HELP!

[NBelviso]

I need help with this homework question. Please explain how to do it.

To get to work, a commenter must cross train tracks. The time the train arrives varies slightly from day to day, but the commuter estimates he’ll get stopped on about 15% of work days. During a certain 5-day work week, what is the probability that he...
a) gets stopped on Monday and again on Tuesday?
b) gets stopped for the first time on Thursday?
c) gets stopped every day?
d) gets stopped at least once during the week?

 

[obh]

Hi NB,

Please try to answer, and I will give you feedback.
Cheers

 

[GretaGarbo]

It is not possible to answer the questions without makeing an additional assumption. So @NBelviso, what is the absolutley most common statistical assumpttion?

(And, please make better and more informative titles.)

 

[NBelviso]

Hi NB,

Please try to answer, and I will give you feedback.
Cheers

a) P(Monday and Tuesday) = (.15)(.15)= 0.0225
b) P(Thursday) = (.15) = .15
c) P(everyday) = (.15)(.15)(.15)(.15)(.15)= 0.000076
d) P(@ least once) = (.85)(.85)(.85)(.85)(.15)= 0.078

thank you for your help

 

[NBelviso]

It is not possible to answer the questions without makeing an additional assumption. So @NBelviso, what is the absolutley most common statistical assumpttion?

(And, please make better and more informative titles.)

Hi,
I’m sorry. I have never posted before and am new at this. Thanks for your feedback.
I am assuming that the events are all independent and therefore the multiplication rule would work here?

 

[GretaGarbo]

Independence, yes!

And then what?

 

[NBelviso]

Do I apply the 15% to every day or the whole week (5 days)? This is my first statistics class ever.

 

[pbl24]

Do I apply the 15% to every day or the whole week (5 days)? This is my first statistics class ever.

That depends on the question. If the days can be considered independent and the probability of being stopped on a given day is 15%, then yes, the multiplication principle would apply. If, however, the probability of being stopped on Tuesday is affected by if you were stopped on Monday, then they are not independent.

 

[NBelviso]

a) P(Monday and Tuesday) = (.15)(.15)= 0.0225
b) P(Thursday) = (.15) = .15
c) P(everyday) = (.15)(.15)(.15)(.15)(.15)= 0.000076
d) P(@ least once) = (.85)(.85)(.85)(.85)(.15)= 0.078

ok, so does this make sense of the days are independent?

 

[Karabiner]

b) P(Thursday) = (.15) = .15

Cave! The question was: "whether he gets stopped for the first time on Thursday?"
I.e. p(not on Monday & not on Tuesday & not on Wednesday & on Thursday).

d) P(@ least once) = (.85)(.85)(.85)(.85)(.15)= 0.078

7.8% seems a bit counter-intuitive? I'd suppose you need "1 - probability(never)"

With kind regards

Karabiner