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Introduction

We can divide mathematics into pure and applied maths. Pure maths is the theory of maths and is very abstract. The work you have covered on algebra is mostly pure maths. Applied maths is all about taking the theory (or pure maths) and applying it to the real world. To be able to do applied maths, you first have to learn pure maths.

But what does this have to do with probability? Well, just as mathematics can be divided into pure maths and applied maths, so statistics can be divided into probability theory and applied statistics. And just as you cannot do applied mathematics without knowing any theory, you cannot do statistics without beginning with some understanding of probability theory. Furthermore, just as it is not possible to describe what arithmetic is without describing what mathematics as a whole is, it is not possible to describe what probability theory is without some understanding of what statistics as a whole is about, and statistics, in its broadest sense, is about 'processes'.

Note: Interesting fact:

Galileo wrote down some ideas about dice games in the seventeenth century. Since then many discussions and papers have been written about probability theory, but it still remains a poorly understood part of mathematics.

A process is how an object changes over time. For example, consider a coin. Now, the coin by itself is not a process; it is simply an object. However, if I was to flip the coin (i.e. putting it through a process), after a certain amount of time (however long it would take to land), it is brought to a final state. We usually refer to this final state as 'heads' or 'tails' based on which side of the coin landed face up, and it is the 'heads' or 'tails' that the statistician (person who studies statistics) is interested in. Without the process there is nothing to examine. Of course, leaving the coin stationary is also a process, but we already know that its final state is going to be the same as its original state, so it is not a particularly interesting process. Usually when we speak of a process, we mean one where the outcome is not yet known, otherwise there is no real point in analyzing it. With this understanding, it is very easy to understand what, precisely, probability theory is.

When we speak of probability theory as a whole, we mean the way in which we quantify the possible outcomes of processes. Then, just as 'applied' mathematics takes the methods of 'pure' mathematics and applies them to real-world situations, applied statistics takes the means and methods of probability theory (i.e. the means and methods used to quantify possible outcomes of events) and applies them to real-world events in some way or another. For example, we might use probability theory to quantify the possible outcomes of the coin-flip above as having a 50% chance of coming up heads and a 50% chance of coming up tails, and then use statistics to apply it a real-world situation by saying that of six coins sitting on a table, the most likely scenario is that three coins will come up heads and three coins will come up tails. This, of course, may not happen, but if we were only able to bet on ONE outcome, we would probably bet on that because it is the most probable. But here, we are already getting ahead of ourselves. So let's back up a little.

To quantify results, we may use a variety of methods, terms, and notations, but a few common ones are:

  • a percentage (for example '50%')
  • a proportion of the total number of outcomes (for example, '5/10')
  • a proportion of 1 (for example, '1/2')

You may notice that all three of the above examples represent the same probability, and in fact ANY method of probability is fundamentally based on the following procedure:

  1. Define a process.
  2. Define the total measure for all outcomes of the process.
  3. Describe the likelihood of each possible outcome of the process with respect to the total measure.

The term 'measure' may be confusing, but one may think of it as a ruler. If we take a ruler that is 1 metre long, then half of that ruler is 50 centimetres, a quarter of that ruler is 25 centimetres, etc. However, the thing to remember is that without the ruler, it makes no sense to talk about proportions of a ruler! Indeed, the three examples above (50%, 5/10, and 1/2) represented the same probability, the only difference was how the total measure (ruler) was defined. If we go back to thinking about it in terms of a ruler '50%' means '50/100', so it means we are using 50 parts of the original 100 parts (centimetres) to quantify the outcome in question. '5/10' means 5 parts out of the original 10 parts (10 centimetre pieces) depict the outcome in question. And in the last example, '1/2' means we are dividing the ruler into two pieces and saying that one of those two pieces represents the outcome in question. But these are all simply different ways to talk about the same 50 centimetres of the original 100 centimetres! In terms of probability theory, we are only interested in proportions of a whole.

