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December 7th, 2017, 05:04 AM   #1
Joined: Dec 2017
From: London

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Tree Diagram and Dependent probability.

Hi so I need some help with this question?

Following a survey, it is found that a train service in Japan is late 2% of
the time. For a particular working week of 5 days, a passenger wishes to
investigate the potential lateness of the train service.
a) Find the probability that the train is not late on any of the 5 days.

b) Calculate the probability that the train is late on fewer than two days.
c) If the train is late on between two and three days, customers who
purchased a weekly train pass are offered a half-refund. Calculate this
d) However, if the train is late on four or more days, customers purchasing
a weekly train pass are offered a free pass for the following week.
Calculate the probability of this occurrence.

any help is greatly appreciated!

Last edited by DecoMate; December 7th, 2017 at 05:25 AM. Reason: new question
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December 7th, 2017, 04:03 PM   #2
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I guess they want you to model this as a binomial distribution with $n=5,~p=0.02$

$P[k]=P[\text{train is late k days out of 5}]=\dbinom{5}{k}p^k(1-p)^{5-k}$

a) $P[\text{train late 0 of 5 days}] = P[0] = (0.8 )^5$

b) $P[\text{train is late < 2 days}]=P[0]+P[1]$

c) $P[\text{half refund}] = P[2]+P[3]$

d) $P[\text{free pass for next week}] = P[4]+P[5]$
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