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					                                              Probability
                3
     Exam Paper

Assessment Standard: 11.4.2


a)    Correctly identify dependent and
      independent events (e.g .: from two way
      contingency tables or Venn - Diagrams)             Collect your Paper 3 Lessons every week!!
      and therefore appreciate when it is
                                                              Guys, both NSC and IEB examinations
      appropriate to calculate the probability
                                                       candidates have the option of writing Paper 3 at the
      of two independent events:
                                                            end of the year! Paper 3 covers additional
      P(A and B) = P(A). P(B)                         mathematics material and is out of 100 marks. Maths
                                                        Paper 3 will really set you apart in the job market,
b) Use Tree and Venn - Diagrams to                    and make studying technical subjects at tertiary level
   solve probability problems                           easier. We have hooked you up with these lessons
   (where events are not necessarily                                  - written by IEB Maths
   independent)                                         Paper 3 examiner Heather Frankiskos. Though the
                                                      lessons apply to both IEB and NSC candidates,where
Over the next two weeks we will give you a              there are differences, we will point them out! The
brief review of PROBABILITY from Grade 11.              lesson this week applies to candidates from both
Remember that about 50% - 60% of the                             examining bodies. Give it a go!
Examination is set on Descriptive Statistics
and Probability. A lot of the Probability was done
in Grade 11.

So recall the following:

•     PROBABILITY is the likelihood of something happening or being true.

•     A PROBABILITY is assigned a value between 0 (the impossible) and 1 (absolutely certain)

                                           even chance


                                       0         
                                                 __          
                                                 2

•     The PROBABILITIES of the possible outcomes in a sample space must sum up to 1.

Let’s see if you can do these basic probabilities without any fancy techniques:

1.    A regular die is rolled. Find:

                                                1
a)    P(1) - (the probability of getting a 1) = _
                                                6
b) P (7) - (the probability of getting a 7) = 0                                                                
                                                                                                        
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                                                            1 1 1
c)    P (3 or 4) - (the probability of getting a 3 or 4 ) =__ + _ = _
                                                           6 6 3
                    1 5
d) P (not a ) = 1- _ = _
                    6 6
              3 1
e) P (even) = _ = _
              6 
. A card is drawn from a normal pack of 5 cards (No jokers included) Find:
               4    1                  A          A
a) P (Ace) = __ = __
              5 13
                 13 1
b) P (Heart) = __ = _
                 5 4                               A           A
                                                                A
                        16 4
c) P (Ace or a Heart) = __ = __
                        5 13          A          A
                          1
                         __
d) P (Ace and a Heart) = 5
             1
e) P (Red) = _
             
                                                    A           A

•     Remember that if A is an event, then A’ is called the complement of A, and means ‘not A’.
      Also P(A) + P(A)’ = 1
               4
So P(Heart) = __
              5
                                     48
      P (Heart)’ = P (not a heart) = __
                                     5
           4 48
      and __ + __ = 1
          5 5
•     Two events which have no outcomes in common are called MUTUALLY EXCLUSIVE events.

      For example:         1) P(Heart) and P(Diamond) are mutually exclusive

                           ) P (roll a 1) and P (roll a 3) are mutually exclusive



                             NB: For mutually exclusive events

                             P(A or B) = P(A) + P (B)




1) P (Heart or Diamond)           = P(Heart) + P(Diamond)
                                    13 13
                                  = __ + __
                                    5 5
                                    6
                                  = __
                                    5
                                   1
                                  =_
                                   
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                                                Probability
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11) P (roll a 1 or roll a 3)          = P (roll a1) + P (roll a 3)
                                       1 1
                                      =_+_
                                       6 6
                                      = /6
                                       1
                                      =_
                                       3
But what would happen if we wanted

    P (King or a Heart)?
                   4
    P (King)    = __
                  5
                13
    P (Heart) = __
                5
Now can you see that the card ‘King of Hearts’ (which is a heart and a king has been counted into both
categories (so counted in twice)

    so P (King or Heart) = P (King) + P (Heart) - P (King of Hearts)
                                  4 13 1
                               = __ + __ - __
                                 5 5 5
                                 16
                               = __
                                 5
                                  4
                               = __
                                 13


                               So if events are NOT mutually exclusive
                           then P (A or B) = P (A) + P (B) - P (A and B)


We have to subtract off the overlap or intersection.