Although there are many ways to define a 'measure', the most common and easiest one to generalize is to use '1' as the total measure. So if we consider the coin-flip, we would say that (assuming the coin was fair) the likelihood of heads is 1/2 (i.e. half of one) and the likelihood of tails is 1/2. On the other hand, if we consider the event of not flipping the coin, then (assuming the coin was originally heads-side-up) the likelihood of heads is now 1, while the likelihood of tails is 0. But we could have also used '14' as the original measure and said that the likelihood of heads or tails on the coin-flip was each '7 out of 14', while on the non-coin-flip the likelihood of heads was '14 out of 14', and the likelihood of tails was '0 out of 14'. Similarly, if we consider the throwing of a (fair) six-sided die, it may be easiest to set the total measure to '6' and say that the likelihood of throwing a '4' is '1 out of the 6', but usually we simply say that it is 1/6, i.e. '1/6 of 1'.

Definitions

There are three important concepts associated with a random experiment: 'outcome', 'sample space', and 'event'. Two examples of experiments will be used to familiarize you with these terms:

  • In Experiment 1 a single die is thrown and the value of the top face after it has come to rest is noted.
  • In Experiment 2 two dice are thrown at the same time and the total of the values of each of the top faces after they have come to rest is noted.

Outcome

An outcome of an experiment is a single result of that experiment.

  • A possible outcome of Experiment 1: the value on the top face is '3'
  • A possible outcome of Experiment 2: the total value on the top faces is '9'

Sample Space

The sample space of an experiment is the complete set of possible outcomes of the experiment.

  • Experiment 1: the sample space is 1,2,3,4,5,6
  • Experiment 2: the sample space is 2,3,4,5,6,7,8,9,10,11,12

Note: Interesting fact:

If you roll two die and add the results, the most common outcome is 7. To understand this, think that there is only one way to get 2 as a result (both dice landing with 1) and there is only way to get 12 as a result (both dice landing with 6). The opposite sides of a six sided dice add up to 7. From this you should be able to work out that there are 12 ways to get a 7.

Event

An event is any set of outcomes of an experiment.

  • A possible event of Experiment 1: an even number being on the top face of the die
  • A possible event of Experiment 2: the numbers on the top face of each die being equal

Random Experiments

The term random experiment or statistical experiment is used to describe any repeatable process, the results of which are analyzed in some way. For example, flipping a coin and noting whether or not it lands heads-up is a random experiment because the process is repeatable. On the other hand, your reading this sentence for the first time and noting whether you understand it is not a random experiment because it is not repeatable (though making a number of random people read it and noting which ones understand it would turn it into a random experiment).

Venn diagrams

A Venn diagram can be used to show the relationship between the possible outcomes of a random experiment and the sample space. The Venn diagram in Figure 1 shows the difference between the universal set, a sample space and events and outcomes as subsets of the sample space.

Figure 1: Diagram to show difference between the universal set and the sample space. The sample space is made up of all possible outcomes of a statistical experiment and an event is a subset of the sample space.
Figure 1 (MG10C17_001.png)

We can draw Venn diagrams for experiments with two and three events. These are shown in Figure 2 and Figure 3. Venn diagrams for experiments with more than three events are more complex and are not covered at this level.

Figure 2: Venn diagram for an experiment with two events
Figure 2 (venn1.png)
Figure 3: Venn diagram for an experiment with three events.
Figure 3 (venn2.png)

Note: Interesting fact:

The Greek, Russian and Latin alphabets can be illustrated using Venn diagrams. All these alphabets have some common letters. The Venn diagram is given below:
Figure 4
Figure 4 (venn_ifact.png)

The union of AA and BB is the set of all elements in AA or in BB (or in both). AorBAorB is also written ABAB. The intersection of AA and BB is the set of all elements in both AA and BB. AandBAandB is also written as ABAB.

Venn diagrams can also be used to indicate the union and intersection between events in a sample space (Figure 5 and Figure 6).

