Probability. Probability is a measure of a chance of a
random event.
Some definitions.
A random experiment: An outcome of an experiment which
cannot be predicted in advance.
Sample space: Set of all possible outcomes of an event. It is
designated by S . n(S ) is the number of outcomes in the
sample space S .
Event: A subset of S is an event space E . Example:
S  {1,2,3,4,5,6} and E  {2,4,6} E  S
A simple event: A single point in a sample space. e.g.
getting a 2 on a dice-throw is a simple event.
A compound event: An event subset E with more than 1
simple event in it. The event of occurring an even number
on a throw of dice is a compound event since E  {2,4,6}
Equally likely event: When one outcome is not preferred
over another rationally.
Exhaustive events: When all possible outcomes of an event
are considered.
Mutually exclusive events: The situation where two or
more events cannot occur simultaneously.
Independent or mutually independent events: Two sets of
event E and E' are independent if the occurrence (or
nonoccurrence) of E has no effect on the occurrence (or
nonoccurrence) of E' , and vice versa.
1. Probability of an event E  p ( E ) 
n( E )
n( S )
2. Probability of a E c  1  p( E ) . The E c event space is
complement to E. One could write this to be ~ E as
well.
3. If two events A and B are mutually exclusive,
A  B   , Null set
Therefore, p ( A  B )  0 .
4. For an event E , 0  p( E )  1.
p( E )  0 implies an improbable event
p( E )  1 implies a certain or sure-event.
5. p( E )  p( E c )  p( E )  p(~ E )  1
6. For A, B  S , A  B is an event space where A or B
or both occur (or at least one of A and one of B
occurs). A  B is an event space where both A and B
must occur.
Since n( A  B)  n( A)  n( B)  n( A  B) we get
p( A  B)  p( A  B)  p( A)  p( B)  p( A  B)
7. We also designate p( A  B)  p( AB )
8. Note that for mutually exclusive events A, B, C etc.
n( A  B  C )  n( A)  n( B)  n(C )
Therefore, p( A  B  C )  p( A  B  C ) 
p( A)  p( B)  p(C )
9.
Examples. Problems.
1. A number is selected from 1 to 50. Find the probability that is a prime
number.
Here, S  {1,2,..,50} and n( S )  50
E  {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47} and n( E )  15 .
Therefore, p( E )  15 / 50  3 /10
2. Two dice are thrown simultaneously. What are the probabilities of
getting:
a. two even numbers
b. getting an even number as the sum of both throws
c. getting a multiple of ‘3’ as a sum
The sample space S  {(1,1), (1,2),...(6,6)} with n( S )  36 .
For (a), E  {(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2)...(6,6)} with
n( E )  9 . Therefore, p( E )  1/ 4
For (b), n( E )  18 . For first throw could be any number between 1
and 6, and given the first number, there is only three number for the
second throw which would make the total even. Therefore, p(E ) =0.5
For (c), the sum could be 3, 6, 9, and 12. The event space for this
would be: E  {(1,2), (2,1), (3,3), (1,5), (5,1), (3,6), (6,3), (6,6)} with
n( E )  8 . Therefore, p( E )  8 / 36  2 / 9
3. What is the probability that a leap year selected at random would
contain 53 Sundays?
A leap year would have 366 days, i.e. 52 weeks (52 Sundays) plus two
additional days. These two consecutive days must be one of the
following: {(S,M), (M,T),(T,W),(W,T),(T,F),(F,S),(S,S)} i.e. n( S )  7 .
Of these, only two pairs offer a Sunday. Therefore, n( E )  2 .
Therefore, p(53 _ Sundays)  2 / 7
4. A bag contains 5 red balls, 6 while balls and 7 black balls. What is the
probability of drawing 3 balls such that (a) one draws 1 red and two
whites, (b) no red.
 18 
 5
For (a) and (b), n(S )    .For (a), n( E | 1red _ ball)    and
3 
1 
 6
 5  6 
n( E | 2white _ balls)    . Therefore, n( E )        75 .
 2
1   2 
 18 
Therefore, p( E )  75 /   75 / 816
3 
For (b), 3 balls would be chosen from white and black balls only.
 13 
Therefore, n(E )    . Therefore, p( E )  286 / 816  143 / 408
3 
5. A box contains 12 light bulbs of which 4 are defective. Three bulbs
are chosen randomly. What is the probability that
a. all the three bulbs are defective
b. at least 2 of the bulbs are chosen defective
c. at most 2 of the bulbs chosen are defective.
 12 
The 3 bulbs could be drawn from a set of 12 bulbs in    220 ways.
3 
 4
For (a), all three bulbs are defective. They could be picked up in   ways
3
or 4 ways. Therefore, n( E )  4 / 220  1/ 55
For (b), “at least” two bulbs defective means two or more defectives.
Therefore, we have to find p(two defectives) + p(three defectives).
 4  8
    
2
1
Now, p (two _ defectives )       48 / 220  12 / 55
220
And p(three _ defectives )  1/ 55 . Therefore, p( at least 2 defectives)=13/55.
Similarly, for (c), “at most” two defectives implies events with 1 defective
 4 8 
    
1
2
+ 2 defectives + 0 defective bulb. p (one _ defective )       28 / 55 .
220
 8
 
3
p(0 defective bulb)=    14 / 55 . Thus, p(at most two defective) = 54/55.
220
Conditional probability.
Probability of an event A given that an event B is occurred
is written as p( A | B) . This is the conditional probability of
the event A given B and is given by
p( A | B) 
p( A  B)
p( B)
S
B
A
p( A  B) 
n( A  B )
n( S )
Therefore, p( A | B) 
and p ( B) 
n( B )
n( S )
n( A  B )
n( B )
Example. 1. A box contains 12 bulbs, 4 of which are defectives. 3 bulbs are
drawn at random from the lot. Find the probability that all three are nondefectives.
p
8 7 6
. .
12 11 10
Example. 2. For a biased coin, assume p( H )  2 / 3 and p(T )  1/ 3 . If heads
appears then a number is selected at random from the set {1, 2, …9}. If tails
appears then a number is selected at random form the set of {1,2, 3, 4, 5}.
Find the probability that an even number appears.
p(even _ number)  p(even _ number | H ) p( H )  p(even _ number | T ) p(T )
Now, p(even _ number | H )  4 / 9
p(even _ number | T )  2 / 5
Therefore, p(even _ number) 
4 2 2 1
   
9 3 5 3
Bayes’ theorem:
Given the definition of conditional probability
p( A | B) 
p( A  B)
p( B)
We can write p( A  B)  p( A | B) p( B)  p( B | A) p( A)
Therefore, p( A | B) 
p( B | A)
p( A)
p( B)
This is Bayes’ theorem.
If p(A) = prior probability of event A before it is
associated with B, and
p( A | B)  posterior probability of A given B, then Bayes’
theorem tells us how to change
Prior 
 Posterior
Posterior = Likelihood function  prior
p( A)
p( A)

1  p( A) p(~ A)
We can express the posterior on ~ A via
Another way to look at it. Let O( A) 
p(~ A | B) 
p( B |~ A)
p(~ A)
p( B)
From these two, we get
p( A | B)
P( B | A) P( A)

p(~ A | B) p( B |~ A) p(~ A)
This is
O( A | B) 
p( B | A)
O( A)
p( B |~ A)
If we define ln O( A)  ev( A) = evidence(A), we get the
Bayes’ theorem in an evidence form (useful in court)
 p ( B | A) 
ev( A | B)  ev( A)  ln 

p
(
B
|~
A
)

