Download The Addition Rules

The Addition Rules
Mutually Exclusive Events
Many problems in probability involve finding the probability of two or more events. For example, when a card is selected
at random from a deck, what is the probability that the card is a king or a queen? In this case, there are two situations to
consider. They are:
The card selected is a king
The card selected is a queen
Now consider another example. When a card is selected from a deck, find the probability that the card is a king or a
diamond. In this case, there are three situations to consider:
The card is a king
The card is a diamond
The card is a king and a diamond. That is, the card is the king of diamonds.
The difference is that in the first example, a card cannot be both a king and a queen at the same time, whereas in the second
example, it is possible for the card selected to be a king and a diamond at the same time. In the first example, we say the
two events are mutually exclusive. In the second example, we say the two events are not mutually exclusive. Two events
then are mutually exclusive if they cannot occur at the same time. In other words, the events have no common outcomes.
EXAMPLE: Which of these events are mutually exclusive?
a. Selecting a card at random from a deck and getting an ace or a club
b. Rolling a die and getting an odd number or a number less than 4
c. Rolling two dice and getting a sum of 7 or 11
SOLUTION:
a. No. The ace of clubs is an outcome of both events.
b. No. One and three are common outcomes.
c. Yes
Addition Rule I
The probability of two or more events occurring can be determined by using the addition rules. The first rule is used when
the events are mutually exclusive.
Addition Rule I: When two events are mutually exclusive, P(A or B) = P(A) + P(B)
EXAMPLE: When a die is rolled, find the probability of getting a 2 or a 3.
SOLUTION:
The problem can be solved by looking at the sample space, which is 1,2, 3, 4, 5, 6. Since there are 2 favorable
outcomes from 6 outcomes, P(2 or 3) = 2/6 = 1/3. Since the events are mutually exclusive, addition rule 1 also can be used:
P(2 or 3) = P(2) + P(3) = 1/6 +1/6 =2/6=1/3
PRACTICE
1.
In a box there are 3 red pens, 5 blue pens, and 2 black pens. If a person selects a pen at random, find the probability
that the pen is:
a.)A blue or a red pen, b.) A red or a black pen.
2.
A small automobile dealer has 4 Buicks, 7 Fords, 3 Chryslers, and 6 Chevrolets. If a car is selected at random, find
the probability that it is:
a.) A Buick or a Chevrolet. b.) A Chrysler or a Chevrolet.
3.
In a model railroader club, 23 members model HO scale, 15 members model N scale, 10 members model G scale,
and 5 members model O scale. If a member is selected at random, find the probability that the member models:
a.) N or G scale. b.) HO or O scale.
4.
A package of candy contains 8 red pieces, 6 white pieces, 2 blue pieces, and 4 green pieces. If a piece is selected at
random, find the probability that it is:
a.) White or green. b.) Blue or red.
5.
On a bookshelf in a classroom there are 6 mathematics books, 5 reading books, 4 science books, and 10 history
books. If a student selects a book at random, find the probability that the book is
a.)A history book or a mathematics book. b.) A reading book or a science book.
Addition Rule II
When two events are not mutually exclusive, you need to add the probabilities of each of the two events and subtract the
probability of the outcomes that are common to both events. In this case, addition rule II can be used.
Addition Rule II: If A and B are two events that are not mutually exclusive, then
P(A or B) = P(A) + P(B) - P(A and B),
where A and B means the number of outcomes that event A and event B have in common.
EXAMPLE: A card is selected at random from a deck of 52 cards. Find the probability that it is a 6 or a diamond.
SOLUTION:
Let A = the event of getting a 6; then P(A) = 4/52 since there are four 6s. Let B = the event of getting a diamond; then P(B)
= 13/52 since there are 13 diamonds. Since there is one card that is both a 6 and a diamond (i.e., the 6 of diamonds), P(A
and B) = 1/52.
Hence, P(A or B) = P(A) + P(B) - P(A and B) =4/52+13/52-1/52=16/52
EXAMPLE: Two dice are rolled; find the probability of getting doubles or a sum of 8.