Figure 5: Venn diagram to show (left) union of two events, AA and BB, in the sample space SS.
Figure 5 (union.png)
Figure 6: Venn diagram to show intersection of two events AA and BB, in the sample space SS. The black region indicates the intersection.
Figure 6 (intersect.png)

We use n(S)n(S) to refer to the number of elements in a set SS, n(X)n(X) for the number of elements in XX, etc.

Exercise 1: Random Experiments

In a box there are pieces of paper with the numbers from 1 to 9 written on them. A piece of paper is drawn from the box and the number on it is noted. Let SS denote the sample space, let PP denote the event 'drawing a prime number', and let EE denote the event 'drawing an even number'. Using appropriate notation, in how many ways is it possible to draw: i) any number? ii) a prime number? iii) an even number? iv) a number that is either prime or even? v) a number that is both prime and even?

Solution
  1. Step 1. Consider the events: :
    • Drawing a prime number: P={2;3;5;7 }P={2;3;5;7 }
    • Drawing an even number: E={2;4;6;8 }E={2;4;6;8 }
  2. Step 2. Draw a diagram :

    Figure 7
    Figure 7 (MG10C17_003.png)

  3. Step 3. Find the union :

    The union of PP and EE is the set of all elements in PP or in EE (or in both). PE=2,3,4,5,6,7,8 PE=2,3,4,5,6,7,8 .

  4. Step 4. Find the intersection :

    The intersection of PP and EE is the set of all elements in both PP and EE. PE=2PE=2.

  5. Step 5. Find the number in each set :
    n ( S ) = 9 n ( P ) = 4 n ( E ) = 4 n ( P E ) = 7 n ( P E ) = 2 n ( S ) = 9 n ( P ) = 4 n ( E ) = 4 n ( P E ) = 7 n ( P E ) = 2
    (1)

Exercise 2

100 people were surveyed to find out which fast food chain (Nandos, Debonairs or Steers) they preferred. The following results were obtained:

  • 50 liked Nandos
  • 66 liked Debonairs
  • 40 liked Steers
  • 27 liked Nandos and Debonairs but not Steers
  • 13 liked Debonairs and Steers but not Nandos
  • 4 liked all three
  • 94 liked at least one
  1. How many people did not like any of the fast food chains?
  2. How many people liked Nandos and Steers, but not Debonairs?

Solution
  1. Step 1. Draw a Venn diagram: The number of people who liked Nandos and Debonairs is 27, so this is the intersection of these two events. The number of people who liked Debonairs and Steers is 13, so the intersection of Debonairs and Steers is 13. We are told that 4 people like all three options, and so this means that there are 4 people in the intersection of all three options. So we can work out that the number of people who like just Debonairs is 66427-13=2266427-13=22 (This is simply the total number who like Debonairs minus the number of people who like Debonairs and Steers, or Debonairs and Nandos or all three). We draw the following diagram to represent the data:
    Figure 8
    Figure 8 (Vennwex1.png)
  2. Step 2. Work out how many people like none: We are told that there were 100 people and that 94 liked at least one. So the number of people that liked none is: 10094=610094=6. This is the answer to a).
  3. Step 3. Work out how many like Nandos and Steers but not Debonairs: We can redraw the part of the Venn diagram that is of interest:
    Figure 9
    Figure 9 (Vennwex2.png)
    Total people who like Nandos: 50
    Of these 27 like both Nandos and Debonairs, and 4 people like all three options. So we find that the total number of people who just like Nandos is: 50274=1950274=19
    Total people who like Steers: 40
    Of these 13 like both Steers and Debonairs, and 4 like all three options. So we find that the total number of people who like just Steers is: 40134=2340134=23
    Now use the identity n(N or S)=n(N)+n(S)n(N and S)n(N or S)=n(N)+n(S)n(N and S) to find the number of people who like Nandos and Steers, but not Debonairs.
    n(N or S) = n(N)+n(S)n(N and S) 28 = 23+19n(N and S) n(N and S) = 14 n(N or S) = n(N)+n(S)n(N and S) 28 = 23+19n(N and S) n(N and S) = 14
    (2)
    The Venn diagram with that represents all this information is given:
    Figure 10
    Figure 10 (Vennwex3.png)

