#### Transcript HSPA Prep Lesson 5 - Probability

```Department Store
A department store is divided into two sections, electronics and furniture. Each section offers a discount
rate; items in the same section sell at the same discount rate, but not necessarily at the same price.
A.) One of the items in the electronics section has an original price \$120 and a sale price of \$102. One of
the items in the furniture section has an original price of \$60 and a sale price of \$48. Determine the
(percent) discount rate in the electronics section and the (percent) discount rate in the furniture section.
B.) Using the discount rate found in part A, find the total amount, including a 6% sales tax, that Andrew paid
for an electric item and a furniture item if the original price of each of these items \$200. Show your
work.
C.) Andrew plans to buy an office lamp from the furniture section and a laptop computer from the electronic
section of this department store. The original price of the laptop computer is \$1,350. The original price
of the office lamp is \$89. Andrew is thinking of two methods to find the total amount he will save (not
including the tax). The two methods are stated below:
Method I: Find the discount on the laptop and find the discount on the lamp.
Find the sum of the two discounts
Method II: Find the sum of the original prices of the two items.
Find the average of the discount rates from part A
Use the average rate to find the discount on the sum.
Lesson 5
Probability of Independent and
Dependent Events
Slide 2
THEORETICAL AND EXPERIMENTAL PROBABILITY
The probability of an event is a number between
0 and 1 that indicates the likelihood the event will occur.
There are two types of probability: theoretical and
experimental.
THEORETICAL AND EXPERIMENTAL PROBABILITY
THE THEORETICAL PROBABILITY OF AN EVENT
The theoretical
probability
of anlikely,
eventthe
is
When
all outcomes
are equally
often simplyprobability
called the probability
of theA event.
theoretical
that an event
will occur is:
P (A) =
number of outcomes in A
total number of outcomes
all possible
outcomes
P (A) = 4
9
outcomes
in event A
You can express a probability as a fraction, a decimal, or a perce
For example:1 , 0.5, or 50%.
2
Finding Probabilities of Events
You roll a six-sided die whose sides are numbered from
1 through 6.
Find the probability of rolling a
4.
SOLUTION
Only one outcome corresponds to rolling a 4.
1
number of ways to roll a 4
=
P (rolling a 4) =
number of ways to roll the die 6
Finding Probabilities of Events
You roll a six-sided die whose sides are numbered from
1 through 6.
Find the probability of rolling an odd number.
SOLUTION
Three outcomes correspond to rolling an odd number:
rolling a 1, 3, or a 5.
number of ways to roll an odd number 3
= 6
P (rolling odd number) =
number of ways to roll the die
=
1
2
Finding Probabilities of Events
You roll a six-sided die whose sides are numbered from
1 through 6.
Find the probability of rolling a number less than
7.
SOLUTION
All six outcomes correspond to rolling a number less than 7.
number of ways to roll less than 7 6
=
= 1
P (rolling less than 7 ) =
number of ways to roll the die
6
There are 52 cards in
a deck. So what are
my chances of picking
an ace?
4
52
How many aces are in a
deck?
How many cards are in a deck?
So I have a
4/52 or 1/13
chance of
drawing an
ace!
P(# or #)
Mutually Exclusive Events
• Mutually exclusive events cannot occur
at the same time
• Cannot draw ace of spaces and king of
hearts
• Cannot draw ace and king
• But drawing a spade and drawing an
ace are not mutually exclusive
Exclusive Events
• Add probabilities of individual events
• Drawing ace of spades or king of hearts
– Probability of ace of spades is 1/52
– Probability of king of hearts is 1/52
– Probability of either ace of spades or king of
hearts is 2/52
Exclusive Events
• Add probabilities of individual events and
subtract probabilities of outcomes
common to both events
Drawing a spade or drawing an ace
– Probability of drawing a spade: 13 outcomes,
so 13/52 = 1/4
– Probability of drawing an ace: 4 outcomes, so
4/52 = 1/13
– Ace of spades is common to both events,
probability is 1/52
– So probability of drawing a spade or an ace is
13/52 + 4/42 – 1/52 = 16/52 = 4/13
Independent and Dependent
Events
• Independent events: if one event occurs, does
not affect the probability of other event
– Drawing cards from two decks
• Dependent events: if one event effects the
outcome of the second event, changing the
probability
– Drawing two cards in succession from same deck
without replacement
Multiplication Rule for Independent
Events
• To get probability of both events
occurring, multiply probabilities of
individual events
• Ace from first deck and spade from
second
– Probability of ace is 4/52 = 1/13
– Probability of spade is 13/52 = 1/4
– Probability of both is 1/13 x 1/4 = 1/52
Conditional Probability
• Probability of second event occurring
given first event has occurred
• Drawing a spade from a deck given you
have previously drawn the ace of spade
– After drawing ace of spades have 51
cards left
– Remaining cards now include only 12
– Conditional probability is then 12/51
Probability Practice Problems
• Suppose you have a bowl of disks
numbered 1 – 15.
• P(even)
even numbers = 2, 4, 6, 8, 10, 12, 14 = 7
total numbers
1 – 15 =
15
Probability Practice Problems
• Suppose you have a bowl of disks
numbered 1 – 15.
• P(even, more than 10)
The “,” indicates “and” (the disk must be
both even and more than 10)
even #’s & #’s greater than 10 = 12, 14
=2
total numbers
1 – 15 =
15
Probability Practice Problems
• Suppose you have a bowl of disks numbered 1
– 15.
