probability – chapter 15 - wordpress.com · 2013-05-21 · ... 2 blue and 5 yellow tickets placed...
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Probability How can probability be found?
One way is using long term relative frequency:
( ), where =# of trials, =# of times occurs,
and ( )=Probability of
n AP A N n A A
N
P A A
Note: the more trials you have the more accurate the probability is.
Sample Space: A list of every possible outcome
Example: a) List the sample space for rolling a die
b) List the sample space for flipping 2 coins
a) {1,2,3,4,5,6} b) {HH, HT, TH, TT}
Theoretical Probability:
#of outcomes in which occurs( )
#of outcomes in the Sample space
n A AP A
n U
Note that 0 ( ) 1P A
a) 1/4 b) 1/9 c) 4/9 d) 1/36 e)1/18 f) 1/5
If the probability of an event is P, in n trials then the number of times the
event should appear is times.
If the probability of getting an A or above in IB Math is 30% and there are
40 students find the number of expected grade of A or above?
How can we solve probability questions:
1. Venn diagrams
2. Tree diagrams
3. Lattice diagrams
4. Frequency Table
5. Another diagram that works
6. Using complementary events
Complementary Events
Lattice diagram: Rolling 2 dice
1 2 3 4 5 6 (Each intersection represents an outcome)
1
2
3
4 Example: What is the P(same number)?
5
6
Tree diagram: Tossing 2 coins
Coin 1 Coin 2 H
H Sample space = {HH, HT, TH, TT}
T
T H
T
Basic Probability Rules: and means multiply
or means add
Example: There are 3 red, 2 blue and 5 yellow tickets placed in a basket.
Determine the probability of getting:
a) a red ticket b) a green ticket
c) not a yellow ticket d) a red or blue ticket
e) a red, yellow or blue ticket f) ( )P R or Y
g) ( )P R h) ( )P R or B
i) ( )P R or Y or B
Solution:
a) 3/10 b) 0 c) 5/10 d) 5/10 e) 1
f) 8/10 g) 7/10 h) 5/10 i) 0
Example: A bag has 20 cards with the numbers 1 to 20 on them.
If a card is drawn at random what is the probability that it is
divisible by 3 or by 4?
3: 3, 6, 9, 12, 15, 18, 4: 4, 8, 12, 16, 20
Therefore:
Venn Diagrams
A Venn diagram is drawn as a rectangle which represents the
universal set U with one or more circles inside representing
subsets. Each area has the number of elements in it. Each subset
has a letter to represent it.
U
A B A = court shoes
B = runners
U = shoe company
100 35 150
20
305
Subsets: B A
A B
Complement: ( ) 1 ( )P A P A
A A
Intersection: A ∩ B
All elements in both A and B A B
Union: A B
All elements in A or B (or both) A B
( ) ( ) ( ) ( )P A B P A P B P A B
Disjoint (Mutually exclusive)
Disjoint sets have no points in common. A B
A ∩ B = 0
( ) ( ) ( )P A B P A P B
Example: If ( ) 0.6, ( ) 0.3,and ( ) 0.2P A P B P A B find:
a) ( )P A B b) ( )P B c) ( )P A B
a)
b)
c)
Example: A coin is tossed 3 times.
a) Draw a tree diagram and write the sample space.
b) P(only one tail)
c) P(at least 2 tails)
d) P(exactly 2 tails in a row)
e) P(exactly 2 tails)
b) 3/8 c) 4/8 d) 2/8 e) 3/8
Conditional Probability
( )P A B is the probability of A occurring given that B has occurred.
A B
4 5 7
What is the probability of ( )P A B ?
5( )
12P A B
What is the formula for conditional probability?
5 is ( )n A B and 12 is ( )n B
Therefore: ( )
( )( )
n A Bn A B
n B or
( )( )
( )
P A BP A B
P B
Another way to write that is ( ) ( ) ( )P A B P A B P B
Note: ( ) ( )P A B P B A
Independent Events do not affect each other so: ( ) ( )P A B P A and
( ) ( )P B A P B if A and B are independent.
If they are independent then ( ) ( ) ( )P A B P A P B
Remember: Mutually exclusive means the events A and B have no
intersection (nothing in common).
Students pick from one of two boxes. Each box contains red and white
shirts. A spinner is used to decide which box to pick from. The probability
of picking from Box A is 3/5. Box A contains 1 R and 4 W shirts. Box B
contains 2 R and 3 W shirts. Determine the probability of the following:
(Draw a tree diagram first)
a) ( )P R b) ( )P W
c) ( )P A R d) ( )P A W
e) ( )P B R f) ( )P B W
Box Shirt
1/5
R
3/5 A
4/5 W
2/5
2/5 B R
3/5 W
a) 3 1 2 2 7
( )5 5 5 5 25
P R b) 3 4 2 3 18
( )5 5 5 5 25
P W
c) ( )P A R
3 1
( ) 35 57( ) 7
25
P A R
P R
d) ( )P A W
3 4
( ) 12 25 518( ) 18 3
25
P A W
P W
e) ( )P B R
2 2
( ) 45 57( ) 7
25
P B R
P R
f) ( )P B W
2 3
( ) 6 15 518( ) 18 3
25
P B W
P W
Example (N01- Non calculator):
The events B and C are dependent, where C is the event “a student
takes Chemistry”, and B is the event “a student takes Biology”. It
is known that
P(C) = 0.4, P(B | C) = 0.6, P(B | C) = 0.5.
(a) Complete the following tree diagram.
(b) Calculate the probability that a student takes Biology.
(c) Given that a student takes Biology, what is the probability
that the student takes Chemistry?
0.4 C
C
B
B
B
B
BiologyChemistry