MATH 131 (062) Finite Mathematics. Chapter 8: Probability. June 3, 2007.
Dr. Raja Latif and Mohammad latif and Abdul Latif and Dr. Raja Mohammad Abdul Latif
(a) At one time, licence plates were issued that consisted of three letters followed by three digits. How many
different licence plates can be issued under this arrangement?
(b) Later on, licence plates were issued that consisted
of three digits followed by three letters. How many different licence plates can be issued under this arrangement?
Solution. (a) The number of licence plate numbers that may be formed is (26) (26) (26) (10) (10) (10) , or
17, 576, 000.
(b) The number licence plate numbers that may be
formed is (10) (10) (10) (26) (26) (26) = 17, 576, 000.
Example.360TAN20. Licence Plate Numbers: In
recent years the state of California issued licence plates
using a combination of one letter of the alphabet followed by three digits, followed by another three letters of the alphabet. How many different licence plates
can be issued using this configuration?
Solution.
(26) (10) (10) (10) (26) (26) (26) = 316, 342, 000.
Example.360TAN21. Exams: An exam consists of
ten true-or-false questions. Assuming that every question
is answered, in how many different ways can a student
complete the exam?
In how many ways the exam be completed if a penalty
is imposed for each incorrect answer, so that a student may
leave some questions unanswered?
Solution. If every question is answered, there are
210 , or 1024, ways.
In the second case, there are 3 ways to answer each
question, and so we have 310 , or 59, 049, ways.
Example.360TAN22. WARRANTY NUMBER: A
warranty identification number for a certain product consists of a letter of the alphabet followed by a five-digit
number. How many possible identification numbers are
there if the first digit of the five-digit number must be
nonzero?
Solution. (26) (9) (10) (10) (10) (10)
= 2, 340, 000.
Example.360TAN24.TELEPHONE NUMBERS: (a)
How many seven-digit telephone numbers are possible if
the first digit must be nonzero?
(b) How many international direct-dialing numbers
are possible if each number consists of a three-digit area
code (the first digit of which must be nonzero) and a number of the type described in part (a)?
Solution. (a) If the first digit must be nonzero,
then there are 9 possible choices for the first digit and
10 possible choices for the remaining 6 digits. Thus,
the number of possible seven-digit numbers is given by
(9) (10) (10) (10) (10) (10) (10) = 9, 000, 000.
Contents
8.1-2:Basic Counting Principle and Permutations;Combinations and Other Counting Principle
Example.360TAN9. Health-Care Plan Options: A
new state employee is offered a choice of ten basic health
plans, three dental plans, and two vision care plans. How
many different health care plans are there to choose from
each category?
Solution. By the multiplication principle, we see
that the number of ways a health-care plan be selected
is (10) (3) (2) = 60.
Example.360TAN10.Code Words: How many three
code words can be constructed from the first ten letters of
the Greek alphabet if no repetitions are allowed?
Solution. Using the multiplication principle, we see
that the number of three-letter code words that can be
formed is
(10) (9) (8) = 720, or 720 ways.
Example.360TAN11. Social Security Number: A social security number has nine digits. How many Social
Security numbers are possible?
Solution. 109 = 1, 000, 000, 000.
Example.360TAN12. Serial Numbers: Computers
manufactured by a certain company have a serial number consisting of a letter of the alphabet followed by a
four-digit number. If all the serial numbers of this type
have been used, how many sets have already been manufactured?
Solution. The number of sets that have already been
manufactured is (26) (10) (10) (10) (10) = 260, 000.
Example.360TAN14. Automobile Selection: An automobile manufacturer has three different subcompact cars
in the line. Customers selecting one of these cars have a
choice of three engine sizes, four body styles, and three
color schemes. How many different selections can a customer make?
Solution. The number of selections a customer can
make is (3) (3) (4) (3) = 108.
Example.360TAN17. ATM CARDS: To gain access
to his account, a customer using an automatic teller machine (ATM) must enter a four-digit code. If repetition
of the same four digits is not allowed (for example, 5555),
How many combinations are there?
