Download PROBABILITY NOTES - 1 SAMPLE SPACES AND EVENTS

PROBABILITY NOTES - 1
SAMPLE SPACES AND EVENTS
Sample point and sample space: A sample point is the simple outcome of a random
experiment. The sample space is the collection of all sample points related to a specified
experiment.
Mutually exclusive outcomes: Outcomes are mutually exclusive if they cannot occur
simultaneously. They are also referred to as disjoint outcomes.
Exhaustive outcomes: Outcomes are exhaustive if they combine to be the entire sample
space, or equivalently, if at least one of the outcomes must occur whenever the
experiment is performed.
Event: Any collection of sample points, or any subset of the sample space is referred to
as an event.
Union of events ( and ) : ( r ) denotes the union of events ( and ) , and consists of
all sample points that are in either ( or ) .
Union of events ( Á ( Á ÀÀÀÁ ( : ( r ( r Ä r ( ~ r ( denotes the union of
~
the events ( Á ( Á ÀÀÀÁ ( , and consists of all sample points that are in at least one of the ( 's.
This definition can be extended to the union of infinitely many events.
Intersection of events ( Á ( Á ÀÀÀÁ ( : ( q ( q Ä q ( ~ q ( denotes the
~
intersection of the events ( Á ( Á ÀÀÀÁ ( , and consists of all sample points that are
simultaneously in all of the ( 's. ( q ) is also denoted ( h ) or () .
Mutually exclusive events ( Á ( Á ÀÀÀÁ ( : Two events are mutually exclusive if they
have no sample points in common, or equivalently, if they have empty intersection. Events
( Á ( Á ÀÀÀÁ ( are mutually exclusive if ( q ( ~ J for all £ , where J denotes the empty
set with no sample points.
Exhaustive events ) Á ) Á ÀÀÀÁ ) : If ) r ) r Ä r ) ~ : , the entire sample space, then
the events ) Á ) Á ÀÀÀÁ ) are referred to as exhaustive events.
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Complement of event (: The complement of event ( consists of all sample points in
c
the sample space that are not in (. The complement is denoted (Á — (Á (Z or ( and is
equal to ¸% ¢ % ¤ (¹.
Subevent (or subset) ( of event ) : If event ) contains all the sample points in event
(, then ( is a subevent of ) , denoted ( ‰ ) . The occurrence of event ( implies that event )
has occurred.
Partition of event (: Events * Á * Á ÀÀÀÁ * form a partition of event ( if ( ~ r *
~
and the * 's are mutually exclusive.
DeMorgan's Laws:
(i) ²( r )³Z ~ (Z q ) Z and
Z
$ r ( % ~ ²( r ( r Ä r ( ³Z ~ (Z q (Z q Ä q (Z ~ q (Z
~
~
(ii) ²( q )³Z ~ (Z r ) Z and
Z
$ q ( % ~ ²( q ( q Ä q ( ³Z ~ (Z r (Z r Ä r (Z ~ r (Z
~
~
Indicator function for event (: The function 0( ²%³ ~ .
if %(
if %¤(
is the indicator
function for event (, where % denotes a sample point.
Example 1: Suppose that an "experiment" consists of tossing a six-faced die. The sample space
of outcomes consists of the set ¸Á Á Á Á Á ¹, each number being a sample point representing
the number of spots that can turn up when the die is tossed. The outcomes and (or more
formally, ¸¹ and ¸¹) are mutually exclusive when tossing a die. The outcomes (sample
points) 1 to 6 are exhaustive for the experiment of tossing a die. The collection ¸Á Á ¹
represents the event of tossing an even number when tossing a die. If ( ~ ¸Á Á ¹ and
) ~ ¸Á Á ¹ then ( r ) ~ ¸Á Á Á Á ¹ and ( q ) ~ ¸¹ . The events ( ~ "a number
less than is tossed" ~ ¸Á Á ¹ and
* ~ "a 4 is tossed" ~ ¸¹ are mutually exclusive since they have no sample points in common
Z
- A q * ~ J. If ( ~ ¸Á Á ¹, then ( ~ ¸Á Á ¹.
