Conditional Probability

Date: 12/31/97 at 14:21:01
From: chris Bonnette
Subject: Conditional probability

Thank you for providing such an interesting way of finding out math 
information. It is much appreciated.

I tutor probability and statistics and people always seem to be very 
confused by conditional probability:

P(AIB) = P(both events)/P(known event occurred) = P(AB)/P(B)

The confusion seems to lie in the use of the multiplication rule.  
If the events are dependent, they seem to feel that they end up in an 
infinite loop:

P(AB) = P(A)*P(BlA)...but P(BlA) = P(BA)/P(A).  It appears that 
everything just cancels and you are left with P(AB) and are no closer 
to finding the solution.  I'm looking for a clear and nonconfusing way 
to initially teach this concept to them so that this confusion doesn't 
even start.  Any suggestions?

Chris

Date: 12/31/97 at 17:21:18
From: Doctor Anthony
Subject: Re: Conditional probability

A picture of the sample space allows you to calculate conditional 
probabilities without too much of the confusing notation.

Example:

A bag contains 5 similar coins except that one is double-headed.  A 
coin is chosen at random and is tossed 4 times.  Each time it lands 
heads. What is the probability that the double-headed coin was chosen?

  Chosen coin is           Chosen coin is
  double-headed.           normal.
  prob = 1/5               prob = 4/5
  ----------------------------------------------------
   (1/5) x 1               (4/5) x (1/2)^4      4 heads obtained. 
  ----------------------------------------------------
    xxxxxx                 xxxxxxxxxxx          other result.
  -----------------------------------------------------


The sample space is restricted to the top line since we are told that 
4 heads were obtained. We can find the probability that we are in the 
lefthand box by putting its value over the total value of the top 
line.

                                       (1/5)
  Prob.double-headed coin  =   --------------------
                                (1/5) + (4/5)(1/16)


                                        1
                           =     -----------------
                                   1   +   1/4

                                      4
                           =       --------
                                    4 + 1

                           =         4/5

So the probability that we have the double-headed coin, on the 
evidence of throwing 4 heads in a row, is 4/5. Of course, even one 
tail would have made it certain that we had a normal coin.

This method allows you to understand what is going on without the 
somewhat opaque notation of conditional probability.

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/