CONDITIONAL PROBABILITY

MARGINAL PROBABILITY

A Probability of only one event that takes places is called a marginal probability.

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JOINT PROBABILITY

 
The probability of occurrence of both events A and B together,denoted by  $P(A \cap B)$   is known as joint probability A and B .

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CONDITIONAL PROBABILITY

The conditional probability of A given B is $ P(A/B)=\frac{P(A \cap B) }{P(B}$   if $ P(B) \neq 0 $  and it is undefined otherwise.A rearrangement of the above definition yields the following: 

MR(Multiplication Rule)

 $ P(A \cap B) $  =$ P(B)P(A/B) \enspace if P(B) \neq 0 $
                         $ P(A)P(B/A) \enspace if P(A) \neq 0 $
                          $0$             otherwise}

Notes:

  $P(A/B) $ means the conditional probability of A and B.
  $ P(B/A) $ means the conditional probability of A and B

    

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 THEOREM 1 If  P(A) >P(B)  then $ P(A/B)>P(B/A)$
proof

 
$ P(A/B)= \frac{P(A \cap B )}{P(B)} $

$ P(B)= \frac{P(A \cap B )}{P(A/B)} $

$  P(B/A)= \frac{P(A \cap B )}{P(A)} $

$ P(A)= \frac{P(A \cap B )}{P(B/A)} $

If P(A) >P(B)  $ \frac{P(A \cap B )}{P(B/A)} > \frac{P(A \cap B )}{P(A/B)}$
and $  P(A/B) > P(B/A)$

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EXAMPLE 1
A box contains  4 bad and 6 good tubes.Two are drawn out from the box at a time.One of them is tested and found to be good.What is the probability that the one is also good?

Solution

               Bad Tubes       Good Tubes        Total

                4                          6                    10   
 

$ P(A \cap B) =  P $                  [both  the  tubes  are  good ]                    

                   $ = \frac{6C_2}{10C_2}$

                    $ =  \frac{\frac{6.5}{1.2}}{\frac{10.9}{1.2}} $         [$ nC_r=\frac{n!}{(n-r)!r!}$]

                    $ = \frac{6.5}{10.9}$             [ cancel  common  term]

                    $ = \frac{1}{3} $                 [ cancel common  term]

 $ P(A \cap B) = \frac{1}{3} $ 

P(A) = P[one tubes  drawn   good]

$ = \frac{6}{10}$ 

$ P(A) = \frac{3}{5} $

That one tube is good, the conditional probability that the other tube is also good is required

$ P(B/A) = \frac{P(A \cap B )}{P(A)} $ 

$ = \frac{\frac{1}{3}}{\frac{3}{5}}  $                      [ $ P(A \cap B) = \frac{1}{3}$ $P(A) = \frac{3}{5} $] 

$= \frac{1}{3} \times \frac{5}{3} $ 

$ P(B/A) = \frac{5}{9}  $ 

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