• P(even or more than 10)
The “or” indicates the disk must be even or more
than 10. You must be careful not to include a
number twice
even #’s – 2, 4, 6, 8, 10, 12, 14 = 7/15
#’s greater than 10 = 11, 12, 13, 14, 15 = 5/15
Since 12 and 14 are common to both sets, you will
subtract 2/15
7/15 + 5/15 – 2/15 = 10/15 = 2/3
Probability Practice Problems
• Suppose you have a bowl of disks numbered 1
– 15. A disk is drawn, replaced, and a second
disk is drawn.
• P(even, even)
Find the probability of each independent event
and multiply
even #’s on first draw – 2, 4, 6, 8, 10, 12, 14 =
7/15
even #’s on second draw – 2, 4, 6, 8, 10, 12, 14
= 7/15
7/15 x 7/15 = 49/225
Probability Practice Problems
• Suppose you have a bowl of disks numbered 1
– 15. A disk is drawn, not replaced, and a
second disk is drawn.
• P(even, even)
Find the probability of each independent event
and multiply
even #’s on first draw – 2, 4, 6, 8, 10, 12, 14 =
7/15
even #’s on second draw – one less even
number than previous set/one less disk from
bowl = 6/14
Independent Events
Whatever happens in one event has absolutely nothing
to do with what will happen next because:
1. The two events are unrelated
OR
2. You repeat an event with an item whose
numbers will not change (eg.: spinners or
dice)
OR
3. You repeat the same activity, but you
REPLACE the item that was removed.
The probability of two independent events, A and B, is equal
to the probability of event A times the probability of event B.
P(A, B) = P(A)  P(B)
Slide 22
Independent Events
Example: Suppose you spin each of these two spinners. What
is the probability of spinning an even number and a vowel?
1
P(even) =
(3 evens out of 6 outcomes)
2
1
(1 vowel out of 5 outcomes)
P(vowel) =
5
1 1 1
 
P(even, vowel) =
2 5 10
1
6
P
S
5
2
O
T
3
4
R
Slide 23
Dependent Event
• What happens the during the second event
depends upon what happened before.
• In other words, the result of the second
event will change because of what
happened first.
The probability of two dependent events, A and B, is equal to the
probability of event A times the probability of event B. However,
the probability of event B now depends on event A.
P(A, B) = P(A)  P(B)
Slide 24
Dependent Event
Example: There are 6 black pens and 8 blue pens in a jar. If you
take a pen without looking and then take another pen without
replacing the first, what is the probability that you will get 2
black pens?
P(black first) =
6
3
or
14
7
5
P(black second) =
(There are 13 pens left and 5 are black)
13
THEREFORE………………………………………………
P(black, black) =
3 5
15

or
7 13
91
Slide 25
TEST YOURSELF
Are these dependent or independent events?
1.
Tossing two dice and getting a 6 on both of them.
2.
You have a bag of marbles: 3 blue, 5 white, and 12 red.
You choose one marble out of the bag, look at it then put it
back. Then you choose another marble.
3.
You have a basket of socks. You need to find the
probability of pulling out a black sock and its matching
black sock without putting the first sock back.
4.
You pick the letter Q from a bag containing all the letters of
the alphabet. You do not put the Q back in the bag before
you pick another tile.
Slide 26
Independent Events
Find the probability
• P(jack, factor of 12)
1
5
5
x
8
5
=
40
1
8
Slide 27
Independent Events
Find the probability
• P(6, not 5)
1
6
5
x
6
5
=
36
Slide 28
Dependent Events
Find the probability
• P(Q, Q)
• All the letters of the
alphabet are in the
bag 1 time
• Do not replace the
letter
1
26
0
x
25
0
=
650
0
Slide 29
Probabilities of Dependent Events
Determining probabilities of dependent events is usually
more complicated than determining them for independent
events.
Since some of the tree diagrams could get very large, we
will focus on a quicker method, multiplication.
Probabilities of Dependent Events
Ex.1) Independent Events:
Spinner #1 is partitioned into three equal sections, coloured
black, white, and grey. Spinner #2 is partitioned into four
equal sections, coloured red, blue, green, and yellow. If
both spinners are spun, what is the probability of getting
black and red?
Since we expect to get black one-third of the time, and we
expect to get red one-quarter of the time, then we expect
to get black and red one-third of one-quarter of the time. . .
Probabilities of Dependent Events
P (black and red )
1

of
3
1


3
1

12
1
4
1
4
Imagine a tree diagram where the first column shows the
three outcomes for Spinner #1, each of which is followed by
the four outcomes for Spinner #2 in the second column.
Three groups of four branches creates 12 possible outcomes.
Probabilities of Dependent Events
Ex.2) Dependent Events:
A bag contains 10 marbles; 5 red, 3 blue, and 2 silver. If you
draw one marble at random and hold it in your left hand, and
then draw a second marble at random and hold it in your
right hand, what is the probability that you are holding two
silver marbles?
It’s easy to determine the probability of the first marble being
silver. However, notice that if you start by getting a silver
marble and then try for the second, the bag will be different.
How? Now, there is only one silver marble in a bag
containing a total of 9 marbles. . .
Probabilities of Dependent Events
2 1
P( silver and silver) 

10 9
1
1


5
9
1

45
color
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