Solution. The number of different selections is
(10) (10) (10) (10) в€’ 10 = 10000 в€’ 10 = 9990.
Example.360TAN19. Licence Plate Numbers: Over
the years the state of California has used different combinations of letters of the alphabet and digits on its automobile licence plates.
1
(b) Using the multiplication principle, we find that the
number of international directdialing numbers is given by
(9) (10) (10) Г— 9, 000, 000 = 8, 100, 000, 000.
Example.373TAN32. How many three-letter permutations can be formed from the first five letters of the alphabet?
Solution. The number of 3в€’ letter permutations is
5!
P (5, 3) =
= 5.4.3 = 60.
2!
Example.373TAN34. In how many ways can five
people line up at a checkout counter in a supermarket?
Solution. The number of different ways they can line
5!
= 120.
up is P (5, 5) =
0!
Example.373TAN37. In how many ways can a member of a hiring committee select 3 of 12 job applicants for
further consideration?
Solution. The number of different ways the 3 candi12.11.10
12!
=
= 220.
dates can be selected is C (12, 3) =
9!3!
3.2.1
Example.373TAN40. Find the number of distinguishable permutations that can be formed from the letters
of the word PHILIPPINES.
Solution.
There are 11 letters in the word
P HILIP P IN ES, 3P s, 1H, 3Is, 1L, 1N, 1E, and 1S.
Therefore, we use the formula for the permutations
n!
11!
of n objects, not all distinct:
=
=
n1 !n2 !...nr !
3!3!
1, 108, 800.
Example.373TAN42. BOOK SELECTION: A student is given a reading list of ten books from which he
must select two for an outside reading requirement. In
how many ways can he make his selections?
Solution. The number of ways the student can select
10!
10.9
the two books is C (10, 2) =
=
= 45.
8!2!
2.1
Example.373TAN46. Seven people arrive at the
ticket counter of a travel agency at the same time. In
how many ways can they line up to purchase their tickets?
Solution. The number of ways they can line up to
purchase their ticket is P (7, 7) = 7! = 5040.
Example.374TAN58. A student taking an examination is required to answer 10 out of 15 questions.
(a) In how many ways can the 10 questions be selected?
(b) In how many ways can the 10 questions be selected
if exactly 2 of the first 3 questions must be answered?
Solution. (a) The number of ways that the 10 questions can be selected from the 15 questions is
15!
C (15, 10) =
= 3003.
10!5!
(b) The number of ways the 10 questions can be selected if exactly 2 of the first 3 questions must be answered
3! 12!
is C (3, 2) .C (12, 8) =
.
= 1485.
1!2! 4!8!
Example.381TAN33. In how many ways can six different compact discs be arranged on a shelf?
Solution. The number of ways the compact discs
can be arranged on a shelf is P (6, 6) = 6! = 720 ways.
Example.381TAN36. How many three-digit numbers can be formed from the numerals in the set
{1, 2, 3, 4, 5} if: (a) Repetitions of digits is not allowed?
(b) Repetitions of digits is allowed?
Solution. (a) If repetition is not allowed, the number
5!
of three-digit numbers is P (5, 3) =
= 60.
2!
(b) If repetition is allowed, the number of three-digit
numbers is (5) (5) (5) = 125.
Example.381TAN37. Find the number of distinguishable permutations that can be formed from the letters
of each word. (a) CINCINNATI
(b) HONOLULU
Solution. (a) Since there is repetition of the letters
C, I, and N, we use the formula for the permutation of
n objects, not all distinct, with n = 10, n1 = 2, n2 = 3,
and n3 = 3. Then the number of permutations that can
10!
be formed is given by
= 50, 400.
2!3!3!
8!
(b) Similarly as in (a) ,
= 5040.
2!2!2!
Example.381TAN40. There are eight seniors and six
juniors in the Math Club at Jefferson High School. In
how many ways can a math team consisting of four seniors and two juniors be selected from the members of the
Math Club? Solution. The math team can be selected in
8! 6!
.
= 1050 ways.
C (8, 4) .C (6, 2) =
4!4! 4!2!