SOLUTION:
Let A = getting doubles; then P(A) =6/36 since there are 6 ways to get doubles and let B = getting a sum of 8. Then P(B) =
5/36 since there are 5 ways to get a sum of 8: (6, 2), (5, 3), (4, 4), (3, 5), and (2, 6). Let (A and B) = the number of ways to
get a double and a sum of 8. There is only one way for this event to occur (4, 4); then P(A and B) = 1/36.
Hence, P(A or B) = P(A) + P(B) - P(A and B) =6/36+5/36-1/36=10/36
EXAMPLE: The probability that a student owns a computer is 0.92, and the probability that a student owns an automobile
is 0.53. If the probability that a student owns both a computer and an automobile is 0.49, find the probability that the
student owns a computer or an automobile.
SOLUTION:
Since P(computer) = 0.92, P(automobile) = 0.53 and P(computer and automobile) = 0.49, P(computer or automobile) =
0.92 + 0.53 - 0.49 = 0.96.
The key word for addition is "or," and it means that one event or the other occurs. If the events are not mutually exclusive,
the probability of the outcomes that the two events have in common must be subtracted from the sum of the probabilities of
the two events. For the mathematical purist, only one addition rule is necessary, and that is
P(A or B) = P(A) + P(B) - P(A and B)
The reason is that when the events are mutually exclusive, P(A and B) is equal to zero because mutually exclusive events
have no outcomes in common.
PRACTICE
1.
2.
3.
4.
5.
When a card is selected at random from a 52-card deck, find the probability that the card is a face card or a spade.
A die is rolled. Find the probability that the result is an even number or a number less than 3.
Two dice are rolled. Find the probability that a number on one die is a six or the sum of the spots is eight.
A coin is tossed and a die is rolled. Find the probability that the coin falls heads up or that there is a 4 on the die.
In a psychology class, there are 15 sophomores and 18 juniors. Six of the sophomores are males and 10 of the juniors
are males. If a student is selected at random, find the probability that the student is
a.
A junior or a male.
b.
A sophomore or a female.
c.
A junior.
Summary
Many times in probability, it is necessary to find the probability of two or more events occurring. In these cases, the
addition rules are used. When the events are mutually exclusive, addition rule I is used, and when the events are not
mutually exclusive, addition rule II is used. If the events are mutually exclusive, they have no outcomes in common. When
the two events are not mutually exclusive, they have some common outcomes. The key word in these problems is "or," and
it means to add.
The Multiplication Rules
Independent and Dependent Events
The multiplication rules can be used to find the probability of two or more events that occur in sequence. For
example, we can find the probability of selecting three jacks from a deck of cards on three sequential draws. Before
explaining the rules, it is necessary to differentiate between independent and dependent events.
Two events, A and B, are said to be independent if the fact that event A occurs does not affect the probability that
event B occurs. For example, if a coin is tossed and then a die is rolled, the outcome of the coin in no way affects or
changes the probability of the outcome of the die. Another example would be selecting a card from a deck, replacing it, and
then selecting a second card from a deck. The outcome of the first card, as long as it is replaced, has no effect on the
probability of the outcome of the second card.
On the other hand, when the occurrence of the first event in some way changes the probability of the occurrence of
the second event, the two events are said to be dependent. For example, suppose a card is selected from a deck and not
replaced, and a second card is selected.
PRACTICE
Determine whether the two events are independent or dependent.
1. Tossing a coin and selecting a card from a deck
2. Drawing a ball from an urn, not replacing it, and then drawing a second ball
3. Tossing one coin and then tossing a second coin
Multiplication Rule I
Before explaining the first multiplication rule, consider the example of tossing two coins. The sample space is HH,
HT, TH, TT. From classical probability theory, it can be determined that the probability of getting two heads is 1/4, since
there is only one way to get two heads and there are four outcomes in the sample space. However, there is another way to
determine the probability of getting two heads. In this case, the probability of getting a head on the first toss is 1/2, and the
probability of getting a head on the second toss is also 1/2. So the probability of getting two heads can be determined by
multiplying 1/2*1/2. This example illustrates the first multiplication rule.
