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Math Help - Probability: The basics with some examples of its application

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    Probability: The basics with some examples of its application

    Probability: The basics with some examples of its application

    I was inspired to create this thread after reading Chris L T521’s “Differential Equations Tutorial”.

    While reading this thread, assumed is a basic knowledge of fractions, percentages and decimal numbers.

    The basic idea of probability is applying a number to the chance of an event occurring. The number we use is always between and including zero and one. Zero representing no chance of an event occurring with one representing the event in question being certain to happen.

    Let’s look at some notation to describe what was just discussed. We shorthand the “probability of event A occurring” as P(A)

    Given the idea that a probability is always between zero and one then 0\leq P(A) \leq 1

    Now we have talked about what a probability is and the restrictions associated let’s define how we find a probability.

    Let’s say we have a universal set (the set of everything in question) of numbers or objects and call it \epsilon . The probability of event A occurring is defined as:

    P(A) = \frac{\text{The number of As in the set}}{\text{The total number of entries in the set}}=\frac{n(A)}{n(\epsilon )}

    Now let’s apply this.


    Example 1: Consider an unbiased die with the numbers one to six.
    a) Find the probability of rolling a 5
    b) Find the probability of rolling a prime number
    c) Find the probability of rolling an 8

    We have \epsilon = \{1,2,3,4,5,6\} and  n(\epsilon ) = 6

    a) P(5) = \frac{\text{The number of 5s in the set}}{\text{The total number of entries in the set}}=\frac{n(5)}{n(\epsilon )} =\frac{1}{6}

    b) P(prime) = \frac{\text{The number of prime numbers in the set}}{\text{The total number of entries in the set}}=\frac{n(prime)}{n(\epsilon )} =\frac{3}{6}


    c) There are no 8’s on the dice in question so

    P(8) = \frac{\text{The number of 8s in the set}}{\text{The total number of entries in the set}}=\frac{n(8)}{n(\epsilon )} =\frac{0}{6}=0


    Example 2: Consider a deck of cards with no jokers.
    a) Find the probability of drawing an ace
    b) Find the probability of drawing a picture card
    c) Find the probability of drawing a red card
    d) Find the probability of drawing a black queen

    As there are 52 cards in a deck so n(\epsilon ) = 52

    a) P(ace) = \frac{\text{The number of aces in the set}}{\text{The total number of entries in the set}}=\frac{n(aces)}{n(\epsilon )} =\frac{4}{52}

    b) P(picture~ card) =\frac{n(picture ~cards)}{n(\epsilon )} =\frac{12}{52}

    c) P(red~ card) =\frac{n(red~ cards)}{n(\epsilon )} =\frac{26}{52}

    d) There are four queens in the deck, one in each suit but we are only after the black ones.

    P(black~ queen) =\frac{n(black ~queens)}{n(\epsilon )} =\frac{2}{52}


    This concludes the first part of a series of posts on probability

    Next I plan on looking at complementary probability, intersection and union of sets and conditional probability.
    Last edited by mash; March 6th 2012 at 12:05 AM. Reason: fixed latex
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    Probability: Intersection and Union of Sets

    Hi again, in my last post I introduced 'the idea of' and 'how to calculate' a probability. I welcome any feedback on this thread at anytime via PM.

    A set is a collection of objects often identified by a certain characteristic. The notation used for listing a set are these curly brackets. \{,\} . Sets are often named for the convenience of not having to re-write or repeat a big sentence or group of numbers.

    Here's some examples.

    The set of numbers on a six sided die (let's call it set A) is A = \{1,2,3,4,5,6\}

    The set of prime numbers between 1 and 10 (let's call it set B) is B = \{2,3,5,7\}



    Intersection of sets. The intersection of sets is a very important idea in probability. It describes objects that are 'shared' between 2 or more sets.

    The notation for intersection is \cap so the intersection between set A and B is written as A\cap B

    Now let's find a solution for A\cap B from the two sets defined above.

    Recall A = \{1,2,3,4,5,6\} and B = \{2,3,5,7\} , the intersection is simply elements of each set that can be found in both. By inspection we can conclude A\cap B = \{2,3,5\} as these elements are in both A and B.