Example.381TAN41. A sample of 4 balls is to be
selected from an urn containing 15 balls numbered 1 to
15. If 6 balls are green, 5 are white, and 4 are black:
(a) How many different samples can be selected?
(b) How many samples can be selected that contain
at least 1 white ball?
Solution. (a) The number of samples can be selected
15!
is C (15, 4) =
= 1365.
4!11!
10!
(b) There are C (10, 4) =
= 210
4!6!
ways of selecting 4 balls none of which are white.
Therefore, there are 1365 в€’ 210, or 1155 ways of selecting 4 balls of which at least one is white.
Example.381TAN42. From a shipment of 60 transistors, 5 of which are defective, a sample of 4 transistors is
selected at random.
(a) In how many different ways can the sample be
selected?
(b) How many samples contain 3 defective transistors?
(c) How many samples do not contain any defective
transistors?
Solution. (a) The number of different ways the sample can be selected is given by
60!
C (60, 4) =
= 487, 365.
56!4!
(b) The number of samples that contain 3 defective
transistors is given by C (5, 3) .C (55, 1)
5! 55!
=
.
= 10 (55) = 550.
3!2! 54!1!
(c) The number of samples that do not contain any
defective transistors is given by
55!
C (55, 4) =
= 341, 055.
51!4!
2
Example.391TAN27. GAME SHOWS: In a television show, the winner is asked to select three prizes from
five different prizes, A, B, C, D and E.
(a) Describe a sample space of possible outcomes (order is not important).
(b) How many points are there in the sample space
corresponding to a selection that includes A?
(c) How many points are there in the sample space to
a selection that includes A and B?
(d) How many points are there in the sample space
correspondingВЅto a selection that includes either A
Вѕ or B?
ABC, ABD, ABE, ACD, ACE,
Solution.(a)
ADE, BCD, BCE, BDE, CDE
(b) 6
(c) 3
(d) 6
8.3: SAMPLE SPACES AND EVENTS
Example.390TAN7-12: Let S = {1, 2, 3, 4, 5, 6} ,
E = {2, 4, 6} , F = {1, 3, 5} , and G = {5, 6} .
(Q7 ) Find the event E в€Є F в€Є G.
Solution.E в€Є F в€Є G = {2, 4, 6} в€Є {1, 3, 5} в€Є {5, 6}
= {1, 2, 3, 4, 5, 6}
(Q8 ) Find the event E ∩ F ∩ G.
Solution. E ∩ F ∩ G =
{2, 4, 6} ∩ {1, 3, 5} ∩ {5, 6} = Φ.
c
(Q9 ) Find the event (E в€Є F в€Є G) .
c
Solution. {1, 2, 3, 4, 5, 6} = О¦.
(Q10 ) Find the event (E ∩ F ∩ G)c .
Solution.(E ∩ F ∩ G)c = {1, 2, 3, 4, 5, 6}
(Q11 ) Are the events E and F mutually exclusive?
8.4: PROBABILITY
Solution. Yes, E ∩ F = Φ, that is, E and F do not
contain any common elements.
Example.402TAN20. QUALITY CONTROL: One
(Q12 ) Are the events F and G mutually exclusive?
light bulb is selected at random from a lot of 120 bulbs, of
Solution. No. 5 is an element of both sets.
which 5% are defective. What is the probability that the
Example.391TAN13-18. Let S be any sample space
light bulb selected is defective?
and E, F, and G be any three events associated with the
Solution.The probability that a defective bulb is choexperiment. Describe the given events using the symbols
6
∪, ∩, and c .
sen is
= 0.05.
120
(Q13 ) The event that E and /or F occurs.
Example.402TAN22. If a ball is selected at random
Solution. E в€Є F
from an urn containing three red balls, two white balls,
(Q14 ) The event that both and F occur.
and five blue balls, what is the probability that it will be
Solution. E ∩ F
a white ball?
(Q15 ) The event that G does not occur.
Solution. The probability that it will be a white ball
Solution. Gc
2
2
=
= 0.2.
is P (W ) =
(Q16 ) The event that E but not F occurs.