If + ~ "a 2 is tossed" ~ ¸¹ and ) ~ "an even number is tossed" ~ ¸Á Á ¹, then + ‰ )À
The events , ~ "a 2 or 4 is tossed" ~ ¸Á ¹ and - ~ "a 6 is tossed" ~ ¸
¹ form a partition
of the event ) ~ "an even number is tossed" ~ ¸Á Á ¹. For the die-tossing experiment, if
( ~ ¸Á Á ¹ and ) ~ ¸Á Á ¹ , then (Z ~ ¸Á Á ¹ and ) Z ~ ¸Á Á ¹, and
( r ) ~ ¸Á Á Á Á ¹ , so that ²( r )³Z ~ ¸¹ ~ (Z q )Z .
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Some rules concerning operations on events:
(i) ( q ²) r ) r Ä r ) ³ ~ ²( q ) ³ r ²( q ) ³ r Ä r ²( q ) ³ and
( r ²) q ) q Ä q ) ³ ~ ²( r ) ³ q ²( r ) ³ q Ä q ²( r ) ³ for any events
(Á ) Á ) Á ÀÀÀÁ )
(ii) If ) Á ) Á ÀÀÀÁ ) are exhaustive events ( r ) ~ : , the entire sample space),
~
then for any event (, ( ~ ( q ²) r ) r Ä r ) ³ . As a special case of this, for any
events ( and ) , ( q ²) r )Z ³ ~ ²( q )³ r ²( q )Z ³ ~ ( , so that ( q ) and
( q ) Z form a partition of event (
(iii) For any event (, ( r (Z ~ : , the entire sample space, and ( q (Z ~ J
(iv) ( q ) Z ~ ¸% ¢ %  ( and % ¤ )¹ is sometimes denoted ( c ) , and consists of
all sample points that are in event ( but not in event )
(v) If ( ‰ ) then ( r ) ~ ) and ( q ) ~ ( .
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PROBABILITY
Probability function: 7 ´ µ or denotes the probability that sample point (or outcome³ occurs; 7 must satisfy the following two conditions:
(i)  7 ´ µ  for each in the sample space, and
(ii) 7 ´ µ b 7 ´ µ b Ä ~ 7 ´ µ ~ ; this definition applies to both finite and
infinite sample spaces;
Uniform probability function: If a sample space has a finite number of sample points, say points - Á Á ÀÀÀÁ , then the probability function is said to be uniform if each sample point
has the same probability of occurring - 7 ´ µ ~ for each ~ Á Á ÀÀÀÁ .
Probability of event (: 7 ´(µ ~ 7 ´ µ , the sum of 7 ´ µ over all sample points in event (.
(
Example 2: In tossing a "fair" die, it is assumed that each of the six faces has the same chance
of of turning up. If this is true, then the probability function 7 ²³ ~ for ~ Á Á Á Á Á is a uniform probability function on the sample space ¸Á Á Á Á Á ¹.