Multiplication Rule I: For two independent events A and B, P(A and B) = P(A) • P{B).
In other words, when two independent events occur in sequence, the probability that both events will occur can be found by
multiplying the probabilities of each individual event.
The word and is the key word and means that both events occur in sequence and to multiply.
EXAMPLE: A coin is tossed and a die is rolled. Find the probability of getting a tail on the coin and a 5 on the die.
SOLUTION: Since P(tail) = 1/2 and P(5) = 1/6; P(tail and 5) = P(tail) * P(5) = 1/2 • 1/6 = 1/12. Note that the events are
independent.
It can also be solved using classical probability. Recall that the sample space for tossing a coin and rolling a die is
HI, H2, H3, H4, H5, H6, Tl, T2, T3, T4, T5, T6
Notice that there are 12 outcomes in the sample space and only one outcome is a tail and a 5; hence, P(tail and 5) = 1/12.
EXAMPLE: An urn contains 2 red balls, 3 green balls, and 5 blue balls. A ball is selected at random and its color is noted.
Then it is replaced and another ball is selected and its color is noted. Find the probability of each of these:
a. Selecting 2 blue balls
b. Selecting a blue ball and then a red ball
c. Selecting a green ball and then a blue ball
SOLUTION:
Since the first ball is being replaced before the second ball is selected, the events are independent.
a.) There are 5 blue balls and a total of 10 balls; therefore, the probability of selecting two blue balls with replacement is
P(blue and blue) = P(blue) • P(blue) = 5/10*5/10=25/100
b.) There are 5 blue balls and 2 red balls, so the probability of selecting a blue ball and then a red ball with replacement is
P(blue and red) = P(blue) • P(red)= 5/10*2/10=10/100
c.) There are 3 green balls and 5 blue balls, so the probability of selecting a green ball and then a blue ball with replacement
is P(green and blue) = P(green) • P(blue)=3/10*5/10=15/100
PRACTICE
1. A card is drawn from a deck, then replaced, and a second card is drawn. Find the probability that two kings are
selected.
2. If 12% of adults are left-handed, find the probability that if three adults are selected at random, all three will be lefthanded.
3. If two people are selected at random, find the probability that both were born in August.
4. A coin is tossed 4 times. Find the probability of getting 4 heads.
5. A die is rolled and a card is selected at random from a deck of 52 cards. Find the probability of getting an odd number
on the die and a club on the card.
Multiplication Rule II
When two sequential events are dependent, a slight variation of the multiplication rule is used to find the probability of
both events occurring. For example, when a card is selected from an ordinary deck of 52 cards the probability of getting a
specific card is 1/52, but the probability of getting a specific card on the second draw is 1/51 since 51 cards remain.
EXAMPLE: Two cards are selected from a deck and the first card is not replaced. Find the probability of getting two
kings.
SOLUTION:
The probability of getting a king on the first draw is 4/52 and the probability of getting a king on the second draw is 3/51,
since there are 3 kings left and 51 cards left. Hence the probability of getting 2 kings when the first card is not replaced is
4/52*3/51=1/221
When the two events A and B are dependent, the probability that the second event B occurs after the first event A has
already occurred is written as P(B|A). This does not mean that B is divided by A; rather, it means and is read as "the
probability that event B occurs given that event A has already occurred." P(B|A) also means the conditional probability that
event B occurs given event A has occurred. The second multiplication rule follows.
Multiplication Rule II:
When two events are dependent, the probability of both events occurring is P(A and B) = P(A) • P(B|A)
EXAMPLE: A box contains 24 toasters, 3 of which are defective. If two toasters are selected and tested, find the
probability that both are defective.
