Counting Principles and Probability
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You will be able to find the number of
possibilities there are given certain conditions
using a variety of counting techniques.
You will be able to determine the probability
of an event occurring using a variety of
counting techniques depending on the
situation described.
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An efficient way of counting is necessary to
handle large masses of statistical data (e.g. the
level of inventory at the end of a given month, or
the number of production runs on a given
machine in a 24 hour period, etc.), and for an
understanding of probability.
Such techniques will enable us to count the
following, without having to list all of the items:
◦ the number of ways,
◦ the number of samples, or
◦ the number of outcomes.
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As an example, we may have an
event E defined as
◦ E = "day of the week“
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We write the "number of outcomes of event E"
as n(E).
So in the example,
◦ n(E)=7, since there are 7 days in the week.
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

Assume you have one red die and one green
die. (a die is the singular version of dice)
If you roll the dice together, what are the
different possibilities for the result?
Use a tree diagram to show the possible
results.


Take the word “Lions” and find out how many
combinations there are of any two letters…..
What if the word of origin was “Tigers”
instead?
Everyone select two numbers……create an ordered pair
How many ordered pairs are possible?
1
2
3
4
5
6
7
8
9
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Is there an easier way to find the number of
possible outcomes given each of these
circumstances?
Suppose that event E1 can result in any one
of n(E1) possible outcomes; and for each
outcome of the eventE1, there are n(E2)
possible outcomes of event E2.
Together there will be n(E1) × n(E2) possible
outcomes of the two events.

That is, if event E is the event that
both E1 and E2 MUST occur, then
n(E ) = n(E 1 ) × n(E 2 )
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If you toss a coin, what are the odds it lands on
heads?
What is the total number of possible outcomes
when a pair of coins is tossed?
The events are described as:
◦ E1 = toss first coin (2 outcomes, so n(E1) = 2.)
◦ E2 = toss second coin (2 outcomes, so n(E2) = 2.)
◦ They are independent, since neither toss affects the
outcome of the other toss.
◦ So n(E) = n(E1) × n(E2) = 2 × 2 = 4

Say we have 3 caps that we could wear with
2 t-shirts and 4 pairs of jeans. How many
different combinations could we choose
from?

We have 2 choices in the first row, 4 in the
second row and 3 in the third row. Together,
we will have
n(E) = n(E1) × n(E2) x n(E3) = 2 × 4 × 3 =
24 combinations
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
The words and in the description of an event
often indicates that the multiplication
principle should be …..
The word or often indicates that the addition
principle should be used.

The addition principle says: if events A and B
can in occur in a and b ways, respectively,
then either event A or event B can occur in
a+ b ways
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
If a coed softball team has 17 members (9
girls and 8 boys), and if one girl and one boy
are going to be selected as captains, then
there are ????? ways of filling the two
positions.
If a single player, who may be either a boy or
girl is going to be selected, then there are
???????? ways of making the selection.

How many different outcomes are there if you roll a pair of
simple dice?

If you draw a card from a standard deck of 52 playing
cards, how many red outcomes are possible?

Find 7!
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If you pick one card at random from a deck of cards, how
many outcomes are clubs? If you pick a club and put it
back, and draw another card, how many outcomes are
black?
How many three letter “words/combinations” are possible
from a seven letter word like “weather”?
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
Mutually Exclusive or Disjoint events occur
when one event precludes another occurring.
For Example:
◦ if the event is selecting a boy OR a girl;
◦ rolling an even versus an odd number;
◦ What might NOT be mutually exclusive events?

