More Probability rules

• Converted set rules   The probabilty  of events satisfies the following rule :
                               .
  
     
  This is easily deduced from the usual formula for sets by after dividing both sides by
  
  The events    are said to be mutually exclusive (disjoint) if as sets they are disjoint and hence   .
  
  The negation of an event  is its complement  and its probabilty is given by  since  is .
  
  There are corresponding rules for union of three or more events, but these are best derived as needed, rather than writen down and memorized.
  
  
  
  
• Some sample calculations . Now we give some examples.
  
  The experiment is to draw a card from a  card standard deck.  
  
  So, .
  
  Consider the event of getting a red card higher than .
  
  In a red suit there are  such cards and there are two red suits,
  so by multiplication principle, we get  sample points.
  
  So the probability of this event is .
  
  The probability of getting one of the face cards, including aces,
  is clearly  and the probability of not getting any face card is
  therefore    or .
  
  
  
  Now suppose we try to pick three cards from the deck (without replacement) then the sample space size becomes  or .
  
  The probability of the event of getting at least one face  
  card is now a bit more complicated to determine,
  since we might have to make cases of getting
  exactly one two or three face cards.
  
  On the other hand, the complement -
  i.e. getting no face cards is simply the count of
  picking three cards from the  non face cards or
    .
  
  Thus, in this case, it is easier to compute the probability from the complement as  or .
  
  It is thus essential to look for simpler ways of calculating the answer.
  
  There are situations where making Venn diagrams also comes in handy.
  
   
  Some poker hands .
  
  Consider the problem of determining the
  
  probability of getting a straight,  
  
  which means  cards in a sequence which
  are not from the same suit .
  
  First, we count the sample space as  or .
  
  Now we think of a sequence.
  
  It starts from the ace or 2 or 3 or ... or 10.
  
  Thus there are 10 types of sequences.
  
  If we ignore the restriction about the suit,
  then a sequence of a given type can be chosen in
     ways (each)  
  
  and if we remove the  forbidden sequences of the chosen type,
  we get .
  
  Thus the count of "straights" of various types is .
  Thus the desired probability is:     =  .
  
  
• If we work on the type three of a kind , we have
     possible triples and each triple can come in four types
  (by omitting one of the four identical value cards),
  thus there are  possible triples.
  
  The remaining two cards need to be mismatched.
  
  So we choose two of the remaining  values in
     ways and each can take one of the four colors, so the total count is .
  Thus the total count is .
  Thus the net probability is     a lot better than above.
  
  Actually, it is possible to allow the remaining two cards to be a pair
  and this will augment the number even more. You should analyze it!
  
  
• Conditional probability . Often, we have made some
  observations and as a result some of the sample points
  from the sample space have been removed.
  
  As a result, our probability of an event changes.
  
  The new probability is called a conditional probability .
  
  Formally we say that
  
  given events  by         we denote
  the probability of  given that  has ocurred  
  
  or .  
  
  For example consider that in a 5 card poker,
  
  we have drawn an ace and   .
  
  Then the probability of a straight is , since no sequence of 5 cards can stretch that far.
  
  On the other hand, if we see a 10 and an ace of different suits,
  then we can fill these in into a straight in  ways and
  there are  total ways of choosing 3 more cards.
  
  Thus the conditional probability is:   .                 
  
  Compare this number with the probability of straights above.
  
  Why is it larger? Should we have expected that?
  
  What should we do differently, if the ace and 10 are from the same suite?
  
  Also note that we are calculating as if the starting 10 and ace are given and we are ignoring the  choices for them. If we throw these in, our probability goes up!
  
• More examples of conditional probability  
  
  Quite often, in order that a certain event happens, another event has to happen first.
  
  For instance, to get to class on time, you need to get up on time
  (with enough time to get ready) and then you need to get into a
  reasonable traffic.
  
  Given that the probability of getting up on time is 80% and then the probability of catching a good traffic pattern is 70%,
  you could argue that the probability of getting to class on time
  is calculated as the product  or 56%.
  
  What happens if you are a little late in getting up,
  say the probability of this is the remaining 20%.
  
  Perhaps, then the probability of getting a good traffic pattern
  is only 40%.
  Now the probability of getting to class on time is  
                                      ( )( )=( ) or only %.
  
  Since, to get to class, you need to get up and fight with the traffic,
  you can now calculate the combined probability of getting to class
  on time as
                                  +  =  or 64%.
  
•  Let us formalize these calculation as follows.
  
  Define events E and L as getting up early and getting up late. Define events G and B as getting a good and bad traffic pattern.  Thus, to get to class on time, you must have the event  or , so the probability desired is that of the event .
  
  Note that the probability of getting a good traffic pattern P(G)
  is not given to us at all, mainly because it changes based on
  the timing.
  
  The probability of G given E i.e. P(G|E)  is 80% while
  probability of G given L i.e. P(G|L) is only 40%.
  
  The probability  is then calculated
  as .
  
  Similarly for the other .
  
  The two events  and  are mutually exclusive,
  so their probabilities are added!
  
  You can imagine a further possibility where you can get up
  so late that the traffic pattern does not matter!
  
  Think what could happen to the probability then.
  
  You should also calculate the probability of missing the class due to bad traffic!