3
+
2
+
5
10
c
Solution. (E ∩ F ) . (Q17 ) The event that none of
Example.402TAN24. A pair of fair dice is cast.
c
the events E, F, and G occurs. (E в€Є F в€Є G) Solution.
What is the probability that (a) The sum of the num(Q18 ) The event that E occurs but neither of the events
bers shown uppermost is less than 5? (b) At least
F or G occurs. Solution. (E ∩ F c ∩ Gc )
one
space S =
ВЅ 6 is cast? Solution. The sample
Вѕ
(1, 1) , (1, 2) , ..., (1, 6) , (2, 1) , (2, 2) ,
Example.391TAN20. Let S = {1, 2, 3} be a sample
..., (2, 6) , ..., (6, 1) , (6, 2) , ..., (6, 6)
space associated with an experiment.
(a) The required event is
(a) List all events of this experiment.
E = {(1, 1) , (1, 2) , (1, 3) , (2, 1) , (2, 2) , (3, 1)}and so
(b) How many subsets of S contain the number 3?
P (E) = P [{(1, 1)}] + P [{(1, 2)}] + ... + P [{(3, 1)}]
Solution.(a) П†, {1} , {2} , {3} , {1, 2} ,
1
1
1
1
{1, 3} , {2, 3} , {1, 2, 3} . (b) 4
(c) 4
=
+
+ ... +
=
(six terms)
36 36
36
6
Example.391TAN22. An experiment consists of se(b) The required event is
lecting a letter at random from the letters in the word
F
= {(1, 6) , (2, 6) , ..., (5, 6) , (6, 1) , ..., (6, 6)} ,
M ASSACHU SET T S and observing the outcomes.
and
so P (F ) = P [{(1, 6)}] + ... + P [{(6, 6)}]
(a) What is an appropriate sample space for this ex1
1
1
11
periment?
=
+
+ ... +
=
(eleven terms)
36 36
36
36
(b) Describe the event ”the letter selected is a vowel”.
Example.402T7A2N25. TRAFFIC LIGHTS: What
Solution. (a) {A, C, E, H, M, S, T, U }
is the probability of arriving at a traffic light when it is red
(b) {A, E, U }
if the red signal is flashed for 30 sec, and the green signal
Example.391TAN23. An experiment consists of tossfor 45 sec?
ing a coin and casting a die and observing the outcomes.(a)
Solution. The probability of arriving at the traffic
Describe an appropriate sample space for this experiment.
30
30
=
= 0.375.
light when it is red is
(b) Describe the event ”a head is tossed and an even num30 + 5 + 45
80
Example.403TAN37. Let S = {s1 , s2 , s3 , s4 , s5 , s6 }
ber is cast”. ½
Вѕ
(H, 1) , (H, 2) , (H, 3) , (H, 4) , (H, 5) , (H, 6) ,
be the sample space associated with an experiment
Sol.:(a)
(T, 1) , (T, 2) , (T, 3) , (T, 4) , (T, 5) , (T, 6)
having the following probability distribution
(b) E = {(H, 2) , (H, 4) , (H, 6)}
Outcome
{s1 } {s2 } {s3 } {s4 } {s5 } {s6 }
1
1
1
1
1
1
Probability
12
4
12
6
3
12
3
1
2
=
3
3
(f ) P (Ac ∪ B c ) = P (Ac ) + P (B c ) − P (Ac ∩ B c ) =
13 11 1
2
+
в€’ = .
24 24 3
3
Example.421TAN17. An exam consists of ten trueor-false questions. If a student guesses at every answer,
what is the probability that he she will answer exactly six
questions correctly? Solution. The number of elements in
10!
, or 210
the sample space is 210 . There are C (10, 6) =
6!4!
ways of answering exactly six questions correctly. There210
210
= 0.205.
fore, the required probability is 10 =
2
1024
Example.421TAN19. QUALITY CONTROL: Two
light bulbs are selected at random from a lot of 24, of
which 4 are defective. What is the probability that:
(a) Both of the light bulbs are defective?
(b) At least 1 of the light bulbs is defective?