The event "an even number is tossed" is ( ~ ¸Á Á ¹, and has probability
7 ´(µ ~ b b ~ . U
Some rules concerning operations on events:
(i) 7 ´Jµ ~ (ii) 7 ´:µ ~ if : is the entire sample space
(iii) if events ( Á ( Á ÀÀÀÁ ( are mutually exclusive then
7 ´ r ( µ ~ 7 ´( r ( r Ä r ( µ ~ 7 ´( µ b 7 ´( µ b Ä b 7 ´( µ ~ 7 ´( µ ;
~
~
this extends to infinitely many mutually exclusive events
(iv) for any event (,  7 ´(µ  (v) if ( ‰ ) then 7 ´(µ  7 ´)µ
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(vi) for any events (, ) and * , 7 ´( r )µ ~ 7 ´(µ b 7 ´)µ c 7 ´( q )µ (since
7 ´(µ b 7 ´)µ counts 7 ´( q )µ twice), and
7 ´( r ) r *µ
~ 7 ´(µ b 7 ´)µ b 7 ´*µ c 7 ´( q )µ c 7 ´( q *µ c 7 ´) q *µ b 7 ´( q ) q *µ ;
(vii) for any events ( Á ( Á ÀÀÀÁ ( , 7 ´ r ( µ  7 ´( µ , with equality holding if
~
~
and only if the events are mutually exclusive
(viii) for any event (, 7 ´(Z µ ~ c 7 ´(µ
(ix) for any events ( and ) , 7 ´(µ ~ 7 ´( q )µ b 7 ´( q )Z µ
(x) for exhaustive events ) Á ) Á ÀÀÀÁ ) , 7 ´ r ) µ ~ ,
~
if ) Á ) Á ÀÀÀÁ ) are exhaustive and mutually exclusive, they form a partition of the
entire sample space, and for any event (,
7 ´(µ ~ 7 ´( q ) µ b 7 ´( q ) µ b Ä b 7 ´( q ) µ ~ 7 ´( q ) µ
~
(xi) if 7 is a uniform probability function on a sample space with points, and if event
( consists of sample points, then 7 ´(µ ~ Example 3: A survey is made to determine the number of households having electric appliances
in a certain city. It is found that 75% have radios (9 ), 65% have irons (0 ), 55% have electric
toasters (; ), 50% have (09 ), 40% have (9; ), 30% have (0; ), and 20% have all three. Find the
probability that a household has at least one of these appliances.
Solution: 7 ´9 r 0 r ; µ ~ 7 ´9µ b 7 ´0µ b 7 ´; µ
c 7 ´9 q 0µ c 7 ´9 q ; µ c 7 ´0 q ; µ b 7 ´9 q 0 q ; µ
~ À b À
b À c À c À c À b À ~ À .
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Example 4: Let 7 ´( q )µ ~ À Á 7 ´(µ ~ À
Á and 7 ´)µ ~ À . Find 7 ´(Z r )Z µ .
Solution: 7 ´(Z r ) Z µ ~ 7 ´²( q )³Z µ ~ c 7 ´( q )µ ~ À .
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CONDITIONAL PROBABILITY AND INDEPENDENCE OF EVENTS
Conditional probability of event ) given event (: If 7 ´(µ € , 7 ´)O(µ ~
7 ´)q(µ
7 ´(µ
is the
conditional probability that event ) occurs given that event ( has occurred.
Bayes rule and Bayes Theorem:
For any events ( and ) with 7 ´(µ € , 7 ´)O(µ ~
7 ´(O)µh7 ´)µ
7 ´(µ
.
If ) Á ) Á ÀÀÀÁ ) form a partition of the entire sample space : , then
7 ´) O(µ ~
7 ´(O) µh7 ´) µ
7 ´(O) µh7 ´) µ
~
for each ~ Á Á ÀÀÀÁ . The values of 7 ´) µ are
called prior probabilities, and the value of 7 ´) O(µ is called a posterior probability.
Independent events ( and ) : If events ( and ) satisfy the relationship
7 ´( q )µ ~ 7 ´(µ h 7 ´)µ , then the events are said to be independent or stochastically
independent or statistically independent. The independence of (non-empty) events ( and ) is
equivalent to 7 ´(O)µ ~ 7 ´(µ or 7 ´)O(µ ~ 7 ´)µ .
Mutually independent events ( Á ( Á ÀÀÀÁ ( : If events ( Á ( Á ÀÀÀÁ ( satisfy the
relationship 7 ´( q ( q Ä q ( µ ~ 7 ´( µ h 7 ´( µÄ7 ´( µ ~ 7 ´( µ , then
~
the events are said to be mutually independent.
Example 5: Suppose the die-tossing experiment is considered again. The sample space is
: ~ ¸Á Á Á Á Á ¹. We define the following events:
( ~ "the number tossed is  " ~ ¸Á Á ¹
) ~ "the number tossed is even" ~ ¸Á Á ¹
* ~ "the number tossed is a or a " ~ ¸Á ¹
+ ~ "the number tossed doesn't start with the letters 'f' or 't'" ~ ¸Á ¹
The conditional probability of ( given ) is
7 ´¸ÁÁ¹q¸ÁÁ
¹µ
7 ´¸ÁÁ
¹µ
7 ´¸¹µ
°
~ 7 ´¸ÁÁ
¹µ ~ ° ~ .
