Hi,
Thanks for the response.
> >
> >The books stated answer is : 0.27
> >Mine is : 0.2450 [ reasoning as follows : 1- P(no ticket at L1) - P(no
> >ticket at L2) - P(no ticket at L3) - P(no ticket at L4) ]
>
> I don't quite understand what you're doing here. Are you maybe
> reacting to the fact that the problem should have said (but didn't)
> that he passes through one and only one of the four locations in any
> trip?
>
I guess i was a bit in a hurry and made an error, my resoning should
have been (assuming that passing through each location is an
independant event):
P(getting a ticket(s)) = 1 - P(getting no ticket at all)
= 1 - P(getting no ticket at L1, L2, L3 and l4)
= 1 - P(no ticket at L1)*P(no ticket at l2)...
= 1 - 0.92 * 0.97 * 0.9 * 0.94
= 0.2450
Yes, this is erronous as you pointed out that he may not pass through
any of L1, L2, L3 or L4. I did not understand the problem. Yes, we
need to assume that the driver needs to pass through only one of the
four locations to get to work.
And thus the answer in the book is correct
> That should have been stated in the problem, but I think we have to
> make that assumption because otherwise the problem can't be solved.
> If he might pass through say L1 and L4 in the same trip, then that
> opens up the possibility that he might pass through _none_ of the
> four locations.
>
> Your book's answer was obtained as follows, using the assumption
> that the driver passes through one and only one of the four:
> P(ticket at L1) = .4 * .2 = .08
> P(ticket at L2) = .3 * .1 = .03
> P(ticket at L3) = .2 * .5 = .10
> P(ticket at L4) = .3 * .2 = .06
> ----
> P(ticket anywhere) = .27
Thank you for your help
G
.
.
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