SOLUTION:
Since there are 3 defective toasters out of 24, the probability that the first toaster is defective is P=3/24. Since the second
toaster is selected from the remaining 23 and there are two defective toasters left, the probability that it is defective is
P=2/23. Hence, the probability that both toasters are defective is
P(D1 and D2) = P(D1) • P(D2|D1) =3/24•2/23=1/92
EXAMPLE: Two cards are drawn without replacement from a deck of 52 cards. Find the probability that both are queens.
SOLUTION: P(Q and Q) = P(Q)P(Q|Q) = 4/52•3/51=1/221
Remember that the key word for the multiplication rule is and. It means to multiply.
When two events are dependent, the probability that the second event occurs must be adjusted for the occurrence of the first
event. For the mathematical purist, only one multiplication rule is necessary for two events, and that is
P(A and B) = P(A) • P(B|A).
The reason is that when the events are independent P(B|A) = P(B) since the occurrence of the first event A has no effect on
the occurrence of the second event B.
PRACTICE
1. In a study, there are 8 guinea pigs; 5 are black and 3 are white. If 2 pigs are selected without replacement, find the
probability that both are white.
2. In a classroom there are 8 freshmen and 6 sophomores. If three students are selected at random for a class project, find
the probability that all 3 are freshmen.
3. Three cards are drawn from a deck of 52 cards without replacement. Find the probability of getting 3 diamonds.
4. A box contains 12 calculators of which 5 are defective. If two calculators are selected without replacement, find the
probability that both are good.
5. A large flashlight has 6 batteries. Three are dead. If two batteries are selected at random and tested, find the probability
that both are dead.
Conditional Probability
Previously, conditional probability was used to find the probability of sequential events occurring when they were
dependent. Recall that P(B\A) means the probability of event B occurring given that event A has already occurred. Another
situation where conditional probability can be used is when additional information about an event is known. Sometimes it
might be known that some outcomes in the sample space have occurred or that some outcomes cannot occur. When
conditions are imposed or known on events, there is a possibility that the probability of the certain event occurring may
change. Consider the following examples.
EXAMPLE: A die is rolled; find the probability of getting a 4 if it is known that an even number occurred when the die
was rolled.
SOLUTION:
If it is known that an even number has occurred, the sample space is reduced to 2, 4, or 6. Hence the probability of getting a
4 is 1/3 since there is one chance in three of getting a 4 if it is known that the result was an even number.
EXAMPLE: Two dice are rolled. Find the probability of getting a sum of 3 if it is known that the sum of the spots on the
dice was less than six.
SOLUTION:
There are 2 ways to get a sum of 3. They are (1, 2) and (2, 1), and there are 10 ways to get a sum less than six.
They are (1, 1), (1, 2), (2, 1), (3, 1), (2, 2), (1, 3), (1, 4), (2, 3), (3, 2), and (4, 1); hence, P(sum of 3|sum less than 6) =2/10
The two previous examples of conditional probability were solved using classical probability and reduced sample spaces;
however, they can be solved by using the following formula for conditional probability.
The conditional probability of two events A and B is
P(A | B) =
P(AB)
P(B)
P(AB) means the probability of the outcomes that events A and B have in common.
€
Lets solve previous example using the formula
for conditional probability.
EXAMPLE: Two dice are rolled. Find the probability of getting a sum of 3 if it is known that the sum of the spots on the
dice was less than six.
SOLUTION
P(AB) means the probability of getting a sum of 3 and sum less then 6. Hence P(AB)=2/36. P(B) means getting a sum less
than 6 and is 10/36. Hence,
2
P(AB) 36 2 10 2 36 2 1
P(A | B) =
=
=
÷
=
⋅
= =
10 36 36 36 10 10 5
P(B)
36
PRACTICE
1.
2.
3.
4.
When two dice are rolled, find the probability that one die is a 6 given that the sum of the spots is 8.
Two coins are€tossed. Find the probability of getting two tails if it is known that one of the coins is a tail.