Events ( and ) are not independent, since ~ 7 ´( q )µ £ 7 ´(µ h 7 ´)µ ~ h ~ ,
or alternatively, events ( and ) are not independent since 7 ´(O)µ £ 7 ´(µ.
7 ´(O)µ ~
7 ´(O*µ ~ £ ~ 7 ´(µ, so that ( and * are not independent.
7 ´)O*µ ~ ~ 7 ´)µ , so that ) and * are independent
(alternatively, 7 ´) q *µ ~ 7 ´¸¹µ ~ ~ h ~ 7 ´)µ h 7 ´*µ).
Both ( and ) are independent of +.
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Some rules concerning conditional probability and independence are:
(i) 7 ´( q )µ ~ 7 ´)O(µ h 7 ´(µ ~ 7 ´(O)µ h 7 ´)µ for any events ( and )
(ii) If ) Á ) Á ÀÀÀÁ ) form a partition of the sample space : , then for any event (,
7 ´(µ ~ 7 ´(O) µ h 7 ´) µ ;
~
as a special case, for any events ( and ) , 7 ´(µ ~ 7 ´(O)µ h 7 ´)µ b 7 ´(O)Z µ h 7 ´)Z µ
(iii) If 7 ´( q ( q Ä q (c µ € , then
7 ´( q ( q Ä q ( µ ~ 7 ´( µ h 7 ´( O( µ h 7 ´( O( q ( µÄ7 ´( O( q ( q Ä q (c µ
(iv) 7 ´(Z O)µ ~ c 7 ´(O)µ
(v) if ( ‰ ) then 7 ´(O)µ ~
7 ´(q)µ
7 ´)µ
7 ´(µ
~ 7 ´)µ , and 7 ´)O(µ ~ (vi) if ( and ) are independent events then (Z and ) are independent events,
( and ) Z are independent events, and (Z and ) Z are independent events
(vii) since 7 ´Jµ ~ 7 ´J q (µ ~ ~ 7 ´Jµ h 7 ´(µ for any event (, it follows that J is
independent of any event (
IMPORTANT NOTE: The following manipulation of event probabilities arises from time to
c
c
time: 7 ´)µ ~ 7 ´)O(µ h 7 ²(³ b 7 ´)O(µ h 7 ²(³ . If we are given conditional probabilities for
c
event ) given some other event ( and its complement (, and if we are given the (unconditional)
probability of event (, then we can find the probability of event ). One of the important aspects
of applying this relationship is the determination of the appropriate events ( and ) .
Example 6: If 7 ´(µ ~ and 7 ´)µ ~ , and 7 ´(O)µ b 7 ´)O(µ ~ ,
find 7 ´( q )µ .
Solution: 7 ´)O(µ ~
7 ´(q)µ
7 ´(µ
~ 7 ´( q )µ and 7 ´(O)µ ~
S ²
b ³ h 7 ´( q )µ ~ S 7 ´( q )µ ~ 2 À
7 ´(q)µ
7 ´)µ
~ 7 ´( q )µ
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Example 7: Three dice have the following probabilities of throwing a "six": Á Á Á
7
respectively. One of the dice is chosen at random and thrown (each is equally likely to be
chosen). A "six" appeared. What is the probability that the die chosen was the first one?
Solution: The event " a 6 is thrown" is denoted by "6"
7 ´die O"
"µ ~
7 ´²die ³q²"
"³µ
7 ´"
" µ
~
7 ´"
"Odie µh7 ´die µ
7 ´"
"µ
~
h 7 ´"
"µ
.
