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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.
Ten's place
One's place
8
1
9
7
2
6
7
3
5
0
2
6
4
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.
P ( N ) = 1− P ( N
C
n(NC )
) = 1− n (S )
= 1−
8 56
=
64 64
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 ) .