SOLUTION: A coin is tossed 3 times. define the events A and B as: A = {num heads > num tails} B = {1st toss is a head} Find P(A|B)

Let's write H for head and T for tail.


Then any outcome of 3 tosses can be written as a word of the length 3 written in this 2-letter alphabet,
consisting of letters H an T.


In all, there are 8 possible outcomes.  They all are listed as such words in the list below


1)   H H H     (*)    (**)   (***)

2)   H H T     (*)    (**)   (***)

3)   H T H     (*)    (**)   (***)

4)   H T T     (*)

5)   T H H            (**)

6)   T H T

7)   T T H

8)   T T T 


Notice that these outcomes have equal probability of   =   each.

By symbol (*), I marked that events where the 1st toss is a head.
There are 4 such events,  nn 1), 2), 3) an 4).
It is the set of events B.


By symbol (**), I marked that events where number of heads > number of tails.
There are 4 such events,  nn 1), 2), 3) an 5).
It is the set of events A.


By the definition of the conditional probability,

    P(A | B)  = P (A ∩ B) / P(B).


The intersection  (A ∩ B)  consists of three events  nn 1), 2) and 3)  in the list;  they are marked by the symbol (***).


Now,  P (A ∩ B) = ;  P(B) = ;  therefore,

      P(A | B)  =   = .


ANSWER.  P(A | B) = .