Mutually exclusive events require a
modification to the addition principle.
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One letter gives you 5 possible ways….
Two letters requires one letter and a second
letter….or 5 x 4 = 20
Three letters…5 x 4 x 3 or 60
Four letters…5 x 4 x 3 x 2 = 120
Five letters…..5 x 4 x 3 x 2 x 1 = 120
A “word” may be composed of one letter, or two,
or three or four or five…..thus..
The total number of words is
5+20+60+120+120 = 325 different possibilities
because they are mutually exclusive
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Factorials are an extension of the multiplication
principle…..remember a five letter word provides
5 x 4 x 3 x 2 x 1 or 120 possible
outcomes…..this can be written as 5!
Likewise a four letter word provides 24 possible
outcomes or 4 x 3 x 2 x 1 = 4!
The exclamation point following a number
indicates a factorial indicating you should
multiply all the integers from 1 to the number of
the factorial.
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An arrangement (or ordering) of a set of
objects is called a permutation. (We can also
arrange just part of the set of objects.)
In a permutation, the order that we arrange
the objects in is important!!
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
Consider arranging 3 letters: A, B, C. How
many ways can this be done?
The possible permutations are
ABC, ACB,
BAC, BCA,
CAB, CBA.


An arrangement (or ordering) of a set of
objects is called a permutation. (We can also
arrange just part of the set of objects.)
In a permutation, the order that we arrange
the objects in is important!!
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

When Arranging ”n ” objects
In general, n distinct objects can be arranged
in n! ways.
Example: If a family of six is asked to sit on a
bench, in how many different ways can they
arrange themselves?

The number of permutations of n distinct
objects taken r at a time where repetitions are
not allowed, is given by:
n
P​ = ​n! / (n−r)!​​
r
Most often, you will see this permutation using this
symbolic approach: nPr
Example: In how many ways can a supermarket manager
display 5 brands of cereals in 3 spaces on a shelf?

The number of different permutations of n objects of which n1 are of
one kind, n2 are of a second kind, ... Nk are of a k-th kind is:
n! / n !×n ​!×n ​!×…×n ​!
​1​
2
3
k
Example: In how many ways can the six letters of the word
"mammal" be arranged in a row?
Solution: Since there are 3 "m"s and 2 "a"s in the word "mammal",
we have:
6! / ​3!2!​​ = 60
There is one "L" in "mammal", but it does not affect the answer,
since 1! = 1.
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There are (n−1)! ways to arrange n distinct
objects in a circle (where the clockwise and
anti-clockwise arrangements are regarded as
distinct.)
Example: In how many ways can 5 people be
arranged in a circle?
Solution: (5−1)!=4!=24 ways
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How many different number-plates for cars can be made if
each number-plate contains four of the digits 0 to 9followed
by a letter A to Z, assuming that
(a) no repetition of digits is allowed?
Solution: There are 10 possible digits (0,1,2,…,9) and we need
to take them 4 at a time. There are 26 letters in the alphabet.
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With no repetition, we have:
10P ​4 ​×26=​10!/ (10−4)!​​×26 =​10!/6! ​×26 =131040

How many different number-plates for cars
can be made if each number-plate contains
four of the digits 0 to 9followed by a letter A
to Z, assuming that repetition of digits is
allowed?
Solution: With repetition, we have simply:
(number of digits from 0000 to 9999)×26
=10,000×26
=260,000
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The life insurance policies of an insurance company are classified
by:
age of the insured:
◦ under 25 years,
◦ between 25 years and 50 years,
◦ over 50 years old;
sex;
marital status:
◦ single or
◦ married.
What is the total number of classifications?
Solution: 3 x 2 x 2 = 12

An ice-cream shop offers 3 types of cones
and five different flavors of ice cream. How
many possible ice cream cone combinations
are there?
Solution: 3 x 5 = 15
How many numbers greater than 1000 can be formed with the
digits 3,4,6,8,9 if a digit cannot occur more than once in a number?

Solution: This is choosing 4 from 5 (any 4 digit number chosen
from 3,4,6,8,9 will be >1000) plus 5 from 5 (any 5digit number will be >1000),
where order is important.
So the number of ways we can arrange the given digits so that our resulting
number is greater than 1000such that no digit occurs more than once, is:
P4 + 5P5 = 5! / (5-4)! + 5! / (5-5)!
5
= 5! / 1! + 5! / 0! = 240

How many different ways can 3 red, 4 yellow
and 2 blue bulbs be arranged in a string of
Christmas tree lights with 9 sockets?
Solution: We use Theorem 3!!
So, 9! / 3! x 4! x 2! = 1260

