# 3.2 Independent and mutually exclusive events  (Page 2/23)

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## Try it

A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Let T be the event of getting the white ball twice, F the event of picking the white ball first, S the event of picking the white ball in the second drawing.

1. Compute P ( T ).
2. Compute P ( T | F ).
3. Are T and F independent?.
4. Are F and S mutually exclusive?
5. Are F and S independent?
1. P ( T ) = $\frac{1}{4}$
2. P ( T | F ) = $\frac{1}{2}$
3. No
4. No
5. Yes

## References

Lopez, Shane, Preety Sidhu. “U.S. Teachers Love Their Lives, but Struggle in the Workplace.” Gallup Wellbeing, 2013. http://www.gallup.com/poll/161516/teachers-love-lives-struggle-workplace.aspx (accessed May 2, 2013).

Data from Gallup. Available online at www.gallup.com/ (accessed May 2, 2013).

## Chapter review

Two events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs. If two events are not independent, then we say that they are dependent.

In sampling with replacement, each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered not to be independent. When events do not share outcomes, they are mutually exclusive of each other.

## Formula review

If A and B are independent, P ( A AND B ) = P ( A ) P ( B ), P ( A | B ) = P ( A ) and P ( B | A ) = P ( B ).

If A and B are mutually exclusive, P ( A OR B ) = P ( A ) + P ( B ) and P ( A AND B ) = 0.

E and F are mutually exclusive events. P ( E ) = 0.4; P ( F ) = 0.5. Find P ( E F ).

J and K are independent events. P ( J | K ) = 0.3. Find P ( J ).

P ( J ) = 0.3

U and V are mutually exclusive events. P ( U ) = 0.26; P ( V ) = 0.37. Find:

1. P ( U AND V ) =
2. P ( U | V ) =
3. P ( U OR V ) =

Q and R are independent events. P ( Q ) = 0.4 and P ( Q  AND  R ) = 0.1. Find P ( R ).

P ( Q AND R ) = P ( Q ) P ( R )

0.1 = (0.4) P ( R )

P ( R ) = 0.25

## Bringing it together

A previous year, the weights of the members of the San Francisco 49ers and the Dallas Cowboys were published in the San Jose Mercury News . The factual data are compiled into [link] .

Shirt# ≤ 210 211–250 251–290 290≤
1–33 21 5 0 0
34–66 6 18 7 4
66–99 6 12 22 5

For the following, suppose that you randomly select one player from the 49ers or Cowboys.

If having a shirt number from one to 33 and weighing at most 210 pounds were independent events, then what should be true about P (Shirt# 1–33|≤ 210 pounds)?

The probability that a male develops some form of cancer in his lifetime is 0.4567. The probability that a male has at least one false positive test result (meaning the test comes back for cancer when the man does not have it) is 0.51. Some of the following questions do not have enough information for you to answer them. Write “not enough information” for those answers. Let C = a man develops cancer in his lifetime and P = man has at least one false positive.

1. P ( C ) = ______
2. P ( P | C ) = ______
3. P ( P | C' ) = ______
4. If a test comes up positive, based upon numerical values, can you assume that man has cancer? Justify numerically and explain why or why not.

1. P ( C ) = 0.4567
2. not enough information
3. not enough information
4. No, because over half (0.51) of men have at least one false positive text

Given events G and H : P ( G ) = 0.43; P ( H ) = 0.26; P ( H AND G ) = 0.14

1. Find P ( H OR G ).
2. Find the probability of the complement of event ( H AND G ).
3. Find the probability of the complement of event ( H OR G ).

Given events J and K : P ( J ) = 0.18; P ( K ) = 0.37; P ( J OR K ) = 0.45

1. Find P ( J AND K ).
2. Find the probability of the complement of event ( J AND K ).
3. Find the probability of the complement of event ( J AND K ).
1. P ( J OR K ) = P ( J ) + P ( K ) − P ( J AND K ); 0.45 = 0.18 + 0.37 - P ( J AND K ); solve to find P ( J AND K ) = 0.10
2. P (NOT ( J AND K )) = 1 - P ( J AND K ) = 1 - 0.10 = 0.90
3. P (NOT ( J OR K )) = 1 - P ( J OR K ) = 1 - 0.45 = 0.55

hi
Kachalla
what's up here ... am new here
Kachalla
sorry question a bit unclear...do you mean how do you analyze quantitative data? If yes, it depends on the specific question(s) you set in the beginning as well as on the data you collected. So the method of data analysis will be dependent on the data collecter and questions asked.
Bheka
how to solve for degree of freedom
saliou
Quantitative data is the data in numeric form. For eg: Income of persons asked is 10,000. This data is quantitative data on the other hand data collected for either make or female is qualitative data.
Rohan
*male
Rohan
Degree of freedom is the unconditionality. For example if you have total number of observations n, and you have to calculate variance, obviously you will need mean for that. Here mean is a condition, without which you cannot calculate variance. Therefore degree of freedom for variance will be n-1.
Rohan
data that is best presented in categories like haircolor, food taste (good, bad, fair, terrible) constitutes qualitative data
Bheka
vegetation types (grasslands, forests etc) qualitative data
Bheka
I don't understand how you solved it can you teach me
solve what?
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ok so you use those two to draw the histogram right.
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yes
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Bheka
To make a histogram, follow these steps: On the vertical axis, place frequencies. Label this axis "Frequency". On the horizontal axis, place the lower value of each interval. ... Draw a bar extending from the lower value of each interval to the lower value of the next interval.
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akeem
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what is random sampling what is sample error
@Nistha Kashyap Random sampling is the selection of random items (or random numbers) from the group. A sample error occurs when the selected samples do not truely represent the whole group. The can happen when most or all of the selected samples are taken from only one section of the group;
Ron
Thus the sample is not truely random.
Ron
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Differences between sample mean & population mean
***keydifferences.com/difference-between-sample-mean-and-population-mean.html
Lucien
Not difference in the formula except the notation, sample mean is denoted by x bar and population mean is denoted by mu symbol. There is formula as well as notation between difference variance and standard deviations
Akash
Likely the difference would be in the result, unless the sample is an exact representation of the population (which is unlikely.)
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what is data
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Data is the very pale android from the Star Trek Enterprise
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