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Lecture 3.5
Contemporary Mathematics
Instruction: Probability with "Not"
Lecture 3.2 presented the Complement Principle for a Sample Space below.
n ( E ) = n ( S ) − n ( EC )
Dividing each term of the equality by n ( S ) , we get:
n(E)
n (S )
= 1−
n ( EC )
n (S )
.
Applying the definition of theoretical probability, we get the Complement Property of
Probability:
P ( E ) = 1− P ( EC )
The Complement Property allows probabilities to be found indirectly. Consider the probability
of a multiple of ten not occurring in an experiment comprised of spinning the two number wheels
below and observing the two-digit numbers that result.
One's place
Ten's place
8
9
1
7
2
6
3
5
4
0
7
2
6
3
5
4
The Multiplication Principle gives n ( S ) : 8 ⋅ 8 = 64 . There are sixty-four possible outcomes,
but only eight outcomes will end in zero (numbers that end in zero are multiples of ten). If N
represents the event that the two-digit number is not a multiple of ten, then N C represents the
event that the two-digit number is a multiple of ten, and the Complement Property finds P ( N )
as below.
n(NC )
8 56
C
P ( N ) = 1− P ( N ) = 1−
= 1−
=
64 64
n (S )
Application Exercise 3.5
Problems
#1
Assume that the probability an astronaut develops a kidney stone during a six-month
space flight is 0.16. What is the probability that an astronaut will not develop a kidney
stone during a six-month space flight?
#2
Assume an actuary accurately assigns an individual a 0.002 probability to die during
the term of a life insurance policy. What is the probability that the individual will live
during the term of the life insurance policy?
#3
Assume that a contestant on a popular game show has a one in eight chance to win
$250,000.00. What is the chance that the contestant will not win $250,000.00.
#4
Assume that a sample space includes three outcomes, i.e., S = {a1 , a2 , a3 } . If
P ( a1 ) = 0.5 and P ( a2 ) = 0.3 , calculate P ( a3 ) .
#5
Assume that a sample space includes six outcomes, i.e., S = {a1 , a2 , a3 , a4 , a5 , a6 } .
If P ( a1 ) = P ( a2 ) = P ( a3 ) = P ( a4 ) = 0.1 , calculate P ( a5 ∪ a6 ) .
#1 P = 0.84
#2 P = 0.998
#3 P = 7/8
#4 P ( a3 ) = 0.2
#5 P ( a5 ∪ a6 ) = 0.6
Assignment 3.5
Problems
#1
A pair of six-sided dice are tossed. Find the probability of not rolling a sum of twelve.
#2
Consider an experiment that requires selecting a natural number from one to onehundred inclusive at random. Given that there are twenty-five prime numbers between
one and one-hundred, find the probability that the number is not a prime number.
#3
Consider an experiment that constructs a number by spinning the two number wheels
below. Find the probability that the number is not divisible by five.
Ten's place
One's place
8
1
9
7
2
6
3
5
4
0
7
2
6
3
5
4
#4
A combination lock consists of three numbers each from zero to fifty-four inclusive.
Find the probability that a combination is not composed of three of the same digits in a
row.
#5
Ten coins are flipped. What is the probability that the result does not include ten tails?
Lecture 3.6
Contemporary Mathematics
Instruction: Probability with "Or"
Lecture 3.2 presented the General Addition Rule below.
n(E ∪ F ) = n(E) + n(F ) − n(E ∩ F )
Dividing each term of the equality by n ( S ) , we get:
n(E ∪ F )
n (S )
=
n (E)
n (S )
+
n(F )
n (S )
−
n(E ∩ F)
n (S )
.
Applying the definition of theoretical probability, we get the General Addition Rule of
Probability:
P (E ∪ F ) = P (E) + P (F ) − P (E ∩ F )
The General Addition Rule of Probability helps find probabilities involving "or." The
conjunction "or" corresponds to the union of events. Consider the probability of a multiple of
three or a multiple of two occurring in an experiment comprised of spinning the two number
wheels below and observing the product of the result.
A
B
8
1
2
7
1
5
3
6
5
4
7
Let W represent the event that the product is a multiple of two. Let H represent the event that the
product is a multiple of three. Thinking of the sample space as a Cartesian product, it is easy to
see that W includes all the outcomes with two, four, six, or eight from dial A as a factor.
Likewise H includes all the outcomes with three or six from dial A as a factor. Consequently,
W ∩ H will include the outcomes with the six from dial A as one factor (multiples of six are
multiples of two and three). The probability of a multiple of three or a multiple of two occurring
is the probability of W ∪ H given by the General Addition Rule of Probability below.
