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Not the First Event and Not the Second — and Not Independent

Date: 07/20/2012 at 22:29:05
From: Brian
Subject: probability word problem

At a card party, two door prizes are to be awarded by drawing two names at
random from a box. The box contains the names of 40 people, including that
of Joe Ramirez.

What is the probability that Joe's name is not drawn for either prize?

I reason that there is a 1/40 chance for Joe's name to be selected on the
first draw, and a 1/39 chance on the second draw. The probability of Joe
getting a prize on one of the draws is the sum of these, or 79/1560. The
probability of him not getting a prize is therefore 1 - 79/1560, or
1481/1560.  

Is this correct? The answer key gives 1482/1560, or 19/20.



Date: 07/21/2012 at 12:59:28
From: Doctor Peterson
Subject: Re: probability word problem

Hi, Brian.

The trouble is that the addition rule you are following doesn't have the
special case that the multiplication rule has, in which you can treat the
event "name is drawn on second draw" as a conditional probability (that
is, ASSUME that the name was not drawn the first time).

To find the actual probability that his name is drawn second, you would
have to do this:

   P(drawn second) = P(not drawn first AND drawn second)
                   = P(not drawn first) * P(drawn second | not first)
                   = 39/40 * 1/39
                   = 1/40

That is, if you don't know what happened on the first draw, it makes no
difference whether you ask about the first or the second draw! That makes
sense when you think about it, but I did it this way to convince you (and
myself), because it's not entirely obvious that it's not a fallacy.

So the probability you calculated is really

   P(drawn first OR drawn second) = P(first) + P(second) - P(both)
                                  = 1/40 + 1/40 - 0
                                  = 1/20
By complementarity,

   P(neither first nor second) = 1 - P(first or second)
                               = 1 - 1/20
                               = 19/20

The usual way is to treat the problem as an "and," using the rule for
non-independent events:

   P(neither first nor second) = P(not first AND not second)
                               = P(not first) * P(not second|not first)
                               = 39/40 * 38/39
                               = 38/40 = 19/20

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 07/21/2012 at 19:13:40
From: Brian
Subject: Thank you (probability word problem)

You identify exactly where I went wrong in your comment that "the addition
rule you are following doesn't have the special case that the
multiplication rule has, in which you can treat the event 'name is drawn
on second draw' as a conditional probability."

Framing the problem this way made the correct solution crystal clear:

   P(drawn second) = P(not drawn first AND drawn second)

Thanks, Dr. Peterson!
Associated Topics:
High School Probability

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