5.2: Three Types of Probability (2024)

  1. Last updated
  2. Save as PDF
  • Page ID
    44211
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\)

    \( \newcommand{\vectorC}[1]{\textbf{#1}}\)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}}\)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}\)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

    Classical Approach to Probability (Theoretical Probability)

    \[P(A) = \dfrac{\text{Number of ways A can occur}}{\text{Number of different outcomes in S}}\]

    The classical approach can only be used if each outcome has equal probability.

    If an experiment consists of flipping a coin twice, compute the probability of getting exactly two heads.

    Solution

    The event of getting exactly two heads is A = {HH}. The number of ways A can occur is 1. The number of different outcomes in S = {HH, HT, TH, TT} is 4. Thus P(A) = ¼.

    If a random experiment consists of rolling a six-sided die, compute the probability of rolling a 4.

    Solution

    The sample space is S = {1, 2, 3, 4, 5, 6}. The event A is that you want is to get a 4, and the event space is A = {4}. Thus, in theory the probability of rolling a 4 would be P(A) = 1/6 = 0.1667.

    Suppose you have an iPhone with the following songs on it: 5 Rolling Stones songs, 7 Beatles songs, 9 Bob Dylan songs, 4 Johnny Cash songs, 2 Carrie Underwood songs, 7 U2 songs, 4 Mariah Carey songs, 7 Bob Marley songs, 6 Bunny Wailer songs, 7 Elton John songs, 5 Led Zeppelin songs, and 4 Dave Matthews Band songs. The different genre that you have are rock from the ‘60s which includes Rolling Stones, Beatles, and Bob Dylan; country includes Johnny Cash and Carrie Underwood; rock of the ‘90s includes U2 and Mariah Carey; Reggae includes Bob Marley and Bunny Wailer; rock of the ‘70s includes Elton John and Led Zeppelin; and bluegrass/rock includes Dave Matthews Band.

    1. What is the probability that you will hear a Johnny Cash song?
    2. What is the probability that you will hear a Bunny Wailer song?
    3. What is the probability that you will hear a song from the ‘60s?
    4. What is the probability that you will hear a Reggae song?
    5. What is the probability that you will hear a song from the ‘90s or a bluegrass/rock song?
    6. What is the probability that you will hear an Elton John or a Carrie Underwood song?
    7. What is the probability that you will hear a country song or a U2 song?
    8. Solution

      a) The way an iPhone works, it randomly picks the next song so you have no idea what the next song will be. Now you would like to calculate the probability that you will hear the type of music or the artist that you are interested in. The sample set is too difficult to write out, but you can figure it from looking at the number in each set and the total number. The total number of songs you have is 67. There are 4 Johnny Cash songs out of the 67 songs. P(Johnny Cash song) = 4/67 = 0.0597

      b) There are 6 Bunny Wailer songs. P(Bunny Wailer) = 6/67 = 0.0896.

      c) There are 5, 7, and 9 songs that are classified as rock from the ‘60s, which is a total of 21. P(rock from the ‘60s) = 21/67 = 0.3134.

      d) There are total of 13 songs that are classified as Reggae. P(Reggae) = 13/67 = 0.1940

      e) There are 7 and 4 songs that are songs from the ‘90s and 4 songs that are bluegrass/rock, for a total of 15. P(rock from the ‘90s or bluegrass/rock) = 15/67 = 0.2239.

      f) There are 7 Elton John songs and 2 Carrie Underwood songs, for a total of 9. P(Elton John or Carrie Underwood song) = 9/67 = 0.1343.

      g) There are 6 country songs and 7 U2 songs, for a total of 13. P(country or U2 song) = 13/67 = 0.1940.

    Empirical Probability (Experimental or Relative Frequency Probability)

    The experiment is performed many times and the number of times that event A occurs is recorded. Then the probability is approximated by finding the relative frequency.

    \[P(A) = \dfrac{\text{Number of ways A occurred}}{\text{Number of times the experiment was repeated}}\]

    Important: The probability of any event A satisfies 0 ≤ P(A) ≤ 1, keep this in mind if the question is asking for a probability, and make sure your answer is a number between 0 and 1. A probability, relative frequency, percentage, and proportion are all different words for the same concept. Probability answers can be given as percentages, decimals, or reduced fractions.