So, if we roll a single die what is the probability of getting a prime number or an even number?
            3 1
P (Prime) = _ = _ {  ; 3; 5}
            6 
           3 1
P (Even) = _ = _ {  ; 4; 6}
           6 
                    1 1 1 5
P (Prime or Even) = _ + __ − _ = _
                      6 6
Clearly the only number not allowed is the number 1.

We can also see this in a Venn-Diagram



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     Exam Paper

                                                                             50


                                     M a t hs            Science

                                            130     80       30



                                                                           10



A random pupil is selected from 50 grade 11s. Find:

a)    Probability that he/she takes Maths
                  10
      P (Maths) = ___ = 0,84
                  50
b) Probability that he /she takes Maths or Science

      P (Maths) + P (Science) - P (Maths and Science)
        10 110 80
      = ___ + ___ - ___
        50 50 50
        40
      = ___
        50        Must subtract off P (M⋂S) - the intersection

      = 0,96
Sometimes we are interested not in one outcome, but in two or three or more of them. For example, we
may toss a coin twice, or select two or 3 cards from a pack or take 3 beads from a bag containing
different colours

Drawing up a TREE DIAGRAM is usually useful to assist with these kinds of problems

Example 1
A coin is tossed and ‘H’ or ‘T’ is recorded. Find the probability of getting two heads and a tail (this means
in any order)




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                                                 Probability
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   Exam Paper
              Toss 1                 Toss 2             Toss 3
                                                        P(H)   1
                                                               _                                            1
                                                                                                            __
                                                                                         (H; H; H) →
                                                                                                            8

                                          P(H)      1
                                                    _
                                                                                                      1
                                                              1
                                                               _                      3   (H; H; T) → __
                                                        P(T)                                               8
                                    1
                                    _
                                    
                                                    1
                                                    _   P(H)   1
                                                               _                                        1
                                                                                                        __
              P(Head)                                                                 3   (H; T; H) →
                                          P(T)                                                           8


                                                               1
                                                               _                                        1
                                                                                                        __
                                                                                          (H; T; T) →
                                                        P(T)                                           8

                                                        P(H)   1
                                                               _                                           1
                                                                                                           __
                                                                                     3   (T; H; H) →
                                                                                                            8
               P(Tail)
                                    1
                                    _     P(H)      1
                                                    _
                                                              1
                                                               _                                        1
                                                                                                        __
                                                                                         (T; H; T) →
                                                        P(T)                                           8

                                                    1
                                                    _   P(H)   1
                                                               _                                        __
                                                                                                        1
                                                                                         (T; T; H) →
                                          P(T)                                                         8

                                                               1
                                                               _                                       1
                                                                                                       __
                                                                                          (T;T; T) →
                                                        P(T)                                          8
                                3
\ P (two heads and a tail) = __ + __ + __ = __
                             8 8 8 8
 Example 2
 Draw two cards from a normal pack with replacement
 1) Find probability of two Kings
 ) Find probability of one King

                         Draw 1           Draw 2
                                                          4
                                                         __                 
                                         P (King)        52        (K;K) = ___
                                                                           69
                                     4
                                    __
                                    52
                         P (King)
                                                         48
                                                         __                      2
                                                                   (K; not K) = ___
                                        P (Not a King) 52                       69
                                         P (King)         4
                                                         __                     2
                P (not a King)                           52        (not K;K) = ___
                                                                               69
                                    48
                                    __
                                    52
                                                         48
                                                         __                         44
                                     P (Not a King)      52        (not K; not K) = ___
                                                                                    69
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     Exam Paper
              1
1) P (K;K) = ___
             169
                   1    1    4
) P (one King) = ___ + ___ = ___ = 0,14
                  169 169 169
Can you see in the previous two examples when we came to toss the coin the second time or draw the
second card - there was no effect of the first result on the second? When this happens the two events
are said to be INDEPENDENT


                                For two INDEPENDENT events
                                P (A and B) = P (A). P (B)


So these are INDEPENDENT EVENTS:

