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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 occur is usually not known. Mathematically, probability theory formulates incomplete knowledge pertaining 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.

In everyday speech, we often refer to probabilities using percentages between 0% and 100%. A probability of 100% (100 out of 100) means that an event is certain, whereas a probability of 0% (0 out of 100) is often taken to mean the event is impossible. However, in certain circumstances, there can be a distinction between logically impossible and occurring with zero probability if the sample space is large enough (i.e. infinite) or in some other way indeterminate. For example, in selecting a random fraction between 0 and 1, the probability of selecting 1/2 is 0, but it is not logically impossible (since there are infinitely many numbers), while trying to predict the exact number of raindrops to fall into a large enough area is also 0 (or close to it), since the possible number of raindrops that might occur at a given time is both very large and difficult to determine.

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: The value of the topmost face of the die (call it 'k') is 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: The value of the topmost face is 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 1

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 drawing 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 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, as was noted in question 3, part (f) above, n(AB)=n(A)+n(B)-n(AB)n(AB)=n(A)+n(B)-n(AB). 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.

A final notion that is important to understand is the notion of complement.

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.

Relative Frequency vs. Probability

There are two approaches to determining the probability associated with any particular event of a random experiment:

  1. determining the total number of possible outcomes and calculating the probability of each outcome using the definition of probability
  2. performing the experiment and calculating the relative frequency of each outcome

Relative frequency is defined as the number of times an event happens in a statistical experiment divided by the number of trials conducted.

It takes a very large number of trials before the relative frequency of obtaining a head on a toss of a coin approaches the probability of obtaining a head on a toss of a coin. For example, the data in Table 1 represent the outcomes of repeating 100 trials of a statistical experiment 100 times, i.e. tossing a coin 100 times.

Table 1: Results of 100 tosses of a fair coin. H means that the coin landed heads-up and T means that the coin landed tails-up.
H T T H H T H H H H
H H H H T H H T T T
T T H T T H T H T H
H H T T H T T H T T
T H H H T T H T T H
H T T T T H T T H H
T T H T T H T T H T
H T T H T T T T H T
T H T T H H H T H T
T T T H H T T T H T

The following two worked examples show that the relative frequency of an event is not necessarily equal to the probability of the same event. Relative frequency should therefore be seen as an approximation to probability.

Exercise 2: Relative Frequency and Probability

Determine the relative frequencies associated with each outcome of the statistical experiment detailed in Table 1.

Solution

  1. Step 1. Identify the different outcomes :

    There are two unique outcomes: H and T.

  2. Step 2. Count how many times each outcome occurs. :
    Table 2
    Outcome Frequency
    H 44
    T 56
  3. Step 3. Determine the total number of trials. :

    The statistical experiment of tossing the coin was performed 100 times. Therefore, there were 100 trials, in total.

  4. Step 4. Calculate the relative frequency of each outcome :
    Probability of H = frequency of outcome number of trials = 44 100 = 0 , 44 Relative Frequency of T = frequency of outcome number of trials = 56 100 = 0 , 56 Probability of H = frequency of outcome number of trials = 44 100 = 0 , 44 Relative Frequency of T = frequency of outcome number of trials = 56 100 = 0 , 56
    (1)

    The relative frequency of the coin landing heads-up is 0,44 and the relative frequency of the coin landing tails-up is 0,56.

Exercise 3: Probability

Determine the probability associated with an evenly weighted coin landing on either of its faces.

Solution

  1. Step 1. Identify the different outcomes :

    There are two unique outcomes: H and T.

  2. Step 2. Determine the total number of outcomes. :

    There are two possible outcomes.

  3. Step 3. Calculate the probability of each outcome :
    Relative Frequency of H = number of favourable outcomes total number of outcomes = 1 2 = 0 , 5 Relative Frequency of T = number of favourable outcomes total number of outcomes = 1 2 = 0 , 5 Relative Frequency of H = number of favourable outcomes total number of outcomes = 1 2 = 0 , 5 Relative Frequency of T = number of favourable outcomes total number of outcomes = 1 2 = 0 , 5
    (2)

    The probability of an evenly weighted coin landing on either face is 0,50,5.

Project Idea

Perform an experiment to show that as the number of trials increases, the relative frequency approaches the probability of a coin toss. Perform 10, 20, 50, 100, 200 trials of tossing a coin.

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
(3)
P ( A B ) = P ( A ) × P ( B ) P ( A B ) = P ( A ) × P ( B )
(4)
P ( A B ) = P ( A ) + P ( B ) - P ( A B ) P ( A B ) = P ( A ) + P ( B ) - P ( A B )
(5)

Exercise 4: 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)=52/52=1. because all cards are black or red!

Exercise 5: 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 )
    (6)
  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
    (7)

    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 1
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

Mutually exclusive events are events, which 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 5 becomes:

P ( A B ) = P ( A ) + P ( B ) P ( A B ) = P ( A ) + P ( B )
(8)

for mutually exclusive events.

Mutually Exclusive Events

Answer the following questions

  1. A box contains coloured blocks. The number of each colour is given in the following table.
    Table 3
    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 4
    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.

Complementary Events

The probability of complementary events refers to the probability associated with events not occurring. 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. This means that

P ( A ' ) = 1 - P ( A ) P ( A ' ) = 1 - P ( A )
(9)

where A' refers to `not A' In other words, the probability of `not A' is equal to one minus the probability of A.