P (W ∪ H ) = P (W ) + P ( H ) − P (W ∩ H ) =
12 6
3 15 5
+
−
=
=
24 24 24 24 8
Lecture 3.6
A classic example of probability with "or" involves drawing a card from a standard deck
of fifty-two playing cards. A standard deck has four suits: hearts, diamonds, clubs, and spades.
Each suit has thirteen cards: ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, and king. Imagine an
experiment that requires the draw of one card from a standard deck. If J represents the event of
drawing a jack, then n ( J ) = 4 because there are four jacks. If D represents the event of drawing
a diamond, then n ( D ) = 13 because there are fourteen diamonds. Since one of the diamonds is a
jack, n ( J ∩ D ) = 1 . What is the probability of drawing a jack or a diamond? Apply the General
Addition Rule of Probability.
P ( J ∪ D) = P ( J ) + P ( D) − P ( J ∩ D)
4 13 1
+ −
52 52 52
16
P( J ∪ D) =
52
4
P( J ∪ D) =
13
P( J ∪ D) =
In the two previous examples, the probability involving "or" involved events not
mutually exclusive. If the events are mutually exclusive, then the intersection of the two events
is empty. Since P ( ∅ ) = 0 , then we have the Addition Rule of Probability for Mutually
Exclusive Events:
If E and F are mutually exclusive, then P ( E ∪ F ) = P ( E ) + P ( F )
To apply this rule, consider the experiment from above involving the draw of a single card from
a deck of playing cards. Let J represent the event of drawing a jack, and let K represent the event
of drawing a king. What is the probability of drawing a jack or a king? Since there are no jacks
that are also kings in the deck, the Addition Rule of Probability for Mutually Exclusive Events
applies. Note n ( J ) = 4 and n ( K ) = 4 , and apply the rule.
P(J ∪ K ) = P(J ) + P(K )
4
4
+
52 52
8
P(J ∪ K ) =
52
2
P(J ∪ K ) =
13
P(J ∪ K ) =
Application Exercise 3.6
Problems
#1
A survey of 1,000 citizens found that 57% of the respondents feel the government
should invest in space exploration. In the same survey, 40% of respondents feel that
space exploration is a waste of tax dollars. Are the events "government should invest in
space exploration" and "space exploration is a waste of tax dollars" mutually exclusive?
#2
Candidates for a particular position must pass a physical exam and a written exam.
Under normal conditions, 69% of the candidates pass the written exam, 64% pass the
physical exam, and 56% pass both exams. What is the probability that a candidate
passes either the physical exam or the written exam but not necessarily both of the
exams?
#3
Suppose a botanist records twenty cases of blight in a sample of 100 plants. Suppose
further that in the same sample the botanist records eighteen cases of invasive pests. If
only four of the plants in the sample suffer from both blight and invasive pests, what is
the probability that a plant suffers from neither blight nor invasive pests?
#4
Consider 1,000 ceramic tiles to be inspected for fissures and discoloration. Tiles pass
inspection if they are free of fissures and free of discoloration. Suppose inspectors find
52 tiles with fissures and 88 with discoloration. Suppose further that the probability
that a tile fails inspection is 0.1. What is the probability that a tile has both a fissure
and discoloration.
#1 Yes, the events are mutually exclusive. Either a person supports government expenditures in
space exploration or does not.
#2 P = 0.77
#3 P = 0.66
#4 P = 0.04
Assignment 3.6
Problems
#1
Consider an experiment involving the toss of two six-sided dice like those shown
below. Let P ( 7 ) represent the probability of rolling a sum of seven. Let P (11)
represent the probability of rolling a sum of eleven. Find P ( 7 ∪ 11) .
#2
Consider the experiment from above. Let E represent the event that the sum of two
dice is even. Let G represent the event that the sum is greater than ten. Find
P(E ∪ G) .
#3
For a standard deck of fifty-two playing cards, if we draw a single card, find the
probability that the card is a red or a jack.
#4
For a standard deck of fifty-two playing cards, if we draw a single card, find the
probability that the card is a club or a face card.
#5
Consider an experiment involving the toss of a coin and of a six-sided die. Let T be the
event that the coin lands "tails" up. Let S be the event that the die lands with the "six"
facing up. Find P (T ∪ S ) .