Solution. (a) Let E denote the event that both
C (4, 2)
=
of the bulbs are defective. Then P (E) =
C
(24, 2)
В¶ Вµ
В¶
Вµ
24!
4.3
4!
/
=
= 0.022.
2!2!
22!2!
24.23
(b) Let F denote the event that none of the bulbs
20! 22!2!
C (20, 2)
=
.
=
are defective. Then P (F ) =
C (24, 2)
18!2! 24!
20 19
.
= 0.6884. Therefore, the probability that at least
24 23
one of the light bulbs is defective is given by 1 в€’ P (F ) =
1 в€’ 0.6884 = 0.3116.
Example.421T7A4N20. A customer at Farm 5 store
picks a sample of 3 oranges at random from a crate containing 60 oranges, of which 4 are rotten. What is the
probability that the sample contains 1 or more rotten oranges? Solution. Let E denote the event that there is
at least one rotten orange in the sample. Then E is the
event that there are no rotten oranges in the sample and
C (56, 3)
166, 320
P (E c ) =
=
≈ 0.81,
C (60, 3)
205, 320
and P (E) = 1 в€’ P (E c ) = 1 в€’ 0.81 = 0.19.
Example.421TAN25. A student studying for a vocabulary test knows the meaning of 12 words from a list
of 20 words. If the test contains 10 words from the study
list, what is the probability that at least 8 of the words on
the test are words that the student knows?
Solution. The probability is given by
C (12, 8) .C (8, 2)
C (12, 9) C (8, 1)
C (12, 10)
+
+
=
C (20, 10)
C (20, 10)
C (20, 10)
(28) (495) + (220) (8) + 66
≈ 0.085.
184, 756
Example.418TAN4. Five people are selected at random. What is the probability that (a) none of the people in
this group have the same birthday?(b) at least two of them
have the same birthday? Solution.(a) We see that in a
group of 5 people,the probability that none of the people
365.364.363.362.361
will have same birthday is
= 0.973
365.365.365.365.365
(b) The probability that at least two of them will have
the same birthday is 1 в€’ 0.973 = 0.027.
Find the probability of the event:
(a) A = {s1 , s2 } , (b) B = {s2 , s4 , s5 , s6 } , (c) C = S.
1
1
1
Solution.(a) P (A) = P (s1 ) + P (s2 ) =
+
=
12 12
6
(b) P (B) = P (s2 ) + P (s4 ) + P (s5 ) + P (s6 )
1 1 1
1
5
= + + +
= . (c) P (C) = 1.
4 6 3 12
6
Example.403TAN39. Consider the composition of
a three-child family in which the children were born at
different times. Assume that a girl is as likely as a boy at
each birth. What is the probability that:
(a) There are two girls and a boy in the family?
(b) The oldest child is a girl?
(c) The oldest child is a girl and the youngest child
is a boy? Solution. Let G denote a female birth and
let B denote a male birth. Then the eight equally likely
outcomes of this experiment are
GGG, GGB, GBG, BGG, BGB, BBG, GBB, BBB.
(a) The event that there are two girls and a boy in the
family is E = {GGB, GBG, BGG} . Since there are three
favorable outcomes, P (E) = 3/8.
(b) The event that the oldest child is a girl is
F = {GGG, GGB, GBG, GBB} .
Since there are 4 favorable outcomes, P (F ) = 1/2.
(c) The event that the oldest child is a girl and the
youngest child is a boy is G = {GGB, GBB} . Since there
are two favorable outcomes, P (F ) = 1/4.
Example.411TAN25. Let E and F be two events
that are mutually exclusive and suppose P (E) = 0.2 and
P (F ) = 0.5. Compute: (a) P (E ∩ F )
(b) P (E в€Є F )
(c) P (E c )
(d) P (E c ∩ F c )
Solution. (a) P (E ∩ F ) = 0 since E and F are mutually exclusive.(b) P (E ∪ F ) =
P (E) + P (F ) − P (E ∩ F ) = 0.2 + 0.5 = 0.7.
(c) P (E c ) = 1 в€’ P (E) = 1 в€’ 0.6 = 0.4.