In how many ways can 5 people be arranged in a
circle such that two people must sit together?
Solution: Regard the 2 people who sit together as
one "unit" and the other 3 people as 3 "units".
Arrange 4 "units" in a circle: So (4−1)!=3!=6 ways
Number of permutations of 2 people who sit
together: 2!=2
So:
6×2=12 ways
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Evaluate 7! / 5!
Solution: 42
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Find 9!
Solution: 362,880

The local pizzeria offers a choice of 2 pizzas
– supreme or vegetarian, 3 sides – chips,
salad or coleslaw, and 4 drinks – juice, coke,
ginger ale, or water. For dinner, I decide to
have 1 pizza, 1 side and 1 drink. How many
possible meals to I have to choose from?
Solution: 2 x 3 x 4 = 24

How many 4-digit numbers can be formed from
the digits 1, 2, 3, 4 if (a) repetitions are allowed?
and (b) repetitions are not allowed?
Solution: When repetitions are allowed, then each
digit provides four options….1, 2, 3 or 4…. Thus
there are 4 x 4 x 4 x 4 or 256 possible outcomes.
When repetitions are not allowed, then each digit
provides one fewer option than the digit before…
or 4 x 3 x 2 x 1 or 24 possible outcomes.
A teacher wants to randomly choose 5 students from the
class of thirty to receive an extra assignment. How many
ways can this be done?
How many ways can it be done where the first person chosen
is given 5 questions to answer, the second is given 4
questions, and so on until the 5th person who gets 1
question to answer?
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
Solution: In the second case, the order that the student are chosen does
matter, and so there are
P5 = n! / (n-r)!
= 30! / 25! = 30 × 29 × 28 × 27 × 26 = 17, 100, 720 ways to choose the
students with those conditions. This is a significantly more as expected.
30
Solution: In the first case, however, the 5 people are chosen, and it doesn’t
matter whether a person is chosen first, second or fifth, they all receive the
same extra work, and so there are actually fewer possibilities….. Further,
none of the four permutation rules apply…..why?
When the order does not matter, then you have something called a
“combination”
𝐶520 = 142,506
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
A combination of n objects taken r at a time
is a selection which does not take into
account the arrangement of the objects. That
is, the order is not important
A permutation is a special kind of
combination….because a permutation
requires order
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
Consider the selection of a set of 4 different letters
from the English alphabet.
Suppose
◦ David selected A, E, R, T;
◦ Karen selected D, E, N, Q; and
◦ John selected R, E, A, T
Note: David and John selected the same set of letters,
even though they selected them in different order.
Hence, these 3 people have selected only 2 different
sets of 4 letters (not 3 sets!!).
Question: How many different sets of 4 letters can be
selected from the alphabet?


There are 26P​4​ ways of arranging any 4 letters chosen from the
alphabet (where the order is important):
P =​ 26! / (26−4)!​​ =​ 26!​​ / 22!​​ =358,800
26
4

But in this question, the order is not important. Any set of 4 letters
chosen can be arranged in 4! ways.
So, the number of different sets of 4 letters is

26

P / 4! = 358,800 / 24 = 14,950
4

The number of ways (or combinations) in
which r objects can be selected from a set
of n objects, where repetition is not allowed,
is denoted by:
n

C = n! / r!(n-r)!
r
Try it: Find the number of ways in
which 3 components can be selected from a
batch of 20 different components.

Solution
◦
C = 20! / 3! ( 20-3)! = 20! / 3! (17!) = 1140
20
3
In how many ways can a group of 4 boys be
selected from 10 if
(a) the eldest boy is included in each
group? Hint: Choose 3 from 9, since the
eldest boy is fixed:
So, C = 9! / 3! (9-3)! = 84

9
3
(b) the eldest boy is excluded? Hint: If
eldest boy is excluded, it is actually
choose 4 boys from 9
So, C = 9! / 4! (9-4)! = 126
9
4
the
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What proportion of all possible groups
contain the eldest boy?
C = 10! / 4! X 6! = 210
10
4
So the proportion of all possible groups
containing the eldest boy is:
​84 / 210 ​=40%

Out of 5 mathematicians and 7 engineers, a
committee consisting of 2 mathematicians
and 3 engineers is to be formed. In how many
ways can this be done if
(a) any mathematician and any engineer
can be included?
(b) one particular engineer must be in the
committee?
(c) two particular mathematicians cannot
be in the committee?

