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

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

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 .

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 .

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 .

NOTE: .

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. .

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 .

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) .

… … 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 nCr

• 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) .