(d) P (E c ∩ F c ) = P [E c ∩ F c ] = P [(E ∪ F )c ]
= 1 в€’ P (E в€Є F ) = 1 в€’ 0.7 = 0.3.
Example.412TAN28. Let S = {s1 , s2 , s3 , s4 , s5 , s6 }
be the sample space associated with an experiment
having the following probability distribution
Outcome
{s1 } {s2 } {s3 } {s4 } {s5 } {s6 }
1
1
1
1
1
1
Probability
3
8
6
6
12
8
If A = {s1 , s2 } and B = {s1 , s5 , s6 } , find:
(a) P (A) , P (B) , (b) P (Ac ) ,
P (B c ) , (c) P (A ∩ B)
(d) P (A ∪ B) , (e) P (Ac ∩ B c ) , (f ) P (Ac ∪ B c ) .
1 1
11
Solution.(a) P (A) = P (s1 ) + P (s2 ) = + =
3 8
24
1
1
1
13
P (B) = P (s1 ) + P (s5 ) + P (s6 ) = +
+ =
3 12 8
24
11
13
(b) P (Ac ) = 1 в€’ P (A) = 1 в€’
=
24
24
13
11
P (B c ) = 1 в€’ P (B) = 1 в€’
=
24
24
1
(c) P (A ∩ B) = P (s1 ) =
3
(d) P (A ∪ B) = P (A) + P (B) − P (A ∩ B)
11 13 1
2
=
+
в€’ =
24 24 3
3
c
(e) P (Ac ∩ B c ) = P [(A ∪ B) ] = 1 −
4
P (Ac ∩ B)
n (Ac ∩ B)
1879
=
=
=
c
c
P (A )
n (A )
2610
0.720 (b) P (B|A) 6= P (B) , so A and B are not independent events. Example.439T7A5N38. MAIL DELIVERY:
Suppose the probability that your mail will be delivered
before 2 p.m. on a delivery day is 0.90. What is the probability that your mail will be delivered before 2 p.m. for
:(a) Two consecutive delivery days? (b) Three consecutive delivery days? Solution. (a) These are independent
events. Therefore, the probability is (0.9) (0.9) = 0.81. (b)
The probability that the mail will be delivered before 2
p.m. is (0.9) (0.9) (0.9) = 0.729. Example.439T7A5N42.
QUALITY CONTROL: Copykwik, Inc has four photocopy
machines A, B, C, and D. The probability that a given ma1
chine will break down on a particular day is P (A) =
,
50
1
1
1
P (B) =
, P (C) =
, P (D) =
. Assuming in60
75
40
dependence, what is the probability on a particular day
that: (a) All four machines will break down?(b) None of
the machines will break down? (c) Exactly one machine
will break down? Solution. The probability that on a
particular day (a) all four machines will break down is
P (A) P (B) P (C) P (D)
1 1 1 1
. . .
= 0.0000001. (b) none of the ma=
50 60 75 40
chines will break down is P (Ac ) P (B c ) P (C c ) P (Dc ) =
49 59 74 39
. . .
= 0.9270473 ≈ 0.927 (c) Exactly one ma50 60 75 40
chine will break down is P (A) P (B c ) P (C c ) P (Dc ) +
P (Ac ) P (B) P (C c ) P (Dc ) +
P (Ac ) P (B c ) P (C) P (Dc )
+
P (Ac ) P (B c ) P (C c ) P (D) =
1 59 74 39 49 1 74 39 49 59 1 39
. . . + . . . + . . .
50 60 75 40 50 60 75 40 50 60 75 40
49 59 74 1
+ . . .
= 0.0189193 + 0.0157127
50 60 75 40
+0.0125277 + 0.2377704 = 0.0709299 ≈ 0.071.
Example.452T7A7N13. In a group of 20 ballpoint
pens on a shelf in the stationery department of the University Store, 2 are known to be defective. If the customer
selects 3 of these pens, what is the probability that: (a) At
least one is defective? (b) No more than one is defective?