5
C x C =
2
7
3
5!
2! X 3!
X
7!
3! X 4!
= 350
C2 x 6C2 =

5

3
C x C =
2
7
3
5!
2! X 3!
X
3!
X
2! X 1!
6!
2! X 4!
7!
3! X 4!
= 150
= 105

Find 10P4

5040

Find C

210


10
4
How many “words” of three letters can be
made from the word insulate?
𝑃38 = 336
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How many different ways can you arrange 7
different cans on a shelf?
7! = 5040
Six different prizes are given by drawing
names from the 69 orchestra members
attending the annual banquet. In how many
ways can the prizes be given if no one can
receive more than one prize?
𝑃669 = 8.63 x 1010
1.
How many different possible “full house” (one pair, one
three of a kind) hands are there in 5 card poker?
SOLUTION:
 The 3 of a kind can be chosen in 13 * (4 choose 3) ways as
there are 13 possible ranks and ( 4 choose 3) ways to choose
3 cards from a given rank.
 The pair (or 2 of a kind) can be chosen in 12 * (4 choose 2)
ways as there are 12 = 13 - 1 possible different ranks and (4
choose 2) ways to choose 2 cards from a given rank.
 In total, there are 13 x 4 x 12 x 6 = 3,744 ways to form a full
house:

C
13
1 ×4
C
3 ×12
C
1 ×4
C
2

2. We choose 12 cards from the usual deck
of 52 playing cards.

◦ How many different ways can this be done?
◦ Solution:
52
C12 = 206,379,406,870
◦ How many ways can it be done if they must all
come from the same suit?
◦ Solutions:
4
C x C
1
13
12
=
52
◦ How many ways can it be done if we need exactly
3 kings and 3 queens?
C×C× C
◦ Solution:
4
3
4
3
44
6
= 112,944,832
◦ How many ways can it be done if all cards must
have different face values?

◦ Solution:
C × 412
13
12
= 218,103,808
3. There are 10 boys and 11 girls at a
school.
◦ How many different ways can the boys each
choose a girl to take to the formal?
◦ Solution:
P
11
10
or
39,916,800
◦ One of the girls doesn’t want to go to the formal,
how many ways are there to make this choice
now?
◦ Solution:
10! 0r 3,628,800
4. A town of 30 people is to choose a
committee of 3 to represent them, how many
different ways can this be done? How many
ways can it be done if one person is to be the
chairperson, one the treasurer and one the
secretary?
Solutions:
30
C
30
P
3
3
or 4,060 and
or 24,360
5.

In a certain electorate there are 6
candidates: labor, liberal, greens, and three
independents. Their names are to be placed
in random order on the ballot paper. In how
many ways can this be done if
(a) the labor candidate comes first?
Solution: 5! or 120
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(b) the liberal candidate comes first?
Solution: 5! or 120

(c) the three independent candidates are
together?
Solution: 3! X 4! or 144
6.
From a class of 30 students, five students
are to be selected to complete a survey. In
how many ways can the choice be made?
C
Solution:
30
or,
30! / 5! (25!) = 142,506
5
7. You are to pick three singers for the
Superbowl, one to sing before the game, one
to sing at the end of the game, and one to sing
at halftime. If there are 12 applicants, how
many different possibilities are there if each
singer can only sing once?
Solution:
C × 3! =
12
3
1,320
8. How many different arrangements of the
word ELLIPSE are possible if

(a) there are no restrictions?
Solution: 7! / 2! X 2!

=
1,260
(b) the arrangement starts with S?
Solution: 6! / 2! X 2!
=
180

(c) both L’s are together?
Solution: 6!/ 2!