Lecture 3.7
Contemporary Mathematics
Instruction: Probability with "And"
Lecture 3.6 discussed the experiment of drawing one card from a standard deck, and
asked, "What is the probability of drawing a jack or a diamond?" This lecture will ask, "What is
the probability of drawing a jack and a diamond." In symbols, if J represents the event that a
jack is drawn and D represents the event that a diamond is drawn, then the question asks, "What
is P ( J ∩ D ) ?" Since only one card in the fifty-two card deck is both a jack and a diamond,
P ( J ∩ D ) = 1 52 . Another way to answer the question is to apply the Special Multiplication
Rule of Probability of Independent Events demonstrated below in particular.
P ( J ∩ D) = P ( J ) ⋅ P ( D)
P ( J ∩ D) =
4 13
⋅
52 52
13
P ( J ∩ D) =
1
52
The reader may notice that the previous lecture gave a rule for probabilities involving "or" that
required addition (and subtraction if the events were not mutually exclusive) while above we see
a probability involving "and" that requires multiplication. It is typically true that probabilities
involving "or" require addition while probabilities involving "and" require multiplication. In the
previous lecture, we also noted, however, that simple addition is not enough to find the
probability of event E or F if the two events are not mutually exclusive (subtraction of the
intersection is also required). Similarly, with probabilities involving "and," we must think
carefully about the types of events involved.
The Special Multiplication Rule of Probability of Independent Events demonstrated
above only applies to the intersection of independent events. Independent events are defined
below.
Events E and F are independent if the occurrence of one has no
affect on the probability of the other.
The Special Multiplication Rule of Probability of Independent Events is stated below in general
terms.
If E and F are independent events, then P ( E ∩ F ) = P ( E ) ⋅ P ( F ) .
Consider an experiment that consists of rolling two six-sided dice, one red and one black.
This experiment involves two tosses: the roll of the red die and the roll of the black die. What is
the probability of rolling a "four" on the red die and rolling a "two" on the black die? Intuitively,
we know that whatever happens with the red die will have no affect on the probabilities of
outcomes associated with the black die, so the Special Multiplication Rule of Probability of
Lecture 3.7
Independent Events applies. Let F represent the event that the red die lands with the "four"
facing up, and let T represent the event that the black die lands with the "two" facing up. Then,
P ( F ∩ T ) = P ( F ) ⋅ P (T )
1 1 1
P (F ∩T ) = ⋅ =
6 6 36
The Special Multiplication Rule of Probability of Independent Events is "special" in the
sense that it only applies to independent events. Consequently, its application depends on
recognizing whether or not two events are independent. Consider our red and black six-sided die
and events F and T from above. Would the probability of T change if we knew F had occurred?
If not, the events are independent. Let's assume we know the red die has rolled a "four," so F has
occurred. In this situation, we call F a given event. The notation P (T | F ) indicates the
probability of event T given that event F has occurred. If P (T | F ) = P (T ) , the events are
independent. Checking to see if this equality is true, is called the Test for Independence.
Events E and F are independent if P ( E | F ) = P ( E ) .
If the test above shows independence, that is, if P ( E | F ) = P ( E ) , then multiplying
P ( E ) by P ( F ) gives P ( E ∩ F ) . If the test, however, fails to show independence, that is, if
P ( E | F ) ≠ P ( E ) , then the Special Multiplication Rule of Probability of Independent Events no
longer applies, and it is necessary to apply the General Multiplication Rule of Probability stated
below.
If E and F are any two events, then P ( E ∩ F ) = P ( E ) ⋅ P ( F | E ) .
The General Multiplication Rule of Probability can be extended for more than two events. The
rule extended to three events is stated below.
If E, F, and G are any three events, then P ( E ∩ F ∩ G ) = P ( E ) ⋅ P ( F | E ) ⋅ P ( G | E ∩ F ) .
The beginning of this lecture examined a single draw from a standard deck of fifty-two
playing cards. Now, we turn to multiple successive draws without replacement. (The phrase
"without replacement" indicates that cards are not returned to the deck once they are drawn.)
What is the probability of drawing two successive kings? Let K1 represent the event that a king
is drawn first. Let K 2 represent the event that the second draw is also a king. Intuitively, we note
that K1 and K 2 are not independent because P ( K 2 ) has two different measures dependent on
the outcome of the first draw. If a king is drawn first, then there are fifty-one cards left and only
three kings remaining, so P ( K 2 ) = 3 51 = 1 17 . If, however, a king is not drawn first, then there
Lecture 3.7
are fifty-one cards left and four kings remaining, so P ( K 2 ) = 4 51 . Accordingly, we will apply
the General Multiplication Rule of Probability, P ( E ∩ F ) = P ( E ) ⋅ P ( F | E ) :
P ( K1 ∩ K 2 ) = P ( K1 ) ⋅ P ( K 2 | K1 )
1
1
4 3
1
P ( K1 ∩ K 2 ) =
⋅
=
.