1) Toss a coin twice
) Toss a die four (or however many) times
3) Toss a coin, roll a die
4) Pick cards from a pack WITH replacement
5) Choose sweets from a bag WITH replacement
Question 1

      P (A) = 0,
      P (B) = 0,5
      P (A∪B) (which means P(A or B)) = 0,6

a)    Are events A and B mutually exclusive? Motivate

b) Are events A and B independent? Motivate

a)    P (A∪B) ≠ P (A) + P (B); so not mutually exclusive

b) P (A∪B) = P(A) + P(B) - P (A and B)

                    0,6 = 0, + 0,5 - P (A and B)

                    P (A and B) = 0,7 - 0,6

                    P (A and B) = 0,1

                    but P (A). P (B) = 0, x 0,5 =0,1 \ are independent




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                                            Probability
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     Exam Paper
Example 2
A weather forecaster classifies all days as either wet or dry. He claims that the probability that 1 September
will be wet is 0,4. If any particular day in September is wet, the probability that the next day is wet is 0,5;
otherwise the probability that the next day is wet is 0,3. Find the probability that:


              Sep                        2 Sep

                                P (wet)    0,5

                              0,4
                                                                           Key
             P (wet)
                                           0,5
                                P (dry)                             ( - 0,4)
                                                                            = P (dry)
                                                                            = 0,6
                                           0,7
             P (dry)                                                ( - 0,3)
                                                                            = P (second day dry)
                                0,6                                         = 0,7


                                P (wet)    0,3


a) The first two days in September are wet
b) September nd is wet

a)    0,4 x 0,5 = 0, [(W;W)]

b) 0,4 x 0,5 + 0,6 x 0,3 [(W;W) or (D; W)]

                = 0,38

and P (W,W) + P (W,D) + P (D, W) + P (D,D)

= 0, + 0, + 0,4 + 0,18

=1

Example 3

P (A) = 0,5 and P (B) = 0,5 and P (A and B) = 0,15

a)    Are events A and B independent?

b) Are events A and B mutually exclusive?
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                                             Probability
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     Exam Paper
a)    P(A) x P (B) = 0,5 x 0,5 = 0,15

      P (A and B) = 0,15

      so 0,15 ≠ 0,15 NOT independent

b) no since P (A or B) = P(A) + P (B) − P (A and B)

                           ≠ P (A) + P (B)
Example 4

Another way that we use to decide on Independence is by using a CONTINGENCY TABLE

4a) Let us look at this table which gives information about Males and Females and whether they prefer
    Red, Blue or Yellow

                                  Red           Blue          Yellow            Total
           Male                    0            40            50               110
           Female                  50            0            0                90
           Total                   70            60            70               00

This is called a two-way x3 Contingency Table . It has  rows and 3 columns.
Can you see that we could get the following from the table?
                 110
1) P (Male) = ___ = 0,55
              00
               60
) P (Blue) = ___ = 0,3
              00
                        40
3) P (Blue and Male) = ___ = 0,
                       00
                       60 110 40 130
4) P (Blue or Male) = ___ + ___ - ___ = ___ = 0,65
                      00 00 00 00
They are NOT mutually exclusive

5) Are the events ‘being Male’ and ‘preferring Blue’ independent or not?

P (Blue and Male) = 0,
                       60 110 33
P (Blue) x P (Male) = ___ x ___ = ___ = 0,165
                      00 00 00
They are NOT independent

6) Are the events ‘being Female’ and ‘preferring Red’ independent or not?
                      50
P (Female and Red) = ___ = 0,5
                     00
                        90    70    63
P (Female) x P (Red) = ___ x ___ = ___ = 0,1575
                       00 00 400
They are NOT independent
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                                          Probability
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  Exam Paper
4b) Research with regard to the effects of a new headache tablet involving 100 males and 80 females
    showed that 60 males and 50 females responded positively to the tablet. Can we conclude that the
    success of the tablet is independent of gender?

