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Chapter review

A tree diagram use branches to show the different outcomes of experiments and makes complex probability questions easy to visualize.

A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events. A Venn diagram is especially helpful for visualizing the OR event, the AND event, and the complement of an event and for understanding conditional probabilities.

The probability that a man develops some form of cancer in his lifetime is 0.4567. The probability that a man 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. Let: C = a man develops cancer in his lifetime; P = man has at least one false positive. Construct a tree diagram of the situation.

This is a tree diagram with two branches. The first branch, labeled Cancer, shows two lines: 0.4567 C and 0.5433 C'. The second branch is labeled False Positive. From C, there are two lines: 0 P and 1 P'. From C', there are two lines: 0.51 P and 0.49 P'.
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Bringing it together

Use the following information to answer the next two exercises. Suppose that you have eight cards. Five are green and three are yellow. The cards are well shuffled.

Suppose that you randomly draw two cards, one at a time, with replacement .
Let G 1 = first card is green
Let G 2 = second card is green

  1. Draw a tree diagram of the situation.
  2. Find P ( G 1 AND G 2 ).
  3. Find P (at least one green).
  4. Find P ( G 2 | G 1 ).
  5. Are G 2 and G 1 independent events? Explain why or why not.
  1. This is a tree diagram with branches showing probabilities of each draw. The first branch shows two lines: 5/8 Green and 3/8 Yellow. The second branch has a set of two lines (5/8 Green and 3/8 Yellow) for each line of the first branch.
  2. P ( GG ) = ( 5 8 ) ( 5 8 ) = 25 64
  3. P (at least one green) = P ( GG ) + P ( GY ) + P ( YG ) = 25 64 + 15 64 + 15 64 = 55 64
  4. P ( G | G ) = 5 8
  5. Yes, they are independent because the first card is placed back in the bag before the second card is drawn; the composition of cards in the bag remains the same from draw one to draw two.
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Suppose that you randomly draw two cards, one at a time, without replacement .
G 1 = first card is green
G 2 = second card is green

  1. Draw a tree diagram of the situation.
  2. Find P ( G 1 AND G 2 ).
  3. Find P (at least one green).
  4. Find P ( G 2 | G 1 ).
  5. Are G 2 and G 1 independent events? Explain why or why not.
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Use the following information to answer the next two exercises. The percent of licensed U.S. drivers (from a recent year) that are female is 48.60. Of the females, 5.03% are age 19 and under; 81.36% are age 20–64; 13.61% are age 65 or over. Of the licensed U.S. male drivers, 5.04% are age 19 and under; 81.43% are age 20–64; 13.53% are age 65 or over.

Complete the following.

  1. Construct a table or a tree diagram of the situation.
  2. Find P (driver is female).
  3. Find P (driver is age 65 or over|driver is female).
  4. Find P (driver is age 65 or over AND female).
  5. In words, explain the difference between the probabilities in part c and part d.
  6. Find P (driver is age 65 or over).
  7. Are being age 65 or over and being female mutually exclusive events? How do you know?
  1. <20 20–64 >64 Totals
    Female 0.0244 0.3954 0.0661 0.486
    Male 0.0259 0.4186 0.0695 0.514
    Totals 0.0503 0.8140 0.1356 1
  2. P ( F ) = 0.486
  3. P (>64| F ) = 0.1361
  4. P (>64 and F ) = P ( F ) P (>64| F ) = (0.486)(0.1361) = 0.0661
  5. P (>64| F ) is the percentage of female drivers who are 65 or older and P (>64 and F ) is the percentage of drivers who are female and 65 or older.
  6. P (> 64 ) = P (>64 and F ) + P (>64 and M ) = 0.1356
  7. No, being female and 65 or older are not mutually exclusive because they can occur at the same time P(>64 and F ) = 0.0661.
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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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