Exercise 6: Probability

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 7: Probability

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

Interpretation of Probability Values

The probability of an event is generally represented as a real number between 0 and 1, inclusive. An impossible event has a probability of exactly 0, and a certain event has a probability of 1, but the converses are not always true: probability 0 events are not always impossible, nor probability 1 events certain. The rather subtle distinction between "certain" and "probability 1" is treated at greater length in the article on "almost surely".

Most probabilities that occur in practice are numbers between 0 and 1, indicating the event's position on the continuum between impossibility and certainty. The closer an event's probability is to 1, the more likely it is to occur.

For example, if two mutually exclusive events are assumed equally probable, such as a flipped or spun coin landing heads-up or tails-up, we can express the probability of each event as "1 in 2", or, equivalently, "50%" or "1/2".

Probabilities are equivalently expressed as odds, which is the ratio of the probability of one event to the probability of all other events. The odds of heads-up, for the tossed/spun coin, are (1/2)/(1 - 1/2), which is equal to 1/1. This is expressed as "1 to 1 odds" and often written "1:1".

Odds a:b for some event are equivalent to probability a/(a+b). For example, 1:1 odds are equivalent to probability 1/2, and 3:2 odds are equivalent to probability 3/5.

Summary

  • Random experiments
  • Outcome
  • Sample space
  • Event
  • Probability models
  • Classical theory Relative frequency vs. probability
  • Probability identities
  • Mutually exclusive events
  • Complementary events

End of Chapter Exercises

  1. A group of 45 children were asked if they eat Frosties and/or Strawberry Pops. 31 eat both and 6 eat only Frosties. What is the probability that a child chosen at random will eat only Strawberry Pops?
    Click here for the solution.
  2. In a group of 42 pupils, all but 3 had a packet of chips or a Fanta or both. If 23 had a packet of chips and 7 of these also had a Fanta, what is the probability that one pupil chosen at random has:
    1. Both chips and Fanta
    2. has only Fanta?
    Click here for the solution.
  3. Use a Venn diagram to work out the following probabilities from a die being rolled:
    1. A multiple of 5 and an odd number
    2. a number that is neither a multiple of 5 nor an odd number
    3. a number which is not a multiple of 5, but is odd.
    Click here for the solution.
  4. A packet has yellow and pink sweets. The probability of taking out a pink sweet is 7/12.
    1. What is the probability of taking out a yellow sweet
    2. If 44 if the sweets are yellow, how many sweets are pink?
    Click here for the solution.
  5. In a car park with 300 cars, there are 190 Opals. What is the probability that the first car to leave the car park is:
    1. an Opal
    2. not an Opal
    Click here for the solution.
  6. Tamara has 18 loose socks in a drawer. Eight of these are orange and two are pink. Calculate the probability that the first sock taken out at random is:
    1. Orange
    2. not orange
    3. pink
    4. not pink
    5. orange or pink
    6. not orange or pink
    Click here for the solution.
  7. A plate contains 9 shortbread cookies, 4 ginger biscuits, 11 chocolate chip cookies and 18 Jambos. If a biscuit is selected at random, what is the probability that:
    1. it is either a ginger biscuit of a Jambo?
    2. it is NOT a shortbread cookie.
    Click here for the solution.
  8. 280 tickets were sold at a raffle. Ingrid bought 15 tickets. What is the probability that Ingrid:
    1. Wins the prize
    2. Does not win the prize?
    Click here for the solution.
  9. The children in a nursery school were classified by hair and eye colour. 44 had red hair and not brown eyes, 14 had brown eyes and red hair, 5 had brown eyes but not red hair and 40 did not have brown eyes or red hair.
    1. How many children were in the school
    2. What is the probility that a child chosen at random has:
      1. Brown eyes
      2. Red hair
    3. A child with brown eyes is chosen randomly. What is the probability that this child will have red hair
    Click here for the solution.
  10. A jar has purple, blue and black sweets in it. The probability that a sweet, chosen at random, will be purple is 1/7 and the probability that it will be black is 3/5.
    1. If I choose a sweet at random what is the probability that it will be:
      1. purple or blue
      2. Black
      3. purple
    2. If there are 70 sweets in the jar how many purple ones are there?
    3. 1/4 if the purple sweets in b) have streaks on them and rest do not. How many purple sweets have streaks?
    Click here for the solution.
  11. For each of the following, draw a Venn diagram to represent the situation and find an example to illustrate the situation.
    1. A sample space in which there are two events that are not mutually exclusive
    2. A sample space in which there are two events that are complementary.
    Click here for the solution.
  12. Use a Venn diagram to prove that the probability of either event A or B occuring is given by: (A and B are not exclusive) P(A or B) = P(A) + P(B) - P(A and B)
    Click here for the solution.
  13. All the clubs are taken out of a pack of cards. The remaining cards are then shuffled and one card chosen. After being chosen, the card is replaced before the next card is chosen.
    1. What is the sample space?
    2. Find a set to represent the event, P, of drawing a picture card.
    3. Find a set for the event, N, of drawing a numbered card.
    4. Represent the above events in a Venn diagram
    5. What description of the sets P and N is suitable? (Hint: Find any elements of P in N and N in P.)
    Click here for the solution.
  14. Thuli has a bag containing five orange, three purple and seven pink blocks. The bag is shaken and a block is withdrawn. The colour of the block is noted and the block is replaced.
    1. What is the sample space for this experiment?
    2. What is the set describing the event of drawing a pink block, P?
    3. Write down a set, O or B, to represent the event of drawing either a orange or a purple block.
    4. Draw a Venn diagram to show the above information.
    Click here for the solution.

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