52 51 221
13
17
If the number of draws increases, the General Multiplication Rule of Probability still
applies. Consider five draws from a standard deck without replacement. What is the probability
of drawing five diamonds? Let D1 , D2 , … , D5 represent the events of drawing diamonds on each
of the five successive draws.
P ( D1 ∩ D2 ∩ D3 ∩ D4 ∩ D5 ) =
13 12 11 10 9
33
⋅ ⋅ ⋅ ⋅
=
≈ 0.000495
52 51 50 49 48 66,640
Application Exercise 3.7
Problems
#1
NASA's primary flight software (PFS) for the space shuttle has over 400,000 computer
words of flight code. A complete test of this much code is infeasible, so NASA uses
redundant systems to guard against software error. To further reduce the chance for a
loss of flight control, NASA uses a backup flight system (BFS), which is independent
of the PFS. Assume the redundant systems reduce the probability of error in the PFS
and BFS to 0.001 each. Assume further that the probability of a loss of flight control
requires both an error in the PFS to occur as well as an error in the BFS. What is the
probability of a loss of flight control?
#2
Nick and Ellen are planning their retirement. Nick is 54 and Ellen is 46. An actuary
says that Nick has a 0.79 probability to be alive in ten years while Ellen has a 0.98
probability to be alive in ten years. What is the probability that both Nick and Ellen
will be alive in ten years?
#3
A quality control procedure for testing motherboards consists of randomly selecting
two circuits from each motherboard. If both circuits are defective, the motherboard is
rejected. Find the probability that a motherboard is rejected if it contains 10,000
circuits and only five are bad.
#4
Consider an industrial process with two stages that either operate or default. The first
stage has a 0.1 probability to default. If the first stage defaults, the second stage has a
0.4 probability to default. If the first stage operates, the second stage only has a 0.05
probability to default. What is the probability that the second stage defaults?
#1 P = 0.000001
1
#3 P =
≈ 0.0000002
4,999,500
#2 P = 0.7742
#4 P = 0.085
Assignment 3.7
Problems
#1
Consider an experiment involving the toss of a coin and a six-sided die. Find the
probability of getting a tail on the coin and an even number on the die.
#2
From a standard deck of fifty-two playing cards, draw one card, note the outcome,
reshuffle the card back into the deck, and draw again. Under these circumstances, find
the probability of drawing two diamonds.
#3
From a standard deck of fifty-two playing cards, draw one card, note the outcome, do
not replace the card, and draw again. Under these circumstances, find the probability of
drawing two diamonds.
#4
Suppose a vase contains ten marbles: five blue, four white, and one purple. Suppose
two marbles are drawn randomly without replacement. Find the probability that both
marbles are blue.
#5
Suppose a vase contains twenty marbles: five blue, four white, one purple, and ten red.
Suppose three marbles are drawn randomly without replacement. Find the probability
that all three marbles are white.
Lecture 3.8
Contemporary Mathematics
Instruction: Conditional Probability
Lecture 3.7 discussed the General Multiplication Rule of Probability stated below.
If E and F are any two events, then P ( E ∩ F ) = P ( E ) ⋅ P ( F | E ) .
P ( F | E ) , the probability of event F given event E, is called a conditional probability.
Conditional probabilities are computed on the assumption that some other event—the given
event—has occurred. This lecture will use the General Multiplication Rule of Probability to
derive a rule for conditional probabilities.
Consider the General Multiplication Rule of Probability:
P (E ∩ F ) = P (E)⋅ P (F | E) .
Divide both sides of the equality by P ( E ) :
P (E ∩ F )
P (E)
P (E ∩ F )
P (E)
=
P (E) ⋅ P (F | E)
P (E)
= P (F | E).
Substituting the definition for probability, we have:
n(E ∩ F )
n (S )
n(E)
= P (F | E)
n (S )
Simplifying, we arrive at a definition for conditional probability stated in the box below.
P (F | E) =
P (F | E) =
P (F | E) =
n(E ∩ F)
n (S )
÷
n(E)
n (S )
n (E ∩ F ) n (S )
⋅
n(E)
n (S )
n(E ∩ F )
n(E)
Lecture 3.8
P (F | E) =
n(F ∩ E)
number of outcomes that describe both F and E
.
=
n(E)
number of outcomes that describe the given event
To arrive at this definition in particular, consider an experiment involving the roll of two
six-sided dice like those pictured in Figure 1 observing the ordered pair of the number of dots
that face up after the roll.