    Union of sets. Equally as important idea of probability is the union of sets. The union describes objects that are in all sets.

    The notation for union is \cup so the union of set A and B is written as A\cup B

    From our example above to find the union of A and B we need list everything (without repetition) that appears in both sets.

    By inspection we can conclude A\cup B = \{1,2,3,4,5,6,7\} as these elements are in both A or B.


    Here's some more examples on intersection and union. I have hidden the answers in case you wanted to have a go yourself.


    1. Consider the sets A = \{1,2,3,4,5,6,7\} and B = \{3,4,5,6,7,8,9\}

    a) find the intersection of these sets
    b) find the union of these sets

    Spoiler:


    By inspection

    a) A\cap B = \{3,4,5,6,7\}

    b) A \cup B = \{1,2,3,4,5,6,7,8,9\}




    2. Consider the set of letters that make up the name "Victoria" and the set of letters that make up the name "Catherine".
    a) find the intersection of these sets
    b) find the union of these sets

    Spoiler:
    Calling these sets 'V' and 'C' respectively.

    No need to repeat any letters in our sets and assuming we are not case sensitive.

    V = \{V,i,c,t,o,r,a \}

    C = \{C,a,t,h,e,r,i,n \}

    a) V \cap C= \{ C,i,a,t,r\}

    b) V \cup C= \{ V,i,c,t,o,r,a,h,e,n\}




    3. Consider the sets M = \{1,4,9,16\} , N = \{2,4,6,8,10\} and P = \{1,2,4,8,16\}

    a) find the intersection of sets M and N
    b) find the intersection of sets M and P
    c) find the intersection of sets M and N and P
    d) find the union of sets M and N
    e) find the union of sets M and P
    f) find the union of sets M and N and P

    Spoiler:


    Once again by inspection

    a) M\cap N = \{4\}

    b) M\cap P = \{1,4,16\}

    c) M\cap N\cap P = \{4\}

    d) M\cup N = \{1,2,4,6,8,9,10,16\}

    e) M\cup P = \{1,2,4,8,9,16\}

    f) M\cup N\cup P = \{1,2,4,6,8,9,10,16\}

    *Notice set P is contained in M\cup N




    Next post will talk about complementary probability.
    Last edited by mash; March 6th 2012 at 12:06 AM. Reason: fixed latex
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    Before we go onto the next topic here's a quick recap on ideas explored in posts #1 and #2.

    Consider a particular set of colours sold by a paint shop denoted by the set

    C = \{\text{Red},\text{Blue},\text{Yellow},\text{Orang  e},\text{Green},\text{Purple}\}

    The following table represents the number of paint cans in stock for each colour.


    \begin{array}{|c|c|}<br />
\text{Colour} & \text{\# in stock} \\ \hline<br />
\text{Red} & 10 \\ \text{Yellow} & 5 \\ \text{Blue} & 20 <br />
\\ \text{Orange} & 1 \\ \text{Green} & 3 \\ \text{Purple} & 1\\ \hline \end{array}

    It is important to note the total number of paint cans in stock is the sum of the second column 10+5+20+1+3+1 = 40

    Now let's find some probabilities


    1. Find the probability of picking a can of Yellow paint
    2. Find the probability of picking a can of Red paint
    3. Find the probability of picking a can of paint which is not Blue

    Spoiler:


    Using some letter abbreviations to make life easier

    1. P(\text{Yellow}) = P(Y)= \frac{5}{40} = \frac{1}{8}

    2. P(R) = \frac{10}{40} = \frac{1}{4}

    3. In this question we need to add together all cans that are not blue

    P(\text{Not B} ) = \frac{10+5+1+3+1}{40} = \frac{20}{40}= \frac{1}{2}




    Well I hope that was easy enough because it's time to introduce Complementary Probability.

    Complementary Probability

    The complement of an event is defined to be everything that is not the event in question.

    In the last set of questions we were asked Find the probability of picking a can of paint which is not Blue. This is the complement of everything that is blue.