Activity: Venn diagrams

Which cellphone networks have you used or are you signed up for (e.g. Vodacom, Mtn or CellC)? Collect this information from your classmates as well. Then use the information to draw a Venn diagram (if you have more than three networks, then choose only the three most popular or draw Venn diagrams for all the combinations). Try to see if you can work out the number of people who use just one network, or the number of people who use all the networks.

Complement and complementary events

A final notion that is important to understand is the notion of complement. Just as in geometry when two angles are called 'complementary' if they added up to 90 degrees, (the two angles 'complement' each other to make a right angle), the complement of a set of outcomes AA is the set of all outcomes in the sample space but not in AA. It is usually denoted A' A' (or sometimes AC AC), and called 'the complement of AA' or simply 'AA-complement'. Because it refers to the set of everything outside of AA, it is also often referred to as 'not-AA'. Thus, by definition, if SS denotes the entire sample space of possible outcomes, and AA is any subset of outcomes that we are interested in (i.e. an event), then AA' = SAA' = S is always true, (i.e. A' A' complements AA to form the entire sample space). So in the exercise above, P'={1,4,6,8,9}P'={1,4,6,8,9}, while E'={1,3,5,7,9}E'={1,3,5,7,9}. So n(P')=n(E')=5n(P')=n(E')=5

The probability of a complementary event refers to the probability associated with the complement of an event, i.e. the probability that something other than the event in question will occur. For example, if P(A)=0,25P(A)=0,25, then the probability of AA not occurring is the probability associated with all other events in SS occurring less the probability of AA occurring.

In theory, it is very easy to calculate complements, since the number of elements in the complement of a set is just the total number of outcomes in the sample space minus the outcomes in that set (in the example above, there were 9 possible outcomes in the sample space, and 4 possible outcomes in each of the sets we were interested in, thus both complements contained 9-4 = 5 elements). Similarly, it is easy to calculate probabilities of complements of events since they are simply the total probability (e.g. 1 if our total measure is 1) minus the probability of the event in question. So,

P ( A ' ) = 1 - P ( A ) P ( A ' ) = 1 - P ( A )
(3)

Sometimes it is much easier to decide the probability of an event occurring by instead calculating the probability that the complementary event will NOT occur. For example, if the process in question was rolling three dice, and the event we were interested in was that at least one of the faces is a one, it is definitely much easier to figure out the probability that not getting a one will not occur than to try to figure out all the possible combinations of three dice where a one does occur!

Exercise 3: Complementary events

If you throw two dice, one red and one blue, what is the probability that at least one of them will be a six?

Solution
  1. Step 1. Work out probability of event 1 :

    To solve that kind of question, work out the probability that there will be no six.

  2. Step 2. Work out probability of event 2 :

    The probability that the red dice will not be a six is 5/6, and that the blue one will not be a six is also 5/6.

  3. Step 3. Probability of neither :

    So the probability that neither will be a six is 5/6×5/6=25/365/6×5/6=25/36.

  4. Step 4. Probability of one :

    So the probability that at least one will be a six is 1-25/36=11/361-25/36=11/36.

Exercise 4: Complementary events

A bag contains three red balls, five white balls, two green balls and four blue balls:

1. Calculate the probability that a red ball will be drawn from the bag.

2. Calculate the probability that a ball which is not red will be drawn

Solution
  1. Step 1. Find event 1 :

    Let R be the event that a red ball is drawn:

    • P(R)-n(R)/n(S)=3/14
    • R and R' are complementary events
  2. Step 2. Find the probabilitys :

    P(R') = 1 - P(R) = 1 -3/14 = 11/14

  3. Step 3. Alternate way to solve it :
    • Alternately P(R') = P(B) + P(W) + P(G)
    • P(R') = 4/14 + 5/14 + 2/14 = 11/14

Probability in Everyday Life

Probability is connected with uncertainty. In any statistical experiment, the outcomes that occur may be known, but exactly which one might not be known. Mathematically, probability theory formulates incomplete knowledge related to the likelihood of an occurrence. For example, a meteorologist might say there is a 60% chance that it will rain tomorrow. This means that in 6 of every 10 times when the world is in the current state, it will rain tomorrow.

