Probability – Chapter 15 - WordPress.com · 2013-05-21 · ... 2 blue and 5 yellow tickets placed in a basket. Determine the probability of getting: a) ... Conditional Probability

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)

Answers:

a) b)

d)

Dependent Events:

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

Solution:

(a)

(A1) (C1)

(b) P(B) = 0.4(0.6) + 0.6 (0.5) = 0.24 + 0.30 (M1)

= 0.54 (A1) (C2)

(c) P(CB) = (= 0.444, 3 sf) (A1) (C1)

[4]

0.4

0.6 0.5

0.5

0.4

0.6

9

4

54.0

24.0

)P(

)P(

B

CB