Your reasoning was incorrect. See below.
On Mon, 16 Feb 2004, gamin wrote (edited):
> [In] studying statistics on my own ... I happened to come across
> this exercise where my answer does not match the one in the book.
> Thank you for the help in advance.
>
> Police plan to enforce speed limits by using radar traps at 4
> different locations within the city limits. The radar traps at each of
> the locations L1, L2, L3 and L4 are operated 40%, 30%, 20% and 30% of
> the time, and if a person who is speeding on his way to work has
> probabilities of 0.2, 0.1, 0.5 and 0.2 respectively, of passing
> through these locations, what is the probability that he will receive
> a speeding ticket ?
>
> 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) ]
[Pedantic quibble: this is not reasoning, this is a report of the
algorithm you used. Doesn't show why you chose it.]
That's equivalent to
{1 - [P(no ticket, L1) + P(none, L2) + P(none, L3) + P(none, L4)]} But
the four events whose probabilities you've added are not mutually
exclusive, because the event {no ticket at L1} includes some of the
event {no ticket at L2}, etc.; hence their sum is greater than
P(no ticket at any Li), hence the complement of that sum is less than
P(receive a ticket at L1 or L2 or L3 or L4 or any combination thereof),
which was asked for.
OTOH, the events {ticket at L1}, {ticket at L2}, ... are all different
events (although more than one of them may occur while driving to work),
so you can add their probabilities:
P(a ticket at all) = P(ticket at L1) + P(...L2) + P(...L3) + P(...L4)
(which also allows for the remote possibility that the speeder actually
receives more than one ticket, being unlucky enough to be caught at more
than one location):
P{ticket(s)} = 40%(0.2) + 30%(0.1) + 20%(0.5) + 30%(0.2)
= 0.08 + 0.03 + 0.10 + 0.06 = 0.27.
Of course, the probability of receiving exactly ONE ticket is somewhat
less than this, and is calculable if one assumes the four events to be
independent, which may be unreasonable. (E.g, independence implies that
P{ticket at L1 AND ticket at L2} = 0.08(0.03) = 0.0024,
P{ticket at L1 AND L2 AND L3} = 0.08(0.03)(0.10) = 0.00024, etc.)
The events may not be independent, however. E.g., suppose there is
only one traffic-control person cruising about; if this person observes
the speed limits, the speeder will be caught at most once. Or, if all
traffic-control persons work only during the evening rush hour, the
speeder will never be caught on the way TO work, if he has the usual
daytime-ish working hours.
HTH. -- DFB.
------------------------------------------------------------
Donald F. Burrill [EMAIL PROTECTED]
56 Sebbins Pond Drive, Bedford, NH 03110 (603) 626-0816
.
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