Another way of referring to probabilities is odds. The odds of an event is defined as the ratio of the probability that the event occurs to the probability that it does not occur. For example, the odds of a coin landing on a given side are 0.50.5=10.50.5=1, usually written "1 to 1" or "1:1". This means that on average, the coin will land on that side as many times as it will land on the other side.

The Simplest Example: Equally Likely Outcomes

We say two outcomes are equally likely if they have an equal chance of happening. For example when a fair coin is tossed, each outcome in the sample space S={heads,tails}S={heads,tails} is equally likely to occur.

Probability is a function of events (since it is not possible to have a single event with two different probabilities occurring), so we usually denote the probability PP of some event EE occurring by P(E)P(E). When all the outcomes are equally likely (in any activity), it is fairly straightforward to count the probability of a certain event occuring. In this case,

P(E) = n(E)/n(S)P(E) = n(E)/n(S)

For example, when you throw a fair die the sample space is S={1;2;3;4;5;6}S={1;2;3;4;5;6} so the total number of possible outcomes n(S)=6n(S)=6.

Event 1: Get a 4

The only possible outcome is a 44, i.e E={4}E={4}. So n(E)=1n(E)=1.

Probability of getting a 4: P(k=4) = n(E)/n(S) = 1/6P(k=4) = n(E)/n(S) = 1/6.

Event 2: Get a number greater than 3

Favourable outcomes: E={4;5;6}E={4;5;6}

Number of favourable outcomes: n(E)=3n(E)=3.

Probability of getting a number greater than 3: P(k>3) = n(E)/n(S) = 3/6 = 1/2P(k>3) = n(E)/n(S) = 3/6 = 1/2.

Exercise 5

A standard deck of cards (without jokers) has 52 cards. There are four sets of cards, called suits. The suit a card belongs to is denoted by a symbol on the card, the four possible symbols being hearts, clubs, spades, and diamonds. In each suit there are 13 cards (4 suits ×13 cards =524 suits ×13 cards =52) consisting of one each of ace, king, queen, jack, and the numbers 2-10.

If we randomly draw a card from the deck, we can the card drawn as a possible outcome. Therefore, there are 52 possible outcomes. We can now look at various events and calculate their probabilities:

  1. Out of the 52 cards, there are 13 clubs. Therefore, if the event of interest is drawing a club, there are 13 favourable outcomes, what is the probability of this event?
  2. There are 4 kings (one of each suit). The probability of drawing a king is?
  3. What is the probability of getting a king or a club?
Solution
  1. Step 1. First question :

    The probability of this event is 1352=141352=14.

  2. Step 2. Second question :

    452=113452=113.

  3. Step 3. Third question :

    This example is slightly more complicated. We cannot simply add together the number of outcomes for each event separately (4 + 13 = 17) as this inadvertently counts one of the outcomes twice (the king of clubs). Why is this so? Well, mathematically if AA and BB are two events, it is actually true that n(AB)=n(A)+n(B)-n(AB)n(AB)=n(A)+n(B)-n(AB) (a more detailed explanation of this identity is provided below, but for now just assume that it is true). In the results involving the throwing of dice, the intersection of any two outcomes was empty (and hence n(AB)=0n(AB)=0) since it is not possible for the top face of a die to have two different values simultaneously. However, in this case, a card can be both a club and a king at the same time (i.e. n(AB)=1n(AB)=1). Therefore, n(AB)=4+13-1=16n(AB)=4+13-1=16. So the correct answer is 16521652.