But 7 ´"
"µ ~ 7 ´²"
"³ q ²die ³µ b 7 ´²"
"³ q ²die ³µ b 7 ´²"
"³ q ²die ³µ
~ 7 ´"
"Odie µ h 7 ´die µ b 7 ´"
"Odie µ h 7 ´die µ b 7 ´"
"Odie µ h 7 ´die µ
h h bb
~ h b h b h ~ S 7 ´die O"
"µ ~ 7 ´"
"µ ~
~ bb . U
²bb³h Example 8: Identical twins come from the same egg and hence are of the same sex. Fraternal
twins have a 50-50 chance of being the same sex. Among twins, the probability of a fraternal set
is and an identical set is ~ c . If the next set of twins are of the same sex, what is the
probability that they are identical?
Solution: Let ( be the event "the next set of twins are of the same sex", and let ) be the event
"the next sets of twins are identical". We are given 7 ´(O)µ ~ Á 7 ´(O) Z µ ~ À
7 ´)µ ~ Á 7 ´) Z µ ~ ~ c . Then 7 ´)O(µ ~
7 ´(q)µ
7 ´(µ
.
But 7 ´( q )µ ~ 7 ´(O)µ h 7 ´)µ ~ , and 7 ´( q )Z µ ~ 7 ´(O)Z µ h 7 ´)Z µ ~ À .
Thus, 7 ´(µ ~ 7 ´( q )µ b 7 ´( q )Z µ ~ b À ~ b À² c ³ ~ À² b ³ , and
7 ´)O(µ ~ À²b³ .
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Example 9: Let events ( and ) be independent. Find the probability, in terms of 7 ´(µ and
7 ´)µ, that exactly one of the events ( and ) occurs.
Solution: 7 ´exactly one of ( and )µ ~ 7 ´²( q ) Z ³ r ²) q (Z ³µ .
Since ( q )Z and ) q (Z are mutually exclusive, it follows that
7 ´exactly one of ( and )µ ~ 7 ´( q ) Z µ b 7 ´) q (Z µ .
Since ( and ) are independent, it follows that ( and ) Z are also independent, as are ) and (Z .
Then
7 ´²( q ) Z ³ r ²) q (Z ³µ ~ 7 ´(µ h 7 ´)Z µ b 7 ´)µ h 7 ´(Z µ
~ 7 ´(µ² c 7 ´)µ³ b 7 ´)µ² c 7 ´(µ³ ~ 7 ´(µ b 7 ´)µ c 7 ´(µ h 7 ´)µ U
COMBINATORIAL PRINCIPLES - PERMUTATIONS AND COMBINATIONS
8
Factorial notation: [ denotes the quantity ² c ³² c ³Ä h ;
[ is defined to be equal to .
Permutations:
(a) Given distinct objects, the number of different ways in which the objects may be ordered
(permuted) is [. For example, the set of 3 letters ¸Á Á ¹ can be ordered in [ ~ ways Á Á Á Á Á .
The number of ways of choosing an ordered subset of size from objects, without replacement
(i.e., after the first object is chosen, the next object is chosen from the remaining c , the next
after that from the remaining c , etc.) is
[
²c³[
~ h ² c ³ h IJ c b ³ , and is denoted 7 or 7Á or 7 ²Á ³.
Using the set ¸Á Á ¹ again, the number of ways of choosing an ordered subset of size is
[
²c³[ ~ ~ - Á Á Á Á Á .
(b) Given objects, of which are of Type 1, are of Type 2, . . . , and ! are of Type !
(! ‚ is an integer), and ~ b b Ä b ! , the number of ways of ordering all objects
(where objects of the same Type are indistinguishable) is
[
[h [Ä! [
, which is sometimes denoted $ Ä %.
!
Combinations:
(a) Given distinct objects, the number of ways of choosing a subset of size  without
replacement (and without regard to the order in which the objects are chosen) is
[
, which is usually denoted " #(or * , *Á or *²Á ³ ) and is read
[h²c³[
" choose ".