=
360
(d) the letters are in alphabetical order?
Solution: 2! X 2!
=4
9.
Seven people sit in a circle. How many ways can this be
done if

(a) there are no restrictions?
Solution:

=
720
(b) two people A and B sit together?
Solution:

7! / 7
2! X 6! =
6
240
(c) three people A, B and C sit together?
Solution:
3! X 5!
5
=
144
10. 5 cards are to be chosen from a standard
52-card deck. In how many ways can this be
done if
 (a) all of the cards are clubs?
Solution:

C5 = 1,287
13
(b) all of the cards are of the same suit?
Solution:
C ×
4
1
13
C
5
=
5,148

(c) there are three clubs and two spades?
Solution:

C
13
3
×
13
C = 22,308
2
(d) there are three of one suit and two of
another?
Solution:
C × C × C × C
4
1
13
3
3
1
13
2
= 267,696
11.

A bag contains 5 red, 6 blue and 4 yellow marbles. Three are drawn
out at random. In how many ways can they be drawn so that
(a) are all blue?
Solution:

C
3
20
=
(b) are all the same color?
Solution:

6
C + C + C
5
3
6
3
4
3
=
120
(c) are all different colors?
Solution:
5 × 6 × 4 = 120
12.
In how many ways can you choose 10
people out of 30 to sit on a bench in order,
if two particular people must be selected
and seated together?
Solution:
2! × C × 9!
28
8
=
13.
You are to pick three singers for the
Superbowl, one to sing before the game,
one to sing at the end of the game, and one
to sing at halftime. If there are 12
applicants, how many different possibilities
are there if each singer can only sing once?
Solution:
C × 3!
12
3
=
1,320
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

If you are rolling a regular die, what is the probability of
rolling an even number?
You are randomly choosing a card from a deck of 52
cards. What is the probability that the card you pick will be
a king?
You are visiting a kennel that has three German
shepherds, four Labrador retrievers, two Chihuahuas,
three poodles, and five West Highland terriers. When you
arrive, the dogs are taking a walk. What is the probability
of seeing a German shepherd first?
Two out of three students in Mr. Allen's class prefer
buying lunch to bringing it. Twenty students prefer buying
lunch. How many students are in Mr. Allen's class?


(Equally Likely Probability Formula): If S is a
finite sample set in which all outcomes are
equally likely and E is an event in S, then the
probability of E, denoted P(E) is the number
of elements in E divided by the number of
elements in S.
Symbolically this is: P(E) = N(E)
N(S)
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
There are three doors on the set of a game show.
Behind only one of the doors is a prize. You pick
a door. Then the host picks a different door and
shows you that it is empty. What is the
probability that the door you picked has the
prize? The host asks if you want to change your
pick to a different door instead, should you do
so?
https://www.google.com/webhp?sourceid=chro
me-instant&ion=1&espv=2&ie=UTF8&safe=active&ssui=on#q=let%27s+make+a+de
al+problem+youtube+monty+hall&safe=strict&s
tart=10

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P(A or B) = p(A) + p(B) – p(A and B) for any
two events A and B;
P( A or B) = p(A) + p(B) for mutually exclusive
events A and B.
If two events are independent then
P(A and B)  P(A) * P(B)
If two events are dependent then
P(A and B)  P(A) * P(B given A)
Conditional
Probability
We will discuss this
more later


A card is drawn at random from a deck of
cards. Find the probability of getting the 3 of
diamond.
P(E) = 1 / 52


A jar contains 3 red marbles, 7 green marbles
and 10 white marbles. If a marble is drawn
from the jar at random, what is the
probability that this marble is white?
P(E) = N (White) / Total Number = 50%
Two dice are rolled, find the probability that the
sum is
a) equal to 1
b) equal to 4

Solution: P(E) = n(E) / n(S) = 3 / 36 = 1 / 12
c) less than 13

Solution:P(E) = n(E) / n(S) = 36 / 36 = 1



A die is rolled and a coin is tossed, find the
probability that the die shows an odd number
and the coin shows a head.
Let E be the event "the die shows an odd
number and the coin shows a head".
P(E) = n(E) / n(S) = 3 / 12 = 1 / 4

Find the number of possible ways to arrange
the letters of the word “Travesty”

How many words are possible from the word
“possibilities”?

