Conditional Probabilities

The conditional probability of event A given that event B has already

occurred is defined as

Note that the conditional probability of B given A is defined as

 General Multiplication Rule
The probability of the intersection of any two events A and B is

If two events A and B are independent if and only if the

if any one of the these equations

is true, then we can conclude that
events A and B are independent.

Otherwise, they are dependent.

 Example 1
Toss a fair coin twice. Define
○ A: head on second toss
○ B: head on first toss
P(A|B) = ½
HH 1/4 P(A|not B) = ½

HT 1/4
P(A) does not A and B are
TH 1/4 change, whether B independent!
happens or not…
TT 1/4
 Example 2
A bowl contains five balls, two red and three blue. Randomly select two
balls, and define
○ A: second candy is red.
○ B: first candy is blue.

P(A|B) =P(2nd red|1st blue)= 2/4 = 1/2

P(A|not B) = P(2nd red|1st red) = 1/4

P(A) does change, A and B are

depending on whether B dependent!
happens or not…
 Example 3: Two Dice
Toss a pair of fair dice. Define
○ A: red die show 1
○ B: green die show 1

P(A|B) = P(A and B)/P(B)

=1/36/1/6=1/6=P(A)

P(A) does not A and B are

change, whether B independent!
happens or not…
 Example 4: Two Dice
Toss a pair of fair dice. Define
○ A: add to 3
○ B: add to 6

P(A|B) = P(A and B)/P(B)

=0/36/5/6=0

P(A) does change A and B are dependent! In

when B happens fact, when B happens, A
can’t
 Example 5
Box 1 contains two red balls and one blue ball. Box 2 contains three blue balls
and one red ball. A coin is tossed. If it falls heads up, box 1 is selected and a
ball is drawn. If it falls tails up, box 2 is selected and a ball is drawn. Find the
probability of selecting a red ball.

Solution: P(red) = (1/2)(2/3) + (1/2)(1/4) = 2/6 + 1/8 = 8/24 + 3/24 = 11/24.

 Example 6

Two fair dice are rolled . What is the conditional probability that the sum of
the two face is 6 given that the two dice are showing different faces?
Solutions:
Let: A-event the two dice are showing different faces .
B-event the sum of the two face is 6 .
 Example 7

In a certain population, 10% of the people can be classified as being high risk for a heart
attack. Three people are randomly selected from this population. What is the probability that
exactly one of the three are high risk?

Define N: not high risk Define H: high risk

P(exactly one high risk) = P(HNN) + P(NHN) + P(NNH)
= P(H)P(N)P(N) + P(N)P(H)P(N) + P(N)P(N)P(H)
= (.1)(.9)(.9) + (.9)(.1)(.9) + (.9)(.9)(.1)= 3(.1)(.9)2 = .243
 Example 8
Suppose we have additional information in the previous example. We know that only 49%
of the population are female. Also, of the female patients, 8% are high risk. A single person
is selected at random. What is the probability that it is a high risk female?

Define F: female Define H: high risk

From the example, P(F) = .49 and P(H|F) = .08. Use the
Multiplicative Rule:

P(high risk female) = P(HF)

= P(F)P(H|F) =.49(.08) = .0392
 Counting Rules
How many ways can it happen? When outcomes are equally likely to occur (like
when tossing a coin or rolling a die), you can use counting rules to find out how
many outcomes are possible and then use that number to find probabilities.

1.Multiplication rule : when outcomes are selected from more than one set or
group, multiply the number of outcomes for each set.

Example 1: How many different meals can be made by pairing up 3 main

courses and 4 side dishes?

