The public transport system is aware that one person in twenty does not buy a ticket for their trip. The manager decides to challenge 200 travelling people as to charger they have a valid ticket or not.
(a) calculate the probabilty that everyone had a ticket,
(b) calculate the probabilty that at most 3 people had no ticket, and
(c) calculate the probabilty that between 7 and 14 (inclusive) had no ticket. I thought binomial? But then 200 is too big for it? Thanks for any help!
Agree that the numbers are enourmous, so I will leave answers in notation form
a) prob of having ticket = 19/20
So prob that 200 people have a ticket = (19/20)^200
b) prob that at most 3 will have no tickets
= prob(3 no ticket) + prob(2 no ticket) + prob(1 no ticket) + prob(0 no ticket)
= C(200,3)(1/20)^3(19/20)^197 + C(200,2)(1/20)^2(19/20)198 + ...
c) C(200,7)(1/20)^7(19/20)^193 + ... + C(200,14)(1/20)^14(19/20)^186
e.g. C(200,14)(1/20)^14(19/20)^186 = .0517578