Year 1
58054.76=63,103(1-r)^1
Year 2
53,410.38=63103(1-r)^2
The solution to the system is (90,40)....x = adults and y = child's...so x = 90 (there were 90 adult tickers) and y = 40 (there were 40 child tickets)...u already had the answer right in front of you
Answer:
Our answer is 0.8172
Step-by-step explanation:
P(doubles on a single roll of pair of dice) =(6/36) =1/6
therefore P(in 3 rolls of pair of dice at least one doubles)=1-P(none of roll shows a double)
=1-(1-1/6)3 =91/216
for 12 players this follows binomial distribution with parameter n=12 and p=91/216
probability that at least 4 of the players will get “doubles” at least once =P(X>=4)
=1-(P(X<=3)
=1-((₁₂ C0)×(91/216)⁰(125/216)¹²+(₁₂ C1)×(91/216)¹(125/216)¹¹+(₁₂ C2)×(91/216)²(125/216)¹⁰+(₁₂ C3)×(91/216)³(125/216)⁹)
=1-0.1828
=0.8172
