3.3 Two Basic Rules of Probability
The following activities and questions relate to material covered in Chapter 3.3 Two Basic Rules of Probability in Introductory Business Statistics (OpenStax).
Data sets for the following questions are available in Excel: 3.3 Data Sets [XLSX].
Questions
- According to the Yukon Bureau of Statistics, the total population of the Yukon territory in the year 2016 was 35,874 people. [1]
- From the 2016 census we see that 32,538 people spoke English most often at home.
- 3,336 people reported speaking a language other than English most often at home.
- Of those who speak another language at home, 0.2% spoke an aboriginal language most often at home.
Let:
- E = speaks English most often, at home.
- E’= Speaks another language most often, at home.
- A = Speaks an aboriginal language most often, at home.
Finish each probability statement. Check your work after answering b. by making sure P(E) + P(E’) = 1.
- P(E’)
- P(E)
- P(A ∩ E′)
- P(A | E′)
- Yukon University is looking to ask the local café to cater some desserts for a faculty meeting. The café makes cookies with chocolate (36%) and nuts (12%) and, of those, 8% contain both chocolate and nuts. Let’s imagine an instructor attending the meeting is allergic to both chocolate and nuts but wants a safe cookie to eat from this selection.
- Find the probability that a cookie contains chocolate or nuts (they cannot eat the cookies).
- Find the probability that a cookie does not contain chocolate or nuts (they can eat the cookie)
- The table below is the roster from newly renamed Edmonton Elks (formerly “Eskimos”).
Weight of Edmonton Elks Roster [2] Jersey Number (J) / Weight (W) ≤200 lbs 201 – 231 lbs 232 – 262 lbs ≥263 lbs 1 – 33 9 6 2 0 34 – 66 4 3 4 7 67 – 99 9 1 1 4 For the following, suppose that you randomly selected one player from the Elks roster.
- Find the probability that the jersey number is from 1 – 33.
- Find the probability that the player weighs at most 200 pounds.
- Find the probability that their jersey number is from 1 – 33 AND the player weighs at most 200 pounds.
- Find the probability that their jersey number is from 1 – 33 OR the player weighs at most 200 pounds.
- Find the probability that their jersey number is from 1 – 33 GIVEN that the player weighs at most 200 pounds.
Solutions
-
- P(E’) = [latex]\dfrac{3,336}{35,874}[/latex] = 0.093 or 9.3%
- P(E) = [latex]\dfrac{32,538}{35,874}[/latex] = 0.907 or 90.7%.
Check your work by making sure P(E) + P(E’) = 1! - P(A ∩ E′) = [latex]\dfrac{7}{35,874}[/latex] = 0.0002 or 0.02%
- P (A | E′) = P(A ∩ E′) / P(E’) = [latex]\dfrac{0.0002}{0.093}[/latex] = 0.0022
- Let:
C = the event that the cookie contains chocolate.
N = the event that the cookie contains nuts.- P(C ∪ N) = P(C) + P(N) – P(C ∩ N) = 0.36 + 0.12 – 0.08 = 0.40
- P(NEITHER chocolate NOR nuts) = 1 – P(C ∪ N) = 1 – 0.40 = 0.60
-
- P(J = 1-33) = [latex]\dfrac{17}{50}[/latex] = 34%
- P(W ≤ 200) = [latex]\dfrac{22}{50}[/latex] = 44%
- P(J = 1-33 ∩ W ≤ 200) = [latex]\dfrac{9}{50}[/latex] = 18%
- P(J = 1-33 ∪ W ≤ 200) = [latex]\dfrac{17}{50} + \dfrac{22}{50}- \dfrac{9}{50} = \dfrac{30}{50}[/latex] = 60%
- P(J = 1-33 | W ≤ 200) = [latex]\dfrac{\text{P}(\text{J}-33\cap\text{W}\leq200)}{\text{P}(\text{W}\leq200)}=\dfrac{0.18}{0.44}[/latex] = 40.9%