10 1prob

Probability Revision

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Discrete Random Variables

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• Outcomes are the possible results of an experiment.

• Sample Space (or Universal Set) is the list of all possible outcomes.
  • We use the symbol $\varepsilon$ to represent the Universal Set.

• Discrete means the outcomes within the sample space are separate and countable
  • as opposed to continuous.

• Continuous means the outcomes can be any Real values between the endpoints - usually resulting from measuring
  • as opposed to counting.

• An event is the set of one or more outcomes being described.

• Probability is the likelihood (chance) of an event happening.
  • Probability is usually expressed as a fraction or decimal between 0 (impossible) and 1 (certain).
  • It is sometimes expressed as a percentage between 0% and 100%.

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Sets and Venn Diagrams

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• A set is a collection of elements (things, people, activities, numbers, ideas, etc)

• Sets are often named with capital letters: A, B, C, etc

• n(A) represents the number of elements in the set called A.

• $\varepsilon$ represents the Universal Set.

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A Venn Diagram is a way of representing one or more sets within the Universal Set (or sample space).

• A rectangle is used to denote the entire sample space (or the Universal Set)

• A circle inside the rectangle denotes a set (a subset of the Universal Set)

• The circle may contain
  • the elements within that set, or
  • the number of elements within the set.

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Example 1

• $\varepsilon = \big\{1,\; 2,\; 3,\; 4,\; 5,\; 6,\; 7,\; 8\big\}$

• $A = \big\{1,\; 3,\; 5,\; 7\big\}$

• $B = \big\{1,\; 2,\; 3,\; 4\big\}$

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Union (OR)

• The Union of two events is $A \cup B$

• It includes all outcomes in either A or B or both.

• $A \cup B = \big\{ 1,\; 2,\; 3,\; 4,\; 5,\; 7 \big\}$

• We sometimes use the word "or" instead of Union

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Intersection (AND)

• The Intersection of two events is $A \cap B$

• It includes only those outcomes which are in both A and B

• $A \cap B = \big\{1,\; 3\big\}$

• We sometimes use the word "and" instead of Intersection

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Complement (NOT)

• The Complement of A is A'
  • not to be confused with "compliment" which is a nice comment about someone

• The Complement of A includes all outcomes that are not in A (or outside of A)

• $A' = \big\{2, \; 4,\; 6,\; 8\big\}$

• We sometimes use the word "not" instead of complement

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NOTE:

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Probability of an Event occurring = Pr(A)

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If each outcome is equally likely then:
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… … $Pr(A) = \dfrac{ \text{Number of outcomes in A}}{\text{Total number of outcomes}}$
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… … $Pr(A) = \dfrac{n(A)}{n(\varepsilon)}$

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All probabilities must be between 0 and 1
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… … $0 \leqslant Pr(A) \leqslant 1$

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… … $Pr(A) = 0 \quad \longrightarrow$ Event is impossible

… … $Pr(A) = 1 \quad \longrightarrow$ Event is certain

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The sum of probabilities of all possible outcomes in the event space is always one.
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… … $Pr(A) + Pr(A') = 1$
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… … An event and its complement covers all possible outcomes in the event space.

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… … If A, B, C don't overlap and A or B or C covers all possible outcomes, then
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… … $Pr(A) + Pr(B) + Pr(C) = 1$
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New notation:

Sometimes we use p(x) instead of Pr(A)

• where p(x) represent the probability of event x happening.

• $\Sigma p(x) = 1$

• recall that $\Sigma$ is the symbol for "Sum"

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Example 2

A bag contains 5 black marbles, 7 red marbles (total = 12)

… a) .. What is the probability that a randomly drawn marble is black?

… b) .. What is the probability that a randomly drawn marble is not black?