"#
is also called a binomial coefficient (and can be defined for any real number
and non-negative integer ). Note that if is an integer and is a non-negative integer, then
" # ~ ²c³Ä²cb³ , and " # ~ " # ~ , and
[
" # ~ " # ~ , and " # ~ " #.
c
c
Using the set ¸Á Á ¹ again, the number of ways of choosing a subset of size without
[
replacement is $ % ~ [h²c³[
~ - these being ¸Á ¹ , ¸Á ¹ Á ¸Á ¹ .
(b) Given objects, of which are of Type 1, are of Type 2, . . . , and ! are of Type !
(! ‚ is an integer), and ~ b b Ä b ! , the number of ways of
choosing a subset of size  (without replacement). with objects of Type 1, objects of
Type 2,..., and ! objects of Type !, where ~ b b Ä b ! is
$ % h $ %Ä$ ! %
!
Binomial Theorem: In the power series expansion of ² b !³5 , the coefficient of ! is
9
B
5
5
$ %, so that ² b !³5 ~ $ % h ! ~ b 5 ! b
~
5 ²5 c³ !
b
5 ²5 c³²5 c³ !
bÄ
If 5 is an integer, then the summation stops at ~ 5 and the series is valid for any real number
!, but if 5 is not an integer, then the series is valid if O!O  .
Multinomial Theorem: In the power series expansion of ²! b ! b Ä b ! ³5 where 5 is a
positive integer, the coefficient of ! h ! Ä! (where b b Ä b ~ 5 )
5
[
is $ Ä % ~ [h5[Ä
[ .
Special note: In questions involving coin flips or dice tossing, it is understood, unless indicated
otherwise, that successive flips or tosses are independent of one another.
In making a random selection of an object from a collection of objects, it is understood that
each object has the same chance of being chosen, . In questions that arise involving choosing objects at random from a total of objects, or in constructing a random permutation of a
collection of objects, it is understood that each of the possible choices or permutations is equally
likely to occur. For instance, if a purse contains one quarter, one dime, one nickel and one
penny, and two coins are chosen, there are $ % ~ possible ways of choosing two coins
without regard to order of choosing - these are Q-D , Q-N , Q-P , D-N , D-P , N-P (the
choice Q-D is regarded as the same as D-Q, etc.). It would be understood that each of the possible ways are equally likely, and each has (uniform) probability of of occurring - the
sample space would consist of the ways of choosing, and each sample point would have
probability . Then, the probability of a particular event occurring would be , where is the
number of sample points in the event. If ( is the event "one of the coins is either a quarter or a
dime", then 7 ´(µ ~ , since event ( consists of the of the sample points
¸Q-D , Q-N , Q-P , D-N , D-P ¹.
Example 10: An ordinary die and a die whose faces have Á Á Á Á Á Á dots are tossed
independently of one another, and the total number of dots on the two dice is recorded as 5 .
Find the probability that 5 ‚ .
Solution: It is assumed that for each die, each face has a probability of turning up. If the
number of dots turning up on die 1 and die 2 are and , respectively, then the tosses that
result in 5 ~ b ‚ are ²Á ³ Á ²Á ³ Á ²Á ³ Á ²Á ³ Á ²Á ³ Á ²Á ³ Á
²Á ³ Á ²Á ³ Á ²Á ³ Á ²Á ³ Á ²
Á ³ Á ²
Á ³ Á ²
Á ³ Á ²
Á ³ - combinations out of a total of
d ~ combinations that can possibly occur. Since each of the ² Á ³
combinations is equally likely, the probability is U
.
Example 11: Three nickels, one dime and two quarters are in a purse. In picking three
10
coins at one time (without replacement), what is the probability of getting a total of at
least 35 cents?
Solution: In order to get at least 35 cents, at least one quarter must be chosen. The possible
choices are 1Q b any 2 of the non-quarters , or 2Q b any 1 of the non-quarters.
The total number of ways of choosing three coins from the six coins is $ % ~ .