                                           Male           Female          Total
                     Positive                60            50              110
                     Not Positive            40            30               70
                     Total                  100            80              180
           100
P (Male) = ___
           180
              80
P (Female) = ___
             180
               110
P (Positive) = ___
               180
                    70
P (not Positive) = ___
                   180
                         60
P (Male and Positive) = ___ = 0,3
                        180
                          100 110
P (Male) x P (Positive) = ___ x ___ = 0,34
                          180 180
Mathematically these are NOT independent

(OR)
                           50
P (Female and Positive) = ___ = 0,7
                          180
                             80 110
P (Female) x P (Positive) = ___ x ___ = 0,716
                            180 180
There is a slight difference so they are still NOT independent

So now we have seen the multiplication law for INDEPENDENT events.

But what happens if we were to draw two cards from a pack and NOT put the first one back before we
took the second one? (This is called WITHOUT replacement)

It is very important to realise that because we do not put the first card back, we are now going to draw the
second card from a smaller sample and the next outcome is influenced by (depends on) what happened in
the first.
                                           4
So if we draw the first card P (King) = __5
Having removed one King and not replaced it means that there are now only 3 Kings in the remaining 51
                                                                    3
cards so the probability of the second card also being a King is __51
So the Probability of two Kings in two successive draws of a pack of cards WITHOUT replacement is
                     4    3
P (King and King) = __ x __
                    5 51
                      1
                  = ___
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                                           Probability
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  Exam Paper
So what if we drew two cards and wanted the probability of one King.? The tree diagram would look as
follows:

                           st Draw                             2nd Draw


                                                 P (King)          3
                                                                  __
                                                                  5
                                                4
                                               __
                                               52
                            P (King)
                                                                  48
                                                                  __
                                                 P (not King)     5


                                                 P (King)          4
                                                                  __
                            P (not King)                          5
                                               48
                                               __
                                               52

                                                                  47
                                                                  __
                                                 P (not King)     5

So the Probability of one King means

      = P (King followed by not King) or P (not King followed by King)

      = 4/5 x 48/51 + 48/5 x 4/51
         3
      = ___
        1

                             4    3     1
Please note:    P (K,K) = __ x __ = ___
                            5 51 1
                                    4 48 16
                P (K, not K)     = __ x __ = ___
                                   5 51 1
                                   48 4       16
                P (not K; K)     = __ x __ = ___
                                   5 51 1
                                   48 47 188
                P (not K; not K) = __ x __ = ___
                                   5 51 1
                  1    16      16 188 1
                ___ ___ ___ ___ ____
now             1 + 1 + 1 + 1 = 1 = 1


A method that we can use to help us organise the relationships between events so that we can make
conclusions about probabilities, mutually exclusive and / or independence is called a VENN-DIAGRAM.
The best way to get to understand how useful they are is with an example or two:
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                                                Probability
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     Exam Paper
Example 1
Given P(A) = 0,7; P (B) = 0,4 and P (A∩B) = 0,8

Find out:
a)    P (A∪B)’

b) whether A and B are independent
      Best way to solve this is draw a picture:



                                            A                      B


                                                  x   0,8 y




                                      now             x + 0,8 = 0,7

                                                      x = 0,4

                                      and             y + 0,8 = 0,4

                                                      y = 0,1



a)    Here P (A∪B) = P (A or B) = 0,4 + 0,8 + 0,1 = 0,8

      so P (A∪B)’ = 1-0,8 = 0,18
b) P (A) x P (B) = 0,7 x 0,4 = 0,8

      P (A∩B) = 0,8

      so they are independent events
Sometimes we get given the picture organiser and we are asked to interpret it.




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                                              Probability
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     Exam Paper

                                                                                   S



                                    R                                    C



                                                  1
                                        108                     54

                                                  36

                                              6            18
                                                                             134

                                                  97
                                                                     J




Example 2
A survey was conducted to determine the preferences for three different music types
Rock (R), Classic (C), Jazz (J)

1) How many people were surveyed?
) How many people preferred Rock (R) only?
3) What percentage of people surveyed like all three types?
4) What percentage liked Rock (R) or Classic (C)?
5) 134 do not like music at all. Is this statement true or false? Motivate your answer.
6) Find P (R’∩C’)



1)    465

) 108
    36
3) ___ = 0,0774 = 7,74%
   465
   108 + 1+ 36 + 6 + 54 + 18 34
4) ____________________ =___ = 0,503 = 50,3%
              46              465
5) False. They may well like other kinds of music - they just don’t like these three.
   97+134 77
6) ______ = ___ = 0,4968
     465    155

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