Probability Models
  1. A bag contains 6 red, 3 blue, 2 green and 1 white balls. A ball is picked at random. What is the probablity that it is:
    1. red
    2. blue or white
    3. not green (hint: think 'complement')
    4. not green or red?
    Click here for the solution.
  2. A card is selected randomly from a pack of 52. What is the probability that it is:
    1. the 2 of hearts
    2. a red card
    3. a picture card
    4. an ace
    5. a number less than 4?
    Click here for the solution.
  3. Even numbers from 2 -100 are written on cards. What is the probability of selecting a multiple of 5, if a card is drawn at random?
    Click here for the solution.

Probability Identities

The following results apply to probabilities, for the sample space SS and two events AA and BB, within SS.

P ( S ) = 1 P ( S ) = 1
(4)
P ( A B ) = P ( A ) × P ( B ) P ( A B ) = P ( A ) × P ( B )
(5)
P ( A B ) = P ( A ) + P ( B ) - P ( A B ) P ( A B ) = P ( A ) + P ( B ) - P ( A B )
(6)

We can demonstrate this last result using a Venn diagram. The union of A and B is the set of all elements in A or in B or in both.

Figure 11
Figure 11 (venn1.png)

The probability of event A occurring is given by P(A)P(A) and the probability of event B occurring is given by P(B)P(B). However, if we look closely at the circle representing either of these events, we notice that the probability includes a small part of the other event. So event A includes a bit of event B and vice versa. This is shown in the following figure:

Figure 12
Figure 12 (identity1.png)
And then we observe that this small bit is simply the intersection of the two events.

So to find the probability of P ( A B )P(AB) we notice the following:

  • We can add P ( A )P(A) and P ( B )P(B)
  • Doing this counts the intersection twice, once in P ( A )P(A) and once in P ( B )P(B).
So if we simply subtract the probability of the intersection, then we will find the total probability of the union: P ( A B ) = P ( A ) + P ( B ) - P ( A B ) P ( A B ) = P ( A ) + P ( B ) - P ( A B )

Exercise 6: Probabilty identities

What is the probability of selecting a black or red card from a pack of 52 cards

Solution
  1. Step 1. Write the answer :

    P(S)=n(E)n(S)=5252=1P(S)=n(E)n(S)=5252=1 because all cards are black or red!

Exercise 7: Probabilty identities

What is the probability of drawing a club or an ace with one single pick from a pack of 52 cards

Solution
  1. Step 1. Identify the identity which describes the situation :
    P ( club ace ) = P ( club ) + P ( ace ) - P ( club ace ) P ( club ace ) = P ( club ) + P ( ace ) - P ( club ace )
    (7)
  2. Step 2. Calculate the answer :
    = 1 4 + 1 13 - 1 4 × 1 13 = 1 4 + 1 13 - 1 52 = 16 52 = 4 13 = 1 4 + 1 13 - 1 4 × 1 13 = 1 4 + 1 13 - 1 52 = 16 52 = 4 13
    (8)

    Notice how we have used P(CA)=P(C )+P(A)-P(CA)P(CA)=P(C )+P(A)-P(CA).

The following video provides a brief summary of some of the work covered so far.

Figure 13
Khan academy video on probability

Probability Identities

Answer the following questions

  1. Rory is target shooting. His probability of hitting the target is 0,70,7. He fires five shots. What is the probability that all five shots miss the center?
    Click here for the solution.
  2. An archer is shooting arrows at a bullseye. The probability that an arrow hits the bullseye is 0,40,4. If she fires three arrows, what is the probability that all the arrows hit the bullseye?
    Click here for the solution.
  3. A dice with the numbers 1,3,5,7,9,11 on it is rolled. Also a fair coin is tossed. What is the probability that:
    1. A tail is tossed and a 9 rolled?
    2. A head is tossed and a 3 rolled?
    Click here for the solution.
  4. Four children take a test. The probability of each one passing is as follows. Sarah: 0,80,8, Kosma: 0,50,5, Heather: 0,60,6, Wendy: 0,90,9. What is the probability that:
    1. all four pass?
    2. all four fail?
    Click here for the solution.
  5. With a single pick from a pack of 52 cards what is the probability that the card will be an ace or a black card?
    Click here for the solution.