If we label the two quarters as Q and Q , then the number of ways of choosing the three coins
so that only Q (and not Q ³ is in the choice is $ % ~ (this is the number of ways of choosing
the other two coins from the three nickels and one dime) - and therefore, the number of choices
that contain only Q (and not Q ) is also 6. The number of ways of choosing the three coins so
that both Q and Q are in the choice is (this is the number of ways of choosing the other coin
from the three nickels and one dime). Thus, the total number of choices for which at least one of
the three coins chosen is a quarter is 16. The probability in question is .
Alternatively, the number of three coin choices that do not contain any quarters is $ % ~ (the
number of ways of choosing the three coins from the non-quarters), so that number of choices
that contain at least one quarter is c ~ .
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Example 12: A and B draw coins in turn without replacement from a bag containing 3 dimes
and 4 nickels. A draws first. It is known that A drew the first dime. Find the probability that A
drew it on the first draw.
Solution: 7 ´( draws dime on first drawO( draws first dimeµ ~
7 ´( draws dime on first drawµ ~ .
7 ´( draws dime on first drawµ
7 ´( draws first dimeµ
Since there only 3 dimes, in order for A to draw the first dime, this must happen on A's first,
second or third draw. Thus,
7 ´( draws first dimeµ ~ 7 ´( draws dime on first drawµ
b 7 ´( draws first dime on second drawµ b 7 ´( draws first dime on third drawµ .
7 ´( draws dime on second drawµ ~ h h ~ , since A's first draw is one of the four nondimes, and B's first draw is one of the three remaining non-dimes after A's draw, and A's second
draw is one of the three dimes of the five remaining coins. In a similar way, 7 ´( draws first
dime on third drawµ ~ h h h h ~ .
Then, 7 ´( draws first dimeµ ~ b b ~ , and
°
7 ´( draws dime on first drawO( draws first dimeµ ~ ° ~ . U
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Example 13: Three people, X, Y and Z, in order, roll an ordinary die. The first one to roll an
even number wins. The game continues until someone rolls an even number. Find the
probability that X will win.
Solution: Since X rolls first, fourth, seventh, etc. until the game ends, the probability that X will
win is the probability that in throwing a die, the first even number will occur on the 1st, or 4th, or
7th, or . . . throw. The probability that the first even number occurs on the -th throw is
² ³c ² ³ ~ ( c odd throws followed by an even throw; because
of independence of successive throws, with ( ~ "throw is even", the probability that the first
even throw occurs on throw is
7 ´(Z q (Z q Ä q (Zc q ( µ ~ 7 ´(Z µ h 7 ´(Z µÄ7 ´(Zc µ h 7 ´( µ
~ ² ³c ² ³ ~ ).
Thus, 7 ´first even throw is on 1st, or 4th, or 7th, or ...µ
~ b b b Ä ~ ² b b b ij ~ .
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Example 14: Urn I contains 7 red and 3 black balls, and Urn II contains 4 red and 5 black balls.
After a randomly selected ball is transferred from Urn I to Urn II, 2 balls are randomly drawn
from Urn II without replacement. Find the probability that both balls drawn from Urn II are red.
Solution: Define the following events:
9 : the ball transferred from Urn I to Urn II is red
) : the ball transferred from Urn I to Urn II is black
9 : two red balls are selected from Urn II after the transfer from Urn I to Urn II .
Since 9 and ) are mutually exclusive,
7 ´9 µ ~ 7 ´9 q ²9 r ) ³µ ~ 7 ´9 q 9 µ b 7 ´9 q ) µ
$ % $ % ~ 7 ´9 O9 µ h 7 ´9 µ b 7 ´9 O) µ h 7 ´) µ ~
$ % h b $ % h ~ . U
Example 15: A calculator has a random number generator button which, when pressed, displays
a random digit Á Á ÀÀÀÁ . The button is pressed four times. Assuming that the numbers
generated are independent of one another, find the probability of obtaining one "0", one "5", and
two "9"'s in any order.
Solution: There are ~ Á four-digit orderings that can arise, from 0-0-0-0 to
9-9-9-9 . From the notes above on permutations, if we have four digits, with one "0", one "5"
[
and two "9"'s, the number of orderings is [h[h[
~ . The probability in question is then
Á
.U
12