Solution. (a) The probability that none of the pens in
C (18, 3)
(18!) / (15!3!)
the sample are defective is
=
=
C (20, 3)
(20!) / (17!3!)
68
18! 17!
.
=
≈ 0.71579.
15! 20!
95
Therefore, the probability that at least one is defective
is given by 1 − 0.71579 ≈ 0.284.
(b) The probability that two are defective is given by
C (2, 2) .C (18, 1)
18/20!
6
=
=
≈ 0.0158. ThereC (20, 3)
17!3!
380
fore, the probability that no more than 1 is defective is
374
6
=
≈ 0.984. Example.452T7A7N14.
given by 1 в€’
380
380
Five people are selected at random. What is the probability that none of the people in this group were born at the
same day of the week?
7.6.5.4.3
≈ 0.150.
Solution. P (E) =
75
8.5: CONDITIONAL PROBABILITY
& 8.6: INDEPENDENT
P (B|Ac ) =
Example.436T7A5N2. Let A and B be two events
in a sample space S such that P (A) = 0.4, P (B) = 0.6,
and P (A ∩ B) = 0.3. Find: (a) P (A|B)
(b) P (B|A) .
P (A ∩ B)
0.3
1
Solution. (a) P (A|B) =
=
= .
P (B)
0.6
2
0.3
3
P (A ∩ B)
=
= .
(b) P (B|A) =
P (A)
0.4
4
Example.436T7A5N3. Let A and B be two events in
a sample space S such that P (A) = 0.6 and P (B |A|) =
0.5. Find P (A ∩ B) .Solution.P (A ∩ B) =
P (A) P (B|A)
=
(0.6) (0.5)
=
0.3.Example.436T7A5N10. If A and B are independent
events and P (A) = 0.35 and P (B) = 0.45, find: (a)
P (A ∩ B)
(b) P (A ∪ B) . Solution. (a) P (A ∩ B) =
P (A) P (B) = (0.35) (0.45) = 0.1575. (b) P (в€ЄB) =
P (A) + P (B) − P (A ∩ B) = 0.35 + 0.45 − 0.1575.
Example.436T7A5N15. A pair of fair dice is cast.
What is the probability that the sum of the numbers falling
uppermost is less than 9, if it is known that one of the
numbers is a 6? Solution. Let A denote the event that
the sum of the numbers is less than 9 and let B denote the
event that one of the numbers is a 6. Then, P (A|B) =
P (A ∩ B)
4/36
4
=
=
. Example.436T7A5N17.
P (B)
11/36
11
A pair of fair dice is cast. Let E denote the event
that the number landing uppermost on the first die
is a 3 and let F denote the event that the sum of
the numbers falling uppermost is 7. Determine whether
E and F are independent events. Solution. E =
{(3, 1) , (3, 2) , (3, 3) , (3, 4) , (3, 5) , (3, 6)}
F = {(1, 6) , (6, 1) , (2, 5) , (5, 2) , (3, 4) , (4, 3)} ; Then
1
E ∩ F = {(3, 4)} .Now, P (E ∩ F ) =
and this is equal
36
µ ¶µ ¶
6
6
1
to P (E) .P (F ) =
=
. So E and F are
36
36
36
independent events. Example.439T7A5N34. STUDENT
FINANCIAL AID: The accompanying data were obtained
from the financial aid office of a certain university:
Finacial Aid NO Finacial Aid
Total
Ugrads
4, 222
3, 898
8, 120
Grads
1, 879
731
2, 610
Total
6, 101
4, 629
10, 730
Let A be the event that a student selected at random
from this university is an undergraduate student and let B
be the event that a student selected at random is receiving
financial aid.(a) Find each of the following probabilities:
P (A) , P (B) , P (A ∩ B) , P (B|A) , and P (B|Ac ) . (b) Are
the events A and B independent events?
8120
Solution.(a) P (A) =
= 0.757; P (B) =
10730
6101
= 0.569;
10730
422
P (A ∩ B)
P (A ∩ B) =
= 0.393; P (B|A) =
=
10730
P (A)
n (A ∩ B)
422
=
= 0.520
n (A)
8120
5
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