a box of 80 tickets containing 4 winning tickets, the probability calculations focus on three outcomes. For one person to win all three winning tickets, the formula involves combinations of choosing from the total tickets. The second scenario examines the likelihood of three different winners among the participants. The third scenario considers the probability of one person winning two tickets while another wins one. The discussion highlights the complexity of calculating these probabilities in a constrained ticket system.
Bachelier
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In a box with 80 tickets, 10 people buy 10 tickets each. If there are 4 winning tickets drawn at random find the probability that
1) one person gets all 3 winning tickets
2) there are 3 different winners
3) some person gets two winners and someone else gets just one

for 1) 10 C 3 * 70 C 7

is the answer = -------------------- where C stands for chooses.
80 C 10

Thnks
 
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Bachelier said:
In a box with 80 tickets, 10 people buy 10 tickets each.

That's 100 tickets :redface: …

have the police been informed?​
 

Thread 'Formal derivation of statement from Peano Arithmetic system'

I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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