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Suppose that 10,000 U.S. licensed drivers are randomly selected.

  • How many would you expect to be male?
  • Using the table or tree diagram from the previous exercise, construct a contingency table of gender versus age group.
  • Using the contingency table, find the probability that out of the age 20 - 64 group, a randomly selected driver is female.
  • 5140
  • 0.49

Approximately 86.5% of Americans commute to work by car, truck or van. Out of that group, 84.6% drive alone and 15.4% drive in a carpool. Approximately 3.9% walk to work and approximately 5.3% take public transportation. ( Source: Bureau of the Census, U.S. Dept. of Commerce. Disregard rounding approximations. )

  • Construct a table or a tree diagram of the situation. Include a branch for all other modes of transportation to work.
  • Assuming that the walkers walk alone, what percent of all commuters travel alone to work?
  • Suppose that 1000 workers are randomly selected. How many would you expect to travel alone to work?
  • Suppose that 1000 workers are randomly selected. How many would you expect to drive in a carpool?

Explain what is wrong with the following statements. Use complete sentences.

  • If there’s a 60% chance of rain on Saturday and a 70% chance of rain on Sunday, then there’s a 130% chance of rain over the weekend.
  • The probability that a baseball player hits a home run is greater than the probability that he gets a successful hit.

Try these multiple choice questions.

The next two questions refer to the following probability tree diagram which shows tossing an unfair coin FOLLOWED BY drawing one bead from a cup containing 3 red ( R size 12{R} {} ), 4 yellow ( Y size 12{Y} {} ) and 5 blue ( B size 12{B} {} ) beads. For the coin, P ( H ) = 2 3 size 12{P \( H \) = { {2} over {3} } } {} and P ( T ) = 1 3 size 12{P \( T \) = { {1} over {3} } } {} where H = "heads" and T = "tails” .

Tree diagram with 2 branches. The first branch consists of 2 lines of H=2/3 and T=1/3. The second branch consists of 2 sets of 3 lines each with the both sets containing R=3/12, Y=4/12, and B=5/12.

Find P(tossing a Head on the coin AND a Red bead)

  • 2 3 size 12{ { {2} over {3} } } {}
  • 5 15 size 12{ { {5} over {"15"} } } {}
  • 6 36 size 12{ { {6} over {"36"} } } {}
  • 5 36 size 12{ { {5} over {"36"} } } {}

C

Find P(Blue bead) .

  • 15 36 size 12{ { {"15"} over {"36"} } } {}
  • 10 36 size 12{ { {"10"} over {"36"} } } {}
  • 10 12 size 12{ { {"10"} over {"12"} } } {}
  • 6 36 size 12{ { {6} over {"32"} } } {}

A

The next three questions refer to the following table of data obtained from www.baseball-almanac.com showing hit information for 4 well known baseball players. Suppose that one hit from the table is randomly selected.

NAME Single Double Triple Home Run TOTAL HITS
Babe Ruth 1517 506 136 714 2873
Jackie Robinson 1054 273 54 137 1518
Ty Cobb 3603 174 295 114 4189
Hank Aaron 2294 624 98 755 3771
TOTAL 8471 1577 583 1720 12351

Find P(hit was made by Babe Ruth) .

  • 1518 2873 size 12{ { {"1518"} over {"2873"} } } {}
  • 2873 12351 size 12{ { {"2873"} over {"12351"} } } {}
  • 583 12351 size 12{ { {"583"} over {"12351"} } } {}
  • 4189 12351 size 12{ { {"4189"} over {"12351"} } } {}

B

Find P(hit was made by Ty Cobb | The hit was a Home Run)

  • 4189 12351 size 12{ { {"4189"} over {"12351"} } } {}
  • 114 1720 size 12{ { {"114"} over {"1720"} } } {}
  • 1720 4189 size 12{ { {"1720"} over {"4189"} } } {}
  • 114 12351 size 12{ { {"114"} over {"12351"} } } {}

B

Are the hit being made by Hank Aaron and the hit being a double independent events?

  • Yes, because P(hit by Hank Aaron | hit is a double) = P(hit by Hank Aaron)
  • No, because P(hit by Hank Aaron | hit is a double) ≠ P(hit is a double)
  • No, because P(hit is by Hank Aaron | hit is a double) ≠ P(hit by Hank Aaron)
  • Yes, because P(hit is by Hank Aaron | hit is a double) = P(hit is a double)

C

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

  • Find P(H or G)
  • Find the probability of the complement of event (H and G)
  • Find the probability of the complement of event (H or G)

  • P(H or G) = P(H) + P(G) − P(H and G) = 0.26 + 0.43 − 0.14 = 0.55
  • P( NOT (H and G) ) = 1 − P(H and G) = 1 − 0.14 = 0.86
  • P( NOT (H or G) ) = 1 − P(H or G) = 1 − 0.55 = 0.45

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Source:  OpenStax, Engr 2113 ece math. OpenStax CNX. Aug 27, 2010 Download for free at http://cnx.org/content/col11224/1.1
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