Figure 1
Let N represent the event that the sum of the two numbers facing up is nine and let T represent
the event that one of the two dice lands with a three facing up. We see,
N = {( 3, 6 ) , ( 4, 5 ) , ( 5, 4 ) , ( 6, 3)} , and
⎧⎪(1, 3) , ( 2, 3) , ( 3, 3) , ( 4, 3) , ( 5, 3) , ( 6, 3) , ⎫⎪
T =⎨
⎬,
3,1
,
3,
2
,
3,
4
,
3,
5
,
3,
6
(
)
(
)
(
)
(
)
(
)
⎩⎪
⎭⎪
To consider P ( N | T ) , first think about the denominator in the definition of theoretical
probability:
number of elements (outcomes) in E
.
P (E) =
number of elements (outcomes) in S
The sample space includes all the possible outcomes of the experiment. The sample space for the
experiment or rolling two six-sided dice includes thirty-six possible outcomes.
⎧(1,1) , (1, 2 ) , (1, 3) , (1, 4 ) , (1, 5 ) , (1, 6 ) , ( 2,1) , ( 2, 2 ) , ( 2, 3) , ( 2, 4 ) , ( 2, 5 ) , ( 2, 6 ) , ⎫
⎪
⎪
S = ⎨( 3,1) , ( 3, 2 ) , ( 3, 3) , ( 3, 4 ) , ( 3, 5 ) , ( 3, 6 ) , ( 4,1) , ( 4, 2 ) , ( 4, 3 ) , ( 4, 4 ) , ( 4, 5 ) , ( 4, 6 ) , ⎬
⎪
⎪
⎩( 5,1) , ( 5, 2 ) , ( 5, 3) , ( 5, 4 ) , ( 5, 5 ) , ( 5, 6 ) , ( 6,1) , ( 6, 2 ) , ( 6, 3) , ( 6, 4 ) , ( 6, 5 ) , ( 6, 6 ) ⎭
For conditional probability, we know some outcomes have occurred, which means we know
some outcomes have not occurred. Thus, the number of possible outcomes has been reduced
from the number of elements in S to the number of elements in T. Adjusting the definition for
this possibility, we might word the definition differently:
Lecture 3.8
P (E) =
number of possible outcomes that describe E
.
number of possible outcomes
For P ( N | T ) , the numerator will equal n ( N ∩ T ) because the only possible outcomes that
describe N | T are those that describe N that are in T. Moreover, the denominator will equal
n (T ) because those are the only possible outcomes when it is known that T has occurred. Thus,
P(N |T ) =
n ( N ∩T )
n (T )
=
2
.
11
Application Exercise 3.8
Problems
#1
#2
#3
Assume P ( A ∩ B ) = 0.6 and P ( B ) = .8 . Calculate P ( A | B ) .
Suppose NASA puts numerous engines through two tests. Assume that 75% of the
engines passed both tests. Assume that 90% passed the first test. What percent of the
engines that passed the first test also passed the second test?
Assume P ( E ) = 0.4 , P ( F ) = 0.3 , P ( E ∩ F ) = 0.1 . Which is greater, P ( E | F ) or
P(F | E) ?
#4
Assume P ( R | S ) = 0.48 and P ( R ∩ S ) = 0.42 . Calculate P ( S ) .
#1 P ( A | B ) = 0.75
#2 P = 0.83
#3 P ( E | F ) = 0.3 > 0.25 = P ( F | E )
#4 P ( S ) = 0.875
Assignment 3.8
Problems
#1
A single six-sided die is cast. Let P ( 6 ) represent the probability that the result is a six.
Let P ( E ) represent the probability that the result is an even number. Find P ( 6 | E ) .
#2
Consider an experiment involving the toss of two six-sided dice observing the sum of
the results of each die. Let P ( 7 ) represent the probability that the sum equals seven.
Let P ( E ) represent the probability that the sum equals an even number and P ( O )
represent the probability that the sum equals an odd number. Find P ( 7 | E ) and
P (7 | O ) .
#3
Consider a single draw from a standard deck of fifty-two playing cards. Let J represent
the event that the card is a jack. Let F represent the event that the card is a face card.
Find P ( J | F ) .
#4
Consider the toss of three coins. Let Z represent the event that there are zero heads.
Let N represent the event that there is one head. Let A represent the event that all the
coins land tails up. Find P ⎡⎣ A | ( Z ∪ N ) ⎤⎦ .
#5
Consider a single draw from a standard deck of fifty-two playing cards. Let R represent
the even that the card is red. Let D represent the event that the card is a diamond. Find
P ( R | D) .