Trinity College20th 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

ExampleActivity:  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?

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

"The combination to the safe was 472"

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.

If the order doesn’t matter, it is a COMBINATION

If the order does matter, it is a PERMUTATION

In Maths we use precise language..

A PERMUTATION IS AN ORDERED COMBINATION.

Permutationsor

ArrangementsArrangementsArrangements

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?

Question 1

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

Combinations

The ( ) Notation

Can also be written as C aswell as nCr 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’.

nr

nr

How to Calculate It.

( )nr = n!

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

r!

Definition! Practical!

Important

( ) = 1

( ) = 1

n0

nn

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 X 8 X 7 X 6 = 126

( ) = 8 X 7 X 6 X 5 = 70

4 X 3 X 2 X 1

9 95 4

4 X 3 X 2 X 1

84

It states that ( ) = ( )

Proof:

The Twin Rulen nr n-r

LHS = =

RHS = = n! = n! = LHS

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

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?

SolutionTwo possible combinations:

(i) 4 teachers and 2 students can be selected in( ) X ( ) ways = 5 X 28 = 140 ways 5

482

(ii) 5 teachers and 1 student can be selected in ( ) X ( ) = 1 X 8 = 8 ways5

381

The total number of possible committees is 140 + 8 = 148 committees

When you have to solve equations the following are very useful.

Equations using (n-c-r)

( ) = 1n1 ( ) = n(n - 1) = n(n - 1)n

22 × 1 2

ExampleSolve for the value of the natural number n such that ( ) = 28. n

2

Solutionn(n - 1) = 28 2

n2 - n = 282

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:

( ) = 28( ) = 28n + 1

2

SOL: (n + 1)(n) = 282

n2 + n - 56 = 0(n + 8)(n - 7) = 0n = -8 and n=7

Reject n = -8 an accept n = 7.

“What are the chances of that

happening?”

Theory of Probability

17th Century GamblingChevalier 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!

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.

Correspondence leads to Theory

Probabilityis 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.

TermsDefinition 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.

2013 Paper 2: Q 1

Binomial Distribution or

Bernoulli TrialsBernoulli TrialsBernoulli 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.

P ( r successes) = ( ) prqn-rnr

probability of failureprobability of failure

probability of successnumber of trials

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

(i) P(1 six) = ( )p1q4 = ( )(1/6)1(5/6)4

=3125/7776

(ii) P( 3sixes) = ( )p3q2 = ( )(1/6)3(5/6)2

=125/3888

P ( r successes) = ( ) prqn-rnr

51

51

53

53

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

P(no six) = ( )p0q5 = ( )(1/6)0(5/6)5

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

50

50

You try: Question 2Given 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

(i) P(no bad apples) = ( )(1/10)0(9/10)6

(ii) P(1 bad apples) = ( )(1/10)1(9/10)5

60

61

= 0.531441

= 0.015640313

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

1 - (0.531441).....from part (i)

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?

Back to Question 1

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:

( ) (1/6)3(5/6)693 = 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

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.

Question 3

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 2P(2 diamonds on the 1st nine draws)

P(diamond on the 10th draw) = 0.25

= ( )(1/4)2(3/4)7 = 0.392

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

Question 4During 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.

SolutionLet p=P(she scores the penalty) = 0.8

Let q=P(she misses the shot) = 0.2

(i) ( )(0.8)0(0.2)4 = 1/62540

(ii) ( )(0.8)3(0.2)1 = 256/62543

(iii) n = 10

P(at least 8 scores) =

P(scores 8) + P(scores 9) + P(scores 10)

( )(0.8)8(0.2)2 + ( )(0.8)9(0.2)1 + ( )(0.8)10(0.2)0 =108

109

1010

ANS = 6,619,136 9,765,625

Exam Questions

2012 Paper 2: Q 4

SOL: 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 1

Step 2 Is she scores four of the shoots she misses two!

( )p4q2 = ( )(0.6)4(0.4)2 64

64

= 972/3125 0.31104

SOL: P(scores once in the first 4 shots)XP(she scores on the 5th shot)

Step 1

Step 2 ( )(0.6)1(0.4)341

0.1536 X 0.6 = 0.09216