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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 ) = 1 4
  2. P ( T | F ) = 1 2
  3. No
  4. No
  5. Yes
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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 ).

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

P ( J ) = 0.3

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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 ) =
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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

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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)?

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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
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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 ).
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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
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Questions & Answers

If a random variable X takes only two values -2 and 1 such that 2 P[X=- 2]=P[X=1]=P, then find variance of x.
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Source:  OpenStax, Introductory statistics. OpenStax CNX. May 06, 2016 Download for free at http://legacy.cnx.org/content/col11562/1.18
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