    We note this where 'B = Blue' and the probability of finding a can that is blue is P(B) then the 'complement of blue' i.e (not being blue) is P(B')

    The following rule can be applied whenever we are asked to find the complement of an event.

    P(B)+P(B')=1 and  P(B')=1-P(B)

    Which means the probability of picking a blue can plus the probability of picking a blue can which is not blue is equal to one. This always is correct as we have picked out every single can!

    So revisiting the question Find the probability of picking a can of paint which is not Blue. and using our new rule we can arrive at our answer somewhat quicker

    3. Find the probability of picking a can of paint which is not Blue

     P(B')=1-P(B) = 1- \frac{20}{40}= 1- \frac{1}{2}= \frac{2}{2}- \frac{1}{2}= \frac{1}{2}


    This can become very helpful when we are given very large sets.

    Now it's time to find some complmentary probabilities, from our table above:

    1. Find the probability of picking a can of paint which is not Yellow
    2. Find the probability of picking a can of paint which is not Red
    3. Find the probability of picking a can of paint which is not Green


    Spoiler:



     P(Y')=1-P(Y) = 1- \frac{5}{40}= 1- \frac{1}{8}= \frac{8}{8}-\frac{1}{8}= \frac{7}{8}

     P(R')=1-P(R) = 1- \frac{10}{40}= 1- \frac{1}{4}= \frac{4}{4}-\frac{1}{4}=\frac{3}{4}

     P(G')=1-P(G) = 1- \frac{3}{40}=  \frac{40}{40}-\frac{3}{40}= \frac{37}{40}




    Next topic will be conditional probability.
    Last edited by mash; March 6th 2012 at 12:06 AM. Reason: fixed latex
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    Conditional Probability


    Thus far we have seen some straight forward examples on how to calculate a probability with a given scenario like a set of numbers or from a table.


    Often in life we need to find the probability of a certain event happening given a set of conditions. For example let's say we have a deck of 52 cards. To find the probability of a certain card being drawn i.e 'find the probability of drawing a king' we can apply the same logic from example 2 in post #1. We get


    P(K) = \frac{\text{The number of Kings in the deck}}{\text{The total number of cards in the deck}}=\frac{4}{52} \approx 0.077


    This is simple enough. But what if there is a condition set on the deck? What if the question was 'find the probability of drawing a king given you draw a red card' ?


    To negotiate this we apply the rule of conditional probability. The notation for using conditional probability is P(A/B) which means find the 'probability of A given B'. The rule is:


    P(A/B)= \frac{\text{The probability of the intersection between A and B }}{\text{The probability of B}}=\frac{P(A\cap B)}{P(B)}


    We can now use this new rule to answer the question 'find the probability of drawing a king given you draw a red card'.


    We need to find P(R) where R represents a red card.


    It is easy to conculde that because half of the cards in a deck is red then P(R) = \frac{26}{52}= \frac{1}{2}


    We need to find P(K\cap R) where K is a king and R is a red card.


    Recall that ths is simply how many red kings are in the deck then P(K\cap R)=\frac{2}{52} = \frac{1}{26}



    In conclusion P(K/R) = \frac{P(K\cap R)}{P( R)}= \frac{\frac{1}{26}}{\frac{1}{2}} = \frac{1}{26}\times \frac{2}{1} = \frac{1}{13}


    Now it's your turn.

    1. Given a fair 6 sided die, find the probability of rolling a 2 given a prime number is rolled.
    2. Given a deck of 52 cards, find the probability of drawing a picture card given a heart number is drawn.
    3. A spinner is divided into even sections and the letters from the word 'Probability' (letters are repeated) are placed in each section. Find the probability of spinnig up an 'i' given a vowel is spun.



    Spoiler:


    1. \frac{\frac{1}{6}}{\frac{3}{6}} = \frac{1}{3}


    2. \frac{\frac{3}{52}}{\frac{13}{52}} = \frac{3}{13}


    3. \frac{\frac{2}{11}}{\frac{4}{11}} = \frac{1}{3}



    Last edited by mash; March 6th 2012 at 12:06 AM. Reason: fixed latex
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