Mutually Exclusive Events

Two events are called mutually exclusive if they cannot be true at the same time.

Examples of mutually exclusive events are:

  1. A die landing on an even number or landing on an odd number.
  2. A student passing or failing an exam
  3. A tossed coin landing on heads or landing on tails

This means that if we examine the elements of the sets that make up AA and BB there will be no elements in common. Therefore, AB=AB= (where refers to the empty set). Since, P(AB)=0P(AB)=0, equation Equation 6 becomes:

P ( A B ) = P ( A ) + P ( B ) P ( A B ) = P ( A ) + P ( B )
(9)

for mutually exclusive events.

We can represent mutually exclusive events on a Venn diagram. In this case, the two circles do not touch each other, but are instead completely separate parts of the sample space.

Figure 14: Venn diagram for mutually exclusive events
Figure 14 (mutualexclusive.png)

Mutually Exclusive Events

  1. A box contains coloured blocks. The number of each colour is given in the following table.
    Table 1
    ColourPurpleOrangeWhitePink
    Number of blocks24324119
    A block is selected randomly. What is the probability that the block will be:
    1. purple
    2. purple or white
    3. pink and orange
    4. not orange?
    Click here for the solution.
  2. A small private school has a class with children of various ages. The table gies the number of pupils of each age in the class.
    Table 2
    3 years female3 years male4 years female4 years male5 years female5 years male
    625746
    If a pupil is selceted at random what is the probability that the pupil will be:
    1. a female
    2. a 4 year old male
    3. aged 3 or 4
    4. aged 3 and 4
    5. not 5
    6. either 3 or female?
    Click here for the solution.
  3. Fiona has 85 labeled discs, which are numbered from 1 to 85. If a disc is selected at random what is the probability that the disc number:
    1. ends with 5
    2. can be multiplied by 3
    3. can be multiplied by 6
    4. is number 65
    5. is not a multiple of 5
    6. is a multiple of 4 or 3
    7. is a multiple of 2 and 6
    8. is number 1?
    Click here for the solution.

Random Experiments

  1. Let SS denote the set of whole numbers from 1 to 16, XX denote the set of even numbers from 1 to 16 and YY denote the set of prime numbers from 1 to 16
    1. Draw a Venn diagram accurately depicting SS, XX and YY.
    2. Find n(S)n(S), n(X)n(X), n(Y)n(Y), n(XY)n(XY), n(XY)n(XY).
    Click here for the solution.
  2. There are 79 Grade 10 learners at school. All of these take either Maths, Geography or History. The number who take Geography is 41, those who take History is 36, and 30 take Maths. The number who take Maths and History is 16; the number who take Geography and History is 6, and there are 8 who take Maths only and 16 who take only History.
    1. Draw a Venn diagram to illustrate all this information.
    2. How many learners take Maths and Geography but not History?
    3. How many learners take Geography only?
    4. How many learners take all three subjects?
    Click here for the solution.
  3. Pieces of paper labelled with the numbers 1 to 12 are placed in a box and the box is shaken. One piece of paper is taken out and then replaced.
    1. What is the sample space, SS?
    2. Write down the set AA, representing the event of taking a piece of paper labelled with a factor of 12.
    3. Write down the set BB, representing the event of taking a piece of paper labelled with a prime number.
    4. Represent AA, BB and SS by means of a Venn diagram.
    5. Find
      1. n(S)n(S)
      2. n(A)n(A)
      3. n(B)n(B)
      4. n(AB)n(AB)
      5. n(AB)n(AB)
    6. Is n(AB)=n(A )+n(B)-n(AB)n(AB)=n(A )+n(B)-n(AB)?
    Click here for the solution.

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