Solution 1:
8!/2! = 40,320/2 = 20,160

Solution 2:
13! / 3! x 4! = 43,243,200




King Arthur and his twelve Knights sat around a
round table deciding who was going to go out
and save the fair maiden. How many
arrangements are possible around their table?
Solution: (13-1)! = 12! = 479,001,600
King Arthur prefers to sit beside Sir Lancelot and
Sir Gawain, so they always sit beside each other.
Now how many ways are there to arrange the
seating?
Solution: (11-1)! x 3! = 10! x 3! =21,772,800




The basketball coach is having tryouts this week
for his travelling summer team. He has 15
players trying out for 10 spots. How many
different ways are there to fill the team?
Solution:
C10 = 3003
15
If the coach wants to assign playing time based
on when each player is selected to the team, how
many ways are there to fill the team?
Solution:
P10 = 15! / 5! = 10,897,286,400
15
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


If you want to draw a card from a standard
deck of cards, what is the probability that you
will draw a card that is even?
Solution:
20 / 52 = 5 / 13 = .3846
What is the probability that you will draw a
card that is both a face card and a diamond?
Solution:
3 / 52




If you draw two cards from a standard deck
what is the probability you draw two Aces?
Solution: 4/52 x 3/51 = 1/13 x 1/17 = 1/221
If you have five playing cards in your hand
from a standard deck of cards, how many
ways are there to arrange them in your hand?
Solution:
5! = 120




What is the probability that you are able to roll a
die three times in a row and get three straight
even numbers?
Solution: ½ x ½ x ½ =1/8
Hummingbirds rarely land, but when they do,
they do it quickly so you have to count them fast.
If seven hummingbirds land simultaneously on a
straight bird feeder, in how many ways can they
arrange themselves?
Solution:
7! = 5040

If there is only room for three of the
hummingbirds to land at a time out of our
seven hummingbirds, how many different
ways can they land?

Solution:
C3

What is the probability of flipping a coin and
it landing on heads four times in a row?

Solution:
7
=
35
½ x ½ x ½ x ½ = 1/16
P( AandB)
P( B | A) 
P( A)
“The Probability of B
given, event A has
occurred”
https://www.khanacademy.org/math/statisticsprobability/probability-library/conditionalprobability-independence/v/calculating-conditionalprobability

For Independent Events, meaning each event
is not affected by any other events.
◦ For example, tossing a coin multiple times…each
event is independent of each other event.

and

For Dependent Events, they can be affected
by previous events ...
◦ For example, drawing marbles out of a bag and not
replacing them affects each subsequent draw….

We know that to find the probability of two
events occcurring we multiply their individual
probabilities….this is the “multiplication
principle for probabilities….

P(A and B) = P(A) x P(B)

Conditional probability changes this for
dependent events to

P(A and B) = P(A) x P(B given A)


So this idea of conditional probability reflects
the fact that dependent events can change
the probability of a subsequent event.
Thus the P(event B given event A) = P(A and B)
P(A)
From this, P(A and B) = P(A) x P(B given A)
To this, P(B I A) = P(A and B)
P(A)
using simple algebra
P( AandB)
P( B | A) 
P( A)
“The Probability of B
given, event A has
occurred”
https://www.khanacademy.org/math/statisticsprobability/probability-library/conditionalprobability-independence/v/calculating-conditionalprobability
Example: Drawing 2 Kings from a Deck
Event A is drawing a King first, and Event B is drawing a King
second.
For the first card the chance of drawing a King is 4 out of 52
(there are 4 Kings in a deck of 52 cards):
P(A) = 4/52
But after removing a King from the deck the probability of
the 2nd card drawn is less likely to be a King (only 3 of the
51 cards left are Kings):
P(B|A) = 3/51
And so:
P(A and B) = P(A) x P(B|A) = (4/52) x (3/51) = 12/2652
= 1/221
So the chance of getting 2 Kings is 1 in 221, or about 0.5%

Try this one…
70% of your friends like Chocolate Ice Cream, and
35% like Chocolate AND like Strawberry.

