Probability trinity college

Trinity College
20th January 2014

Introduce..

The Fundamental Principle of
Counting

Operations
•The result of an operation is an outcome.
For example, if we throw a die, one
possible outcome is 5.
If we throw a die there are six possible
outcomes: 1, 2, 3, 4, 5 or 6.

Suppose one operation has m possible
outcomes and that a second operation has n
outcomes. The number of possible outcomes
when preforming the first operation followed
by the second operation is m × n.
Performing one operation and another
operation means we multiply the number of
possible outcomes.
m×n

Activity:

Example

draw two cards from a standard deck of 52 cards
without replacing the cards

There are 52 ways to draw the first card.

There are 51 ways to draw the second card.

There are 52 X 51 = 2,652 ways to draw the two
cards.

Whats the Difference?

"The combination to
the safe was 472"

"My fruit salad
is a combination
of apples,
grapes and
bananas"

We don't care what order the fruits are in, they could
also be "bananas, grapes and apples" or "grapes,
apples and bananas", its the same fruit salad.
Now we do care about the order. "724" would not
work, nor would "247". It has to be exactly 4-7-2.

In Maths we use precise
language..
If the order doesn’t matter, it is a
COMBINATION
If the order does matter, it is a
PERMUTATION
A PERMUTATION IS AN ORDERED COMBINATION.

Permutations
or
Arrangements
Arrangements
Arrangements

Question 1
There are 6 different books, including a
science book, on a shelf.
In how many different ways can the six
books be arranged on a shelf?
In how many ways can the 6 books be
arranged if the science book is always on
the extreme left?

Solution
6! = 6 X 5 X 4 X 3 X 2 X 1= 720
1 X 5! = 1 X 5 X 4 X 3 X 2 X 1 = 120

Question 2
In how many ways can the letters of the
word SCOTLAND be arranged in a line?
In how many of theses arrangements can
two vowels come together?
How many of these arrangements begin
with S and end with the two vowels?

Solutions
8! = 40320
7! X 2 = 10080 ways
5! x 2 = 240 ways

Question 3
How many different four digit numbers
greater than 6,000 can be formed using
the digits 1, 2, 3, 4, 5, 6, 7 if :
(i) no digit can be repeated
(ii) repetitions are allowed?

Solution
(i) 2 X 5 X 4 X 3 = 120
(ii) 2 X 6 X 6 X 6 = 432

n
r

The ( ) Notation
n

Can also be written as C aswell as nCr
r
and C(n,r).
It gives the number of ways of choosing r
objects from n different objects.
It is pronounced ‘n-c-r’ or ‘n-choose-r’.

How to Calculate It.
n) =
(r

n!
r! (n - r)!

Definition!

n) = n(n - 1)(n - 2)...(n - r +1)
(r
r!

Practical!

Question 1
In how many ways can a team of 5
players be chosen from 9 players?
In how many ways can this be done if a
certain player must be selected in each
team?

Solution
9 ) = ( 9 ) = 9 X 8 X 7 X 6 = 126
(5
4
4X3X2X1

8
( 4 ) = 8 X 7 X 6 X 5 = 70
4X 3X2X1

The Twin Rule
n
n
It states that ( r = ( n-r
)
)

Proof:
LHS =
RHS =

=
= n!
(n - r)!(n - (n - r))!

=

n!
(n - r)!r!

= LHS

Question 2
In how many ways can a committee of
six be formed from 5 teachers and 8
students if there are to be more teachers
than students on each committee?

Solution

Two possible combinations:

(i) 4 teachers and 2 students can be selected in
( 5 ) X ( 8) ways = 5 X 28 = 140 ways
4
2
(ii) 5 teachers and 1 student can be selected in
( 5 ) X ( 8 ) = 1 X 8 = 8 ways
3
1
The total number of possible committees
is 140 + 8 = 148 committees

Equations using (n-c-r)
When you have to solve equations the following are
very useful.
n
( )=1
1

n
(2 ) = n(n - 1) = n(n - 1)
2×1
2

Example
Solve for the value of the natural number
n
n such that ( ) = 28. 2

Solution

n(n - 1) = 28
2

n2 - n = 28
2
n2 - n = 28 -> n2 - n - 28 = 0
(n - 8)(n + 7) = 0
n=8

n=-7

Reject n = - 7 is not a natural number.
Therefore n = 8.

Find the value of n E N
in the following:
(n 2+ 1 ) = 28
SOL: (n + 1)(n) = 28
( ) = 28
2
n2 + n - 56 = 0
(n + 8)(n - 7) = 0
n = -8 and n=7
Reject n = -8 an accept n = 7.

“What are the
chances of that
happening?”
Theory of Probability

17th Century Gambling
Chevalier de Mere gambled frequently to increase his
wealth.
He bet on a roll of a die that at least one six would
appear in four rolls.
Tired of this approach he decided to change his bet to
make it more challenging.
He bet that he would get a total of 12, a double 6, on 24
rolls of two dice.

The old method was most profitable!

Correspondence leads to Theory
He asked his friend Blaise Pascal why this was the case.
Pascal worked it out and found that the probability of
winning was only 49.1% with the new method compared
to 51.8% using the old approach!
Pascal wrote a letter to Pierre De Fermat, who wrote
back.
They exchanged their mathematical principles and
problems and are credited with the founding of
probability theory.

Probability
is really about dealing with the unknown
in a systematic way, by scoping out the
most likely scenarios, or having a backup
plan in case those most likely scenarios
don’t happen.
Life is a sequence of unpredictable events,
but probability can be used to help predict
the likelihood of certain events occurring.