Solution:

… … … Let B = Black Marble (the event that a black marble is drawn)
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… … … $Pr(B) = \dfrac{5}{12}$

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.. b) .. What is the probability that a randomly drawn marble is not black?
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… … … Since B = Black Marble,
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… … … B'= the complement of drawing a black marble (ie the event the marble is not black)
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… … … $Pr(B) + Pr(B') = 1$
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… … … $Pr(B') = 1 - Pr(B)$
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… … … $Pr(B') = 1 - \dfrac{5}{12}$
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… … … $Pr(B') = \dfrac{7}{12}$

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The Addition Rule of Probability

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For any events A and B:

… … $Pr(A \cup B) = Pr(A) + Pr(B) – Pr(A \cap B)$

… … We have to subtract $Pr(A \cap B)$ because otherwise we would be counting it twice.

In the above Venn Diagram, we have

… … … … $Pr(A \cup B) = \dfrac{6}{8} = \dfrac{3}{4}$

… … … … $Pr(A) = \dfrac{4}{8} = \dfrac{1}{2}$

… … … … $Pr(B) = \dfrac{4}{8} = \dfrac{1}{2}$

… … … … $Pr(A \cap B) = \dfrac{2}{8} = \dfrac{1}{4}$
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… … $Pr(A \cup B) = Pr(A) + Pr(B) – Pr(A \cap B)$
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… … $Pr(A \cup B) = \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{4}$
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… … $Pr(A \cup B) = \dfrac{3}{4}$

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Mutually Exclusive Events

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Two events are mutually exclusive if they cannot both occur at the same time

… … $A \cap B = \big\{ \big\}$

so

… … $Pr(A \cap B) = 0$

Therefore:

… …. $Pr(A \cup B) = Pr(A) + Pr(B)$

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For example, if

• A = the current month is May
• B = the current month is July

• Then A and B are mutually exclusive. They cannot both be true at the same time.

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Independent Events

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Two events are independent if the outcome of one does not influence the outcome of the other.

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Two events are dependent if the outcome of one does influence the outcome of the other.
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For Example:

… …. Tossing one coin and then a second coin:
… … … … the outcomes are independent.

… … Drawing one card from a deck of playing cards and then drawing a second card without replacing the first one:
… … … … the outcomes are dependent.

… … … … because the probabilities for the second card are different depending on which card was drawn first.
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If A and B are independent then

… …. $Pr(A \cap B) = Pr(A) \times Pr(B)$

Also

If $Pr(A \cap B) = Pr(A) \times Pr(B)$ then

… … the events A and B are independent.

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Example 3

An eight-sided dice has possible outcomes $\big\{1,\; 2,\; 3,\; 4,\; 5,\; 6,\; 7,\; 8\big\}$

If $A = \big\{1,\; 2\big\} \text{ and } B = \big\{2,\; 3,\; 4,\; 5\big\}$

Show that A and B are independent.

Solution:

… … $Pr(A) = \dfrac{2}{8} = \dfrac{1}{4}$

… … $Pr(B) = \dfrac{4}{8} = \dfrac{1}{2}$
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… … $Pr(A) \times Pr(B) = \dfrac{1}{4} \times \dfrac{1}{2} = \dfrac{1}{8}$
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… … and

… … $A \cap B = \big\{ 2 \big\}$

… … $Pr(A \cap B) = \dfrac{1}{8}$
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… … Since $Pr(A \cap B) = Pr(A) \times Pr(B)$ we have that A and B are independent.

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Karnaugh Maps and Probability Tables

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These are both laid out the same except that

• a Karnaugh Map shows the number of possible outcomes in each cell
• a Probability Table shows the probability of that cell.

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Example 4

A bag of 20 beads contains 8 red beads and 12 blue beads.

• Six of the red beads are cubes, the rest are spheres.
• Six of the blue beads are cubes, the rest are spheres.

Draw a Karnaugh Map and a Probability Table for this situation.

Solution:

… … Let A = red … then A' = blue
… … Let B = cube … then B' = sphere

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Conditional Probability

Conditional probability is the probability of one event, given that another event has already happened.
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The probability of A, given that B has happened is:
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… … $Pr(A \, | \, B) = \dfrac{Pr(A \cap B)}{Pr(B)}$

or

… … $Pr(A \cap B) = Pr(A \, | \, B) \times Pr(B)$

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Example 5

It has been noticed that in a particular town

• The probability of rain during the day is 0.4
• The probability there are clouds at dawn is 0.3
• The probability that there are clouds at dawn and it rains during the day is 0.2

Find
… a) .. the probability that there is rain, given that there were clouds at dawn
… b) .. the probability that there were clouds at dawn, given that it rains.