What percent of those who like Chocolate also
like Strawberry?
P(Strawberry|Chocolate) = P(C and S)
P(C)

0.35 / 0.7 = 50%

https://www.youtube.com/watch?v=WmcoWd
8Uv-0
You are off to soccer, and want to be the
Goalkeeper, but that depends who is the Coach
today:
◦ with Coach Sam the probability of being Goalkeeper
is 0.5
◦ with Coach Alex the probability of being Goalkeeper
is 0.3

Sam is Coach more often ... about 6 out of every
10 games (a probability of 0.6).

So, what is the probability you will be a
Goalkeeper today?

Let’s look at this as a tree diagram….

A math teacher gave his class two tests. 25%
of the class passed both tests and 42% of the
class passed the first test. What percent of
those who passed the first test also passed
the second test?
P(second I first) = P(1st & 2nd)
P(1st)
= .25/.42 = .60

Solution:


A jar contains black and white marbles. Two
marbles are chosen without replacement. The
probability of selecting a black marble and
then a white marble is 0.34, and the
probability of selecting a black marble on the
first draw is 0.47. What is the probability of
selecting a white marble on the second draw,
given that the first marble drawn was black?
P(W I B) = P(B & W)/P(B) = .34/.47 =.72= 72%

The probability that it is Friday and that a
student is absent is 0.03. Since there are 5
school days in a week, the probability that it
is Friday is 0.2. What is the probability that a
student is absent given that today is Friday?
P(A I F) = P(F & A) = .03/0.2 = .15 = 15%
P(F)

At Kennedy Middle School, the probability
that a student takes Technology and Spanish
is 0.087. The probability that a student takes
Technology is 0.68. What is the probability
that a student takes Spanish given that the
student is taking Technology?
P(S I T) = P(T & S) = .087/.68 = .13 = 13%
P(T)

In a school 65% of students like pork chops.
Two students from the school are picked at
random. Given that at least one of them likes
pork chops, what is the probability that both
like pork chops?

48.15%




Donald, the quarter back, has 2 wide receivers.
He throws to Goofy three out of five plays and
Goofy drops the ball 90% of the time. Donald
throws to Pluto two out of five plays and Pluto is
able to catch the ball 70% of the time.
a) Find the probability that the ball is dropped by
either Goofy or Pluto.
b) If a ball is dropped, then the probability that it
as Goofy who dropped it is_____
c) Given that the pass is caught, what is the
probability that it was Goofy who caught it?

A) 0.66

B) 0.818

C) 0.176

Darrell has just left his social studies class
and bumped into his friend Carla. Darrell
told Carla that the social studies teacher gave
a ten-question true or false quiz today.
When Carla asks about the quiz, Darrell says
he found it easy and thinks that four of the
answers are false. When Carla takes the quiz,
in how many ways can she select four
questions to mark false?

The director of the music department is
holding tryouts for the school’s jazz band.
There are seven students competing for three
saxophone positions, 8 for two piano spots,
five for two percussion spots and 12 for three
places as guitarists. In how many ways can
she make her selections for the jazz band?

A binomial random variable is the number of
successes x in n repeated trials of
abinomial experiment.
The probability distribution of
abinomial random variable is called
a binomialdistribution. Suppose we flip a coin
two times and count the number of heads
(successes).

https://www.khanacademy.org/math/probabi
lity/binomial-probability-a2/binomialprobability-a2i/v/probability-of-making-2shots-in-6-attempts

The formula for the binomial distribution is
shown below: where P(x) is the probability of
x successes out of N trials, N is the number
of trials, and π is theprobability of success on
a given trial. Applying this to the coin flip
example, If you flip a coin twice, what is
the probability of getting one or more heads?

We have a coin and after toss we can have
only two outcomes, Head or Tail. Suppose we
toss the coin 10 times. If we want all of the
ten outcomes to be head then what will be
the probability for this?
In one toss for head to come as outcome the
1
probability is
2
Hence, for ten tosses, the probability is
Or,
1
1024
1 10
( )
2

If a coin is tossed ten times then what will be
the probability for getting 7 heads and 3 tails
in 10 tosses?