Terms
Definition

Example

An EXPERIMENT is a situation involving chance or
probability that leads to results called outcomes.

What color would we land on?

A TRIAL is the act of doing an
experiment in P!

Spinning the spinner.

The set or list of all possible outcomes in a trial is
called the SAMPLE SPACE.

Possible outcomes are Green, Blue, Red
and Yellow.

An OUTCOME is one of the possible
results of a trial.

Green.

An EVENT is the occurrence of one or
more specific outcomes

One event in this experiment is
landing on blue.

Binomial Distribution
or
Bernoulli Trials
Bernoulli Trials
Bernoulli Trials

Experiment that
satisfies:
A fixed number, n, of trials are carried out.
Each trial has two possible outcomes: success or
failure.
The trials are independent.
The probability of success in each trial is constant.
The probability of a success is generally called p.
The probability of a failure is q, where p + q = 1.

number of trials

probability of success
n
r

P ( r successes) = ( ) p q
probability of failure

r n-r

Question 1
An unbiased die is thrown 5 times. Find
the probability of obtaining
(i) 1 six
(ii) 3 sixes
(iii) at least 1 six.

Solution - Step 1
Let p = P(success i.e. getting a six) = 1/6.
Let q = P(failure i.e. NOT getting a six) = 5/6.
The number of trials: n = 5.

Solution - Step 2
n

P ( r successes) = ( r ) p q

r n-r

(i) P(1 six) = (
(ii) P( 3sixes)

5 1 4
1 )p q

5
1
4
)(1/6) (5/6)
1

=(

=3125/7776
5 3 2
= ( 3)p q
=125/3888

=(

5
)(1/6)3(5/6)2
3

(iii) P(at least 1 six) = 1 - P(no six)
P(no six) = (

5 0 5
)p q
0

=(

5)(1/6)0(5/6)5
0

P(at least 1 six) = 1 - (5/6)
= 4651/7776

5

You try: Question 2
Given that 10% of apples are bad, find the
probability that in a box containing 6
apples there is
(i) no bad apple
(ii) just one bad apple
(iii) at least one bad apple

Solution - Step 1
Let p = P(bad apple) = 1/10
Let q= P(good apple) = 9/10
Number of trials: n = 6

Solution - Step 2
6
0
6
(i) P(no bad apples) = ( 0 )(1/10) (9/10)

= 0.531441

(ii) P(1 bad apples) = (

6
1
5
)(1/10) (9/10)
1
=

0.015640313

(iii) P(at least 1 bad apples) = 1 - P(no bad apples)
1 - (0.531441).....from

part (i)

Back to Question 1
We worked out the probability of getting
3 sixes wen a dice is thrown 5 times. If the
same dice is thrown continuously until a
six appears for the 4th time, how do we
find the probability that the 4th six will
appear on the 10th throw?

For a six to appear on the 10th throw,
(i) we need to get 3 sixes on the first nine
throws and then
(ii) get the 4th on the 10th throw!

three sixes on the first nine throws is
given by:
9
3
6
( 3 ) (1/6) (5/6)

= 0.13

P(six on the 10th throw) = 1/6
Thus P(4th six on the 10th throw) = 0.13 X
1/6 = 0.0217

Question 3
A card is drawn at random from a
normal deck of playing cards and then
replaced. The process is repeated until the
third diamond is appears. Find the
probability that this happens when the
tenth card is drawn.

Solution - Step 1
P(drawing a diamond is) = 1/4
The probability of drawing a third
diamond on the 10th draw is:
P(drawing 2 diamonds in the first nine draws)
X
P(drawing a diamond on the 10th draw)

Solution - Step 2
P(2 diamonds on the 1st nine draws)
9
= ( )(1/4)2(3/4)7
2

= 0.3

P(diamond on the 10th draw) = 0.25
Thus P(third diamond on the 10th draw)
= 0.25 X 0.3 = 0.075

Question 4
During match Sarah takes a number of penalty shots. The shots
are independent of each other and her probability of scoring with
each shot is 4/5.
(i) Find the probability that Sarah misses each of her first four
shots.
(ii) Find the probability that Sarah scores exactly three of his
first four penalty shots.
(iii) It Sarah takes 10 penalty shots during the match, find the
probability that she scores at least 8 of them.

Solution
Let p=P(she scores the penalty) = 0.8
Let q=P(she misses the shot) = 0.2
4 )(0.8)0(0.2)4 = 1/625
(i) (
0
4
(ii) ( 3 )(0.8)3(0.2)1 = 256/625

(iii)

n = 10

P(at least 8 scores) =
P(scores 8)
+ P(scores 9)
+ P(scores 10)
10
10
10 9
( )(0.8)8(0.2)2 + ( )(0.8) (0.2)1 + ( )(0.8)10(0.2)0 =
10
8
9
ANS = 6,619,136
9,765,625

SOL:
Step 1

Let p=P(that she scores her free throw) = 0.6
Let q=P(she misses the free throw) = 0.4
The number of trials: n = 6

Step 2

Is she scores four of the shoots she misses two!
( 6 )p4q2 = ( 6 )(0.6)4(0.4)2
4
4

= 972/3125
0.31104

SOL:
Step 1

Step 2

P(scores once in the first 4 shots)
X
P(she scores on the 5th shot)
4
1
3
( )(0.6) (0.4)
1
0.1536

X

0.6 = 0.09216

Probability trinity college

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