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Chapter 6
Lesson 6.6
Probability
6.6 General Probability Rules
Example 1: Suppose I will pick two cards from a
standard deck. This can be done two ways:
1)PickSampling
a card atwithout
random,replacement
replace the–card,
the then
pickevents
a second
arecard
typically dependent events.
2) Pick a card at random, do NOT replace, then
replacement – the events
pickSampling
a secondwith
card.
are typically independent events.
If I pick two cards from a standard deck
without replacement,
is the
probability
Probability ofwhat
a spade
given
I drew a
that I select twospade
spades?
on the first card.
Are the events E1 = first card is a spade and E2 =
second card is a spade independent? NO
P(E1 and E2) =
P(E1) × P(E2|E1) =
13 12 1
× =
52 51 17
General Rule for Multiplication
For any two events E and F,
P(AÇ B) = P(A)× P(B| A)
Here is a process to use when calculating the
intersection of two or more events.
P (A  B )
Yes
P (A)  P (B )
Ask yourself, “ Are these
events independent?”
No
P (A)  P (B | A)
There are seven girls and eight boys in a
math class. The teacher selects two
students at random to answer questions on
the board. What is the probability that
both students are girls?
Are these events independent? NO
7 6
P(G1  G2 ) 

 .2
15 14
Example 2: Suppose the manufacturer of
a certain brand of light bulbs made 10,000
of these bulbs and 500 are defective.
You randomly pick a package of two
such bulbs off the shelf of a store. What is the
probability that both bulbs are defective?
Are the events E1 = the first bulb is defective
and E2 = the second bulb is defective
independent?
To answer this question, let’s explore
probabilities
of these of
twoselecting
events? a
Whatthe
would
be the probability
defective light bulb? 500/10,000 = .05
Light Bulbs Continued . . .
What would be the probability of selecting a
defective light bulb?
500/10,000 = .05
Having selected one defective bulb, what is the
probability of selecting another without
replacement?
499/9999 = .0499
These values are so close to each other that
when rounded to three decimal places they are
both .050. For all practical purposes, we can
treat them as being independent.
Light Bulbs Continued . . .
What is the probability that both
bulbs are defective?
Are the selections independent?
We can assume independence.
P(defective  defective) 
(0.05)(0.05) = .0025
Light Bulbs Revisited . . .
A certain brand of light bulbs are defective five
percent of the time. You randomly pick a package
of two such bulbs off the shelf of a store. What
is the probability that exactly one bulb is
defective?
Let D1 = first light bulb is defective
D2 = second light bulb is defective

P (exactly one defective)  P D1  D
C
2
  D
C
1
 D2
= (.05)(.95) + (.95)(.05) = .095

General Rule for Addition
Since the intersection is added in twice, we
For any twosubtract
events Eout
andthe
F, intersection.
P (E  F )  P (E )  P (F )  P (E  F )
E
F
Musical styles other than rock and
pop are becoming more popular. A
survey of college students finds that
the probability they like country
music is .40. The probability that they
liked jazz is .30 and that they liked
both is .10. What is the probability
that they like country or jazz?
P(CountryÈ Jazz) =
.4 + .3 -.1 = .6
Here is a process to use when calculating the
union of two or more events.
P (E  F )
Ask yourself, “Are the
events mutually exclusive?”
In some problems, the
Yes intersection of the two events is
No
given (see previous example).
P (E )  P (F )
In some problems,
P (E )  P (F )  P (E  F )
the intersection of
the two events is not
given, but we know
If independent
that the events are
independent.
P (E )  P (F )
Suppose two six-sided dice are rolled (one
white and one red). What is the probability
that the white die lands on 6 or the red
Howdie
can
lands on 1?
you find
Let
A = white die landing on 6
B = red die landing on 1
Are A and B disjoint?
the
probability
of A and B?
NO
P (A  B )  P (A)  P (B )  P (A  B )
1 1  1  1  11
     
6 6  6  6  36
An electronics store sells DVD players made by
one of two brands. Customers can also purchase
Thiswarranties
can happenfor
in one
two
ways: The
extended
the of
DVD
player.
following
areextended
given:
1) Theyprobabilities
purchased the
warranty
player
Let B1 = eventand
thatBrand
brand 11 DVD
is purchased
OR2 is purchased
B2 = event that brand
2) E
They
purchased the extended warranty
= event that extended warranty is purchased
and Brand 2 DVD player
P(B1) = .7
P(B2) = .3
P(E|B1) = .2
P(E|B2) = .4
If a DVD customer is selected at random, what
is the probability that they purchased the
extended warranty?
E  E  B1   E  B2 
DVD Player Continued . . .
Let
B1 = event that brand 1 is purchased
B2 = event that brand 2 is purchased
E = event that extended warranty is purchased
P(B1) = .7
P(B2) = .3
P(E|B1) = .2
P(E|B2) = .4
If a DVD customer
selected
at events
random, what
Theseisare
disjoint
is the probability that they purchased the
extendedUse
warranty?
the General Multiplication Rule:
E  E  B1   E  B2 
P E   P E  B1   P E  B2 
P E   P E | B1   P (B1 )  P E | B2   P (B2 )
DVD Player Continued . . .
Let
B1 = event that brand 1 is purchased
B2 = event that brand 2 is purchased
E = event that extended warranty is purchased
P(B1) = .7
P(B2) = .3
P(E|B1) = .2
P(E|B2) = .4
If a DVD customer is selected at random, what
is the probability that they purchased the
extended warranty?
P E   P E | B1   P (B1 )  P E | B2   P (B2 )
P(E) = (.2)(.7) + (.4)(.3) = .26
Practice with Homework
• Pg.372: #59-63odd, 64, 65, 76, 77