For 7

For 3
1 7
heads, it will be ( ) ….
2
1 3
tails, it will be ( ) …
2

But still one more point is needed to get the
complete solution. Out of 10 tosses, which are
those seven who will give heads?
The number of possible events will be 107𝐶

Hence, the probability to get 7 heads out of 10

tosses of a coin will be, P =
 Or……
15
128
7
10𝐶
*
1 7
1 3
( ) *( )
2
2

In a store, out of all the people who came
there thirty percent bought a shirt. If four
people came in the store together then find
the probability of one of them buying a shirt.



The probability of buying a shirt will be 0.3
and not buying a shirt is given as 1 - 0.3 =
0.7
Now, if one person will buy a shirt out of
four, then the probability for this event will
be: 14𝐶 ∗ 0.3 ∗ (0.7)3
Or, 0.2916 or around 29%

In a hospital sixty percent of patients are
dying of a disease. If on a certain day, eight
patients got admitted in the hospital for that
disease what are the chances of three to
survive?


The probability of the patient dying because
of the disease is 0.6. So, the probability of
the patient being cured is 1 - 0.6 = 0.4.
Out of the 8 patients, the probability survival
of 3 patients will be,
3
 8𝐶

∗ (0.4)3 ∗ (0.6)5
Or……. 0.2787 or around 27.9%



Suppose we are throwing a die three times. Find
the probability of finding a multiple of 3 in one
of the throws.
There are two cases of getting a multiple of 3,
that is, 3 and 6. Hence, the probability of getting
2
1
a multiple of three will be =
6
3
The probability of not getting a multiple of 3 is
1
2
1- =
3

3
Hence, the probability of getting a multiple of 3
in one out of three events will be,
𝟏
 𝟑𝑪
𝟏
𝟑
𝟐
𝟑
∗ ( )𝟏 * ( )𝟐 =
𝟐
𝟗


In a restaurant seventy percent of people order
Chinese food and thirty percent opt for Italian
food. A group of three persons enter the
restaurant. Find the probability of at least two of
them ordering for Italian food.
The probability of ordering Chinese food is 0.7
and the probability of ordering Italian food is 0.3.
Now, if at least two of them are ordering Italian
food then it implies that either two or three will
order Italian food.
Probability for two ordering Italian food,

P(2 of 3) = 23𝐶 ∗ (0.3)2 ∗ (0.7)1 = 0.189 or around
19% will order Italian food

But all three COULD order Italian, so you must
consider this probability as well.

Probability for all three ordering Italian food…
P(3 of 3) = 33𝐶 ∗ (0.3)3 ∗ (0.7)0 = 0.081 or around
8.1%

So all together, .189 + .081 = 0.27 or 27% is the
probability that at least two will order Italian food.


In an exam only ten percent students can
qualify. If a group of 4 students have
appeared, find the probability that at most
one student will qualify?




For at most one student to qualify, either 1
student will qualify or none of the 4 will qualify.
Probability for a student to qualify= 0.1
Probability for a student to disqualify= 0.9
The probability of qualification of,
P(none) = 04𝐶 ∗ (0.1)0 ∗ (0.9)4 = .656
P(1 of 4) = 14𝐶 ∗ (0.1)1 ∗ (0.9)3 = .073
Hence, the probability of at the most one student
to qualify out of four will be, .656 + .073 = .729


A spinner is divided into 12 sections that are
each equally likely to occur. The sections are
lettered from A to L. Phillip will spin the spinner
3 times. What is the probability that the spinner
will land on the letter G exactly 1 out of 3 times?
Solution: P(G) = 1/12 and the P(NG) = 11/12
One possibility is 1/12 x 11/12 x 11/12 = 121/1728
because there are three spins…the G can show up in three
spots so 121/1728 x 3 will give you 121/576



Measures of Central Tendency
Standard Deviation
Normal Distribution

http://www.mathplayground.com/SMP_Proba
bility.html

http://www.regentsprep.org/regents/math/al
gebra/MultipleChoiceReview/Probability.htm