Write Me An Assignment Of Probability

Reading Assignment
An Introduction to Statistical Methods and Data Analysis, (see your Course Schedule).
Notation
Often students in introductory statistics courses struggle with probability due to getting caught up and/or confused with general notation used in describing events and the associated probability. To put it simply, the notation is shorthand to keep one from continually needing to write out long phrases to explain what is taking place. For instance, if one were to consider the toss of a fair coin the common theme is that there is a 1/2 chance, or 0.5 probability, of the coin coming up Tails. But how does one write this event?
 Begin by identifying the outcome event of interest: "Getting a Tail when we toss a fair coin."
 Using a single letter or word to represent this outcome of interest: 'Tail' or 'T' for instance.
 Stating your interest in the probability of this outcome: P(Tail) or P(T) which is read, "Probability of getting a Tail when we toss a fair coin."
Now instead of writing, "The probability is 1/2 of getting a Tail when we toss a fair coin", we can use probability notation to write P(T) = 1/2 or 0.5.
When you read a statistics text book a common lettering system uses the beginning of the alphabet. That is, the authors use 'A', 'B', etc. to define outcome events of interest. For the above example you could have stated, "Let A represent the outcome of getting a Tail when a fair coin is tossed." Using the steps defined above we would have P(A) = 1/2.
As you can see, the lettering can become convoluted! Just remember that the key is to identify what your outcome event of interest is. Try this: Apply the steps above to write out a probability statement for random selecting a female employee from a company where 35% of the employees are female.
Now let's complicate things. Consider again a fair coin. We stated P(T) was the probability we get a Tail when we toss the coin. How would you write the probability statement if the outcome of interest was getting two tails when the coin was tossed twice? Applying the steps:
 Getting a Tail on both tosses of a fair coin.
 With two trials (i.e. two tosses of the coin) where 'T' represents Tail we can write this as T,T where the first 'T' represents outcome of first toss and second 'T' the outcome of second toss.
 P(T,T)
Three Interpretations of Probability
What is probability? There are three interpretations:
1. Classical Interpretation of Probability
Event is a collection of outcomes, denoted by A, B, C, etc. The probability that event E occurs is denoted by P(E). When all outcomes are equally likely, then:
\[P(E)=\frac{number\ of\ outcomes\ in\ E}{number\ of\ possible\ outcomes}\]
For example, flip a coin. What is the chance of getting a head (H)?
Answer:
 Is the coin fair?
 If the coin is fair, then
\(P(H)=\frac{1}{2}\)
2. Relative Frequency Concept of Probability (Empirical Approach)
Flip the coin (since we don't know whether it is fair or not) a very large number of times and count the number of H out of the total number of flips.
\[P(E) \approx \frac{number\ of\ outcomes\ in\ E}{number\ of\ possible\ outcomes}\]
For example, if we flip the given coin 10,000 times and observe 4555 heads and 5445 tails, then for that coin, P(H) 0.4555.
Note: means approximate.
3. Subjective Probability
Subjective probability reflects personal belief which involves personal judgment, information, intuition, etc.
For example, what is P (you will get an A in a certain course)? Each student may have a different answer to the question.
Examples for Using the Classical Approach to Find the Probability
Find the probability that exactly one head appears in two flips of a fair coin.
Answer: The possible outcomes are listed as: {(H, H), (H, T), (T, H), (T, T)}.
Note that the four outcomes are of equal probability since the coin is fair.
\[P(getting\ exactly\ one\ H\ in\ two\ flips\ of\ a\ fair\ coin)=P\{(H,T),(T,H)\}=\frac{2}{4}=\frac{1}{2}\]
Find the probability that two heads appear in two flips of a fair coin.
Answer:\(P(getting\ two\ H\ in\ two\ flips\ of\ a\ fair\ coin)=P\{(H,H)\}=\frac{1}{4}\)
Note that 0 ≤ P(A) ≤ 1
P(A) = 0 means that the event A is impossible and P(A) = 1 means that the event A is sure to occur.Find the probability that the sum of two faces is greater than or equal to 10 when one rolls a pair of fair dice.
Answer: The possible outcomes of the experiment are:
{ (1,1) (2,1) (3,1) (4,1) (5,1) (6,1)
(1,2) (2,2) (3,2) (4,2) (5,2) (6,2)
(1,3) (2,3) (3,3) (4,3) (5,3) (6,3)
(1,4) (2,4) (3,4) (4,4) (5,4) (6,4)
(1,5) (2,5) (3,5) (4,5) (5,5) (6,5)
(1,6) (2,6) (3,6) (4,6) (5,6) (6,6) }If S denotes the sum of the points in the two faces:
\[ \begin {align} P(S\ greater\ than\ or\ equal\ to\ 10)& =P(S=10)+P(S=11)+P(S=12)\\
& = P\{(4,6),(5,5),(6,4)\}+P\{(5,6),(6,5)\}+P\{(6,6)\} \\
& = \frac{3}{36}+\frac{2}{36}+\frac{1}{36} \\
& =\frac{1}{6} \\
\end {align} \]
Probability Law (Set Operations)
1. Union
A or B also written as \(A \cup B\) = outcomes in A or B or both (shaded part)
This type of diagram that represents sets operations is called a Venn Diagram.
2. Intersection
A and B also written as \(A \cap B\) = outcomes in both A and B (shaded part)
3. Complement
A^{c} also written as \( \bar{A}\) = outcomes not in A (shaded part)
A and B are called mutually exclusive (disjoint) if the occurrence of outcomes in A excludes the occurrence of outcomes in B. In other words there are no elements in \(A \cap B\) and thus \(P(A \cap B)=0\). One example of two mutually exclusive events is that A and \( \bar{A}\) are mutually exclusive.
NOTATION REMINDER:
1. \(A \cup B\) is equivalent to A or B.
2. \(A \cap B\) is equivalent to A and B
3. \( \bar{A}\) is eqivalent to A^{c}
Probability Properties
 \(0 \leq P(A) \leq 1\)
 \(P(A)=1P(\bar{A})\)
 \(P(A \cup B)=P(A)+P(B)P(A \cap B)\).
The above Venn diagram shows why the third property is correct.
Given P(A) = 0.6, P(B) = 0.5, and \(P(A\cap B)=0.2\).
Use the information given above to work out your answer to the questions below, then click the graphic to the left to compare answers.
Find \(P(\bar{A})\).
Find \(P(\bar{A})\).
Answer:\(P(\bar{A})=1P(A)=0.4\)
Find \(P(A \cap \bar{B})\).
Find \(P(A \cap \bar{B})\).
Answer:\(P(A \cap \bar{B})=P(A)P(A\cap B)=0.60.2=0.4\)
Find \(P(B \cap \bar{A})\).
Find \(P(B \cap \bar{A})\).
Answer:\(P(B \cap \bar{A})=P(B)P(A\cap B)=0.50.2=0.3\)
Find \(P(A \cup B)\).
Find \(P(A \cup B)\).
Answer:\(P(A \cup B)=P(A)+P(B)P(A \cap B)=0.6+0.50.2=0.9\)
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