Solution:

… … … Let R = the event that it rains during the day

… … … Let C = the event that there were clouds at dawn

… … … From the question, we know:

… … … … $Pr(R) = 0.4$

… … … … $Pr(C) = 0.3$

… … … … $Pr(R \cap C) = 0.2$
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… a) .. the probability that there is rain, given that there were clouds at dawn

… … … $Pr(R \, | \, C) = \dfrac{Pr(R \cap C)}{Pr(C)}$

… … … $Pr(R \, | \, C) = \dfrac{0.2}{0.3}$

… … … $Pr(R \, | \, C) = \dfrac{2}{3}$
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… b) .. the probability that there were clouds at dawn, given that it rains.

… … … $Pr(C \, | \, R) = \dfrac{Pr(C \cap R)}{Pr(R)}$

… … … $Pr(C \, | \, R) = \dfrac{0.2}{0.4}$

… … … $Pr(C \, | \, R) = \dfrac{1}{2}$

… Notice that

… … … $Pr (R \, | \, C) \neq Pr (C \, | \, R)$
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In general:

… … $Pr (A \, | \, B)$ is not equal to $Pr (B \, | \, A)$

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Tree Diagrams

Tree diagrams are useful for understanding both dependent and independent events

• The probability of each outcome goes on the middle of each branch

• The probability of an outcome is the product of the probabilities on the branches leading to that result.

• If more than one outcome meets the criteria for an event then
  • the probability of that event is the sum of the probabilities of the relevant outcomes.

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Example 6

Nadine's car starts 70% of the time. If her car starts, she has an 80% chance of arriving at work on time.

If her car doesn't start, she has a 50% chance of arriving at work on time.

Find the probability that Nadine arrives on time.

Solution:

… … Let C = Car Starts … so C' = car doesn't start

… … Let O = On time … so O' = not on time)

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… … The Probability that Nadine arrives on time is $Pr(O)$
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… … $Pr(O) = Pr(CO) + Pr(C'O)$
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… … $Pr(O) = 0.56 + 0.15$
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… … $Pr(O) = 0.71$

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Combinations

Combinations are the number of ways that items can be selected from a set where the order they are selected in is not important.

For example, to select 3 letters out of $\big\{a,\; b,\; c,\; d\big\}$ where the order is not important,
… … there are 4 possible combinations. $\big\{abc,\; abd,\; acd,\; bcd\big\}$

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The number of Combinations of n objects, selected r at a time, is given by:
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… … $^nC_r = \left( \begin{matrix} n \\ r \\ \end{matrix} \right) = \dfrac{n!}{(n-r)! \,r!}$
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… … … … $^nC_r \text{ and } \left( \begin{matrix} n \\ r \\ \end{matrix} \right)$ are alternate notations for n combinations taken r at a time.

… … Recall that $n! = n(n - 1)(n - 2) \dots 3 \times 2 \times 1$

… … … … where n can be any Natural Number (positive integer)
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… … Eg: $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

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… … NOTE: To make the Combinations rule work correctly, we define $0! = 1$

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The Classpad Calculator can find both n! and ⁿC_r

• In the Virtual Keyboard, ADV tab (use the down arrow below the tabs)

… … Enter: nCr(4, 3) to get ⁴C₃

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Example 7

A committee of 6 people is to be selected from a group of 10.
How many combinations are possible?

Solution:

… … A total of 10 people involved so $n = 10$
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… … We select 6 at a time so $r = 6$
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… … $^{10}C_6 = \left( \begin{matrix} 10 \\ 6 \\ \end{matrix} \right) = \dfrac{10!}{4! \, 6!} = 210$
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… … or use the calculator by entering nCr(10, 6)

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