A card is selected from a deck. Find the probability that it is an ace given that it is a black card.
Three dice are rolled. Find the probability of getting three twos if it isknown that
Summary
When two events occur in sequence, the probability that both events occur can be found by using one of the multiplication
rules. When two events are independent, the probability that the first event occurs does not affect or change the probability
of the second event occurring. If the events are independent, multiplication rule I is used. When the two events are
dependent, the probability of the second event occurring is changed after the first event occurs. If the events are dependent,
multiplication rule II is used. The key word for using the multiplication rule is "and." Conditional probability is used when
additional information is known about the probability of an event.
PRACTICE
1. In a box there are 3 red pens, 5 blue pens, and 2 black pens. If a person selects a pen at random, find the probability
that the pen is: a.)A blue or a red pen, b.) A red or a black pen.
2. A small automobile dealer has 4 Buicks, 7 Fords, 3 Chryslers, and 6 Chevrolets. If a car is selected at random, find the
probability that it is: a.) A Buick or a Chevrolet. b.) A Chrysler or a Chevrolet.
3. In a model railroader club, 23 members model HO scale, 15 members model N scale, 10 members model G scale, and
5 members model O scale. If a member is selected at random, find the probability that the member models : a.) N or G
scale. b.) HO or O scale.
4. A package of candy contains 8 red pieces, 6 white pieces, 2 blue pieces, and 4 green pieces. If a piece is selected at
random, find the probability that it is: a.) White or green. b.) Blue or red.
5. On a bookshelf in a classroom there are 6 mathematics books, 5 reading books, 4 science books, and 10 history books.
If a student selects a book at random, find the probability that the book is a.)A history book or a mathematics book. b.)
A reading book or a science book.
6. When a card is selected at random from a 52-card deck, find the probability that the card is a face card or a spade.
7. A die is rolled. Find the probability that the result is an even number or a number less than 3.
8. Two dice are rolled. Find the probability that a number on one die is a six or the sum of the spots is eight.
9. A coin is tossed and a die is rolled. Find the probability that the coin falls heads up or that there is a 4 on the die.
10. In a psychology class, there are 15 sophomores and 18 juniors. Six of the sophomores are males and 10 of the juniors
are males. If a student is selected at random, find the probability that the student is
a.
A junior or a male.
b.
A sophomore or a female.
c.
A junior.
11. Determine whether the two events are independent or dependent.
a. Tossing a coin and selecting a card from a deck
b. Drawing a ball from an urn, not replacing it, and then drawing a second ball
c. Tossing one coin and then tossing a second coin
12. A card is drawn from a deck, then replaced, and a second card is drawn. Find the probability that two kings are
selected.
13. If 12% of adults are left-handed, find the probability that if three adults are selected at random, all three will be lefthanded.
14. If two people are selected at random, find the probability that both were born in August.
15. A coin is tossed 4 times. Find the probability of getting 4 heads.
16. A die is rolled and a card is selected at random from a deck of 52 cards. Find the probability of getting an odd number
on the die and a club on the card.
17. In a study, there are 8 guinea pigs; 5 are black and 3 are white. If 2 pigs are selected without replacement, find the
probability that both are white.
18. In a classroom there are 8 freshmen and 6 sophomores. If three students are selected at random for a class project, find
the probability that all 3 are freshmen.
19. Three cards are drawn from a deck of 52 cards without replacement. Find the probability of getting 3 diamonds.
20. A box contains 12 calculators of which 5 are defective. If two calculators are selected without replacement, find the
probability that both are good.
21. A large flashlight has 6 batteries. Three are dead. If two batteries are selected at random and tested, find the probability
that both are dead.
22.
23.
24.
25.
When two dice are rolled, find the probability that one die is a 6 given that the sum of the spots is 8.
Two coins are tossed. Find the probability of getting two tails if it is known that one of the coins is a tail.
A card is selected from a deck. Find the probability that it is an ace given that it is a black card.
Three dice are rolled. Find the probability of getting three twos if it is known that