Answer: 3 x 4 = 12 meals (What if you have 2 dessert options? Then you can
make 2 x 3 x 4 = 24 meals!)
 Counting Rules (1. Multiplication rule )
In general if a task can be performed in n1 ways, a second task in n2 ways and a
third task in n3 ways ….. , then the total number of distinct ways of performing
all tasks together is n1× n2× n3………

Example 2: How many 4 or 5 digit telephone numbers are possible, assuming

the first is not zero?
Answer : 9 × 10 × 10 × 10 + 9 × 10 × 10 × 10 × 10 = 99 000
…………………………………………………………………………………………………………
Example 3: Toss two coin ?
Answer : m*n=2*2=4
…………………………………………………………………………………………………………
Example 4: Throw two dice ?
Answer: m*n=6*6=36
 Counting Rules (1. Multiplication rule )
Example 5:
(a) How many different car number plates are Possible with 3 letters followed
by 3 digit ?
Answer : 26*26*26*10*10*10
(b) How many of these number plates begin with ABC
Answer :
(c) If a plate is chosen at random, what is the probability that it begins with
ABC?)
Answer: P(ABC) =
 Counting Rules (2. Permutation Rule )
when outcomes are selected from only one set and the order that they are
selected does matter, the total number of ways r outcomes can be chosen from n
outcomes is

The number of permutations of n objects using all of them is n!

 Counting Rules (2. Permutation Rule )
Example 1: In how many ways can 5 people line up in a queue?
Answer: 5! = 120
Example2: Suppose that a class president, vice-president, secretary, and
treasurer are to be randomly selected out of a group of 12 students nominated
and that the order in which they are picked determines which office they will
hold. What is the probability of getting a specific set of class officers?

Since there are 11,880 possible sets of class officers and they are all equally
likely, the probability of getting a specific set of class officers is
 Counting Rules (2. Permutation Rule )
Example 3:How many three -letter codes are there using letters A,B,C and D if
no letter can be repeated?
Answer : Note ,the order does matter P4,3

Example 4: A club has 12 members. How many ways could a president, vice-
president and treasurer be appointed.

Answer: P12,3
 Counting Rules (2. Permutation Rule )
 Permutations with repeated elements
If a bag contains some objects in which 𝒎𝟏 are of type 1, 𝒎𝟐 are of type 2, .... 𝒎𝒌
are of type k. The number of permutation is:

Example 5: How many ways can you permute the letters: B A N A N A ?

Of the 6 letters, there are 3 A's, 2 N's, and 1 B.
The 2 N's could be rearranged in 2! = 2 different ways.
The 3 A's could be rearranged in 3! = 6 different ways.
So we need to divide 6! by both 6 and 2. The number of ways to rearrange the
letters in BANANA is
 Counting Rules (3.Combination rule )
when outcomes are selected from only one set and the order that they are
selected does not matter, the total number of ways r outcomes can be chosen
from n outcomes is

Example 1: Three members of a 5-person committee must be chosen to form a

subcommittee. How many different subcommittees could be formed?
Answer:
5! 5(4)(3)(2)1 5(4)
The order of C 
5
   10
3!(5  3)! 3(2)(1)(2)1 (2)1
3
the choice is
not important!
 Counting Rules (3.Combination rule )
Example 2: Suppose there are 8 students in a group and that 5 of them must be
selected to form a basketball team. How many different teams could be formed?
What is the probability of ending up with one specific team?
Answer: Use the combination rule with n = 8 and r = 5 as shown below.

56 teams are possible and they are all equally is players are picked randomly, so
the probability of ending up with one specific team is
Suppose YOU were one of the 8 people to be selected for the team. What is
the probability that you would be selected to be on the team?
Answer : Since you must be on the team, we only need to select the other 4
players from the remaining 7, giving a total of C7,4 = 35 teams.
So, the probability of YOU being on the team is
 Counting Rules (3.Combination rule )
Example 3: From a club of 12 members, how many ways are there of
selecting a committee of three?
Answer: C12,3 = = 220

Example 4: A group consists of 8 boys and 5 girls: (a) How many ways can
you select 2 boys and 2 girls? (b) How many ways can you select a committee
of 4 containing at least 2 boys?
Answer:
to be continue…..