Math 141, Sem. I, 1998-99 (R.A. Brualdi)
Handout 22 (November 30, 1998)
Topic: Probability II
EXAMPLE 1
Suppose we draw a card at random from an ordinary deck of 52 cards.
Let
A= event card is a spade
B= event card is a heart. Then
Prob[A] = 13/52
Prob[B] = 13/52.
Since there are 26 cards which are either spades or hearts, the event "A
or B" (the card is either a spade or a heart) has probability
Prob[A or B] =26/52.
This can also be expressed as:
Prob[A or B] = 26/52 = 13/52 +13/52 = Prob[A] + Prob[B].
Now let
C = event the card is an ace, so that Prob[C]=4/52. Then since there are
16 cards which are either spades or aces,
Prob[A or C] = 16/52 but this does not equal Prob[A]+Prob[C]=17/52.
The reason is that the events A and C can occur simultaneously (the ace of
spades), that is, they are not *mutually exclusive*. The formula
Prob[A or B] =Prob[A]+Prob[B]
holds only for mutually exclusive events.
For non-mutually exclusive events, like A and C, the more general formula
Prob[A or C] = Prob[A]+Prob[C]-Prob[A and C]
holds: Prob[A or C] = 13/52 + 4/52 - 1/52 = 16/52.
EXAMPLE 2
Consider an urn contain 3 identical red balls, 4 identical white balls,
and 5 identical blue balls. If we draw a ball at random from the urn, then
we assign probabilities to each of the three colors being drawn by:
Prob[Red]=3/12, Prob[White]=4/12, Prob[Blue]=5/12
(the colors are not equally likely; we do not use 1/3, 1/3, 1/3.)
Suppose we draw two balls in succession from the urn. Now the results
are (sample space):
RR WR BR
RW WW BW
RB WB BB.
Again these results are not equally likely.
Our computations are based on whether or not the first ball is replaced
before the second is drawn.
WITH REPLACEMENT:
We could say that there are 12x12 possible outcomes. Of these, 3x3 would
give a Red followed by a Red ball. So
Prob[RR]=3x3/12x12 = 3/12 x 3/12; the latter two numbers can be
interpreted as
3/12: probability first ball is Red
3/12: probability second ball is Red.
This suggests Prob[A and B] = Prob[A]xProb[B].
WITHOUT REPLACEMENT:
We could say that there are 12x11 possible outcomes. Of these, 3x2 would
give a Red followed by a Red ball. So
Prob[RR] = 3x2/12x11 = 3/12 x 2/11; the latter two numbers can be
interpreted as
3/12: probability first ball is Red
2/11; probability second is red, IF the first is Red.
This suggest Prob[A and B] = Prob[A]xProb[B IF A occurs].
In the case of replacement, the second drawing is *independent* of the
first:
Prob[A and B] = Prob[A] x Prob [B] provided A and B are independent.
In the second case of no replacement, the second drawing is *dependent* on
the first:
Prob[A and B] = Prob[A]X Prob[B IF A occurs].
In class we will see how tree diagrams are useful to organize one's
reasoning with problems such as these.
Remark: Another useful rule is:
Prob[A] = 1 - Prob[not A].
For instance, if one tosses a die two times in succession, the
probability of getting a sum of at least 3 equals 1 minus the probability
of getting less than 3 (i.e. snake eyes) and so equals
1 - 1/36 = 35/36.
Exercises:
1. Compute the remaining probabilities in the above example (with and
without replacement)
2. Page 186: 1 to 12.
3. Page 192: 1 to 12.