    Suppose that the experiment is rolling a die. Compute the probability of rolling a 4.

    Solution

    The sample space is S = {1, 2, 3, 4, 5, 6}. The event A is that you want is to get a 4, and the event space is A = {4}. To do this, roll a die 10 times. When you do that, you get 4 two times. Based on this experiment, the probability of getting a 4 is 2 out of 10 or 1/5 = 0.2. To get more accuracy, repeat the experiment more times. It is easiest to put this information in a table, where n represents the number of times the experiment is repeated. When you put the number of 4s found divisible by the number of times you repeat the experiment, this is the relative frequency. See the last column in Figure 4-3.

    Figure 4-3: Trials for Die Experiment n Number of 4s Relative Frequency 10 2 0.2 50 6 0.12 100 18 0.18 500 81 0.162 1,000 163 0.163

    Notice that as n increased, the relative frequency seems to approach a number; it looks like it is approaching 0.163. You can say that the probability of getting a 4 is approximately 0.163. If you want more accuracy, then increase n even more by rolling the die more times.

    These probabilities are called experimental probabilities since they are found by actually doing the experiment or simulation. They come about from the relative frequencies and give an approximation of the true probability.

    The approximate probability of an event \(A\), notated as \(P(A)\), is

    \[P(A) = \frac{\text{Number of ways A occurred}}{\text{Number of times the experiment was repeated}}\]

    For the event of getting a 4, the probability would be P(Roll a 4) = \(\frac{\text{Number of times A occurred}}{\text{Number of times the experiment was repeated}}\) = 0.163

    “‘What was that voice?’ shouted Arthur.

    ‘I don't know,’ yelled Ford, ‘I don't know. It sounded like a measurement of probability.’

    ‘Probability? What do you mean?’

    ‘Probability. You know, like two to one, three to one, five to four against. It said two to the power of one hundred thousand to one against. That's pretty improbable you know.’”

    (Adams, 2002)

    Law of Large Numbers: as n increases, the relative frequency tends towards the theoretical probability

    Figure 4-4 shows a graph of experimental probabilities as n gets larger and larger. The dashed yellow line is the theoretical probability of rolling a four of 1/6 \(\neq\) 0.1667. Note the x-axis is in a log scale.

    Note that the more times you roll the die, the closer the experimental probability gets to the theoretical probability.

    5.2: Three Types of Probability (2)

    Figure 4-4

    You can compute experimental probabilities whenever it is not possible to calculate probabilities using other means. An example is if you want to find the probability that a family has 5 children, you would have to actually look at many families, and count how many have 5 children. Then you could calculate the probability. Another example is if you want to figure out if a die is fair. You would have to roll the die many times and count how often each side comes up. Make sure you repeat an experiment many times, because otherwise you will not be able to estimate the true probability of 5 children. This is due to the law of large numbers, since the more times we repeat the experiment, the closer the experimental probabilities will get to the theoretical probabilities. For difficult theoretical probabilities, we can run computer simulations that can run an experiment repeatedly many times very quickly and come up with accurate estimates of the theoretical probability.

    A fitness center coach kept track of members over the last year. They recorded if the person stretched before they exercised, and whether they sustained an injury. The following contingency table shows their results. Select one member at random and find the following probabilities.

    InjuryNo InjuryStretched 52 270 Did Not Stretch 21 57
    1. Compute the probability that a member sustained an injury.
    2. Compute the probability that a member did not stretch.
    3. Compute the probability that a member sustained an injury and did not stretch.
    Solution

    a) Find the totals for each row, column, and grand total.

    Injury No Injury Total
    Stretched 52 270 322
    Did Not Stretch 21 57 78
    Total 73 327 400

    Next, find the relative frequencies by dividing each number by the total of 400.

    Injury No Injury Total
    Stretched 0.13 0.675 0.805
    Did Not Stretch 0.0525 0.1425 0.195
    Total 0.1825 0.8175 1

    Using the definition of a probability we get P(Injury) = \(\frac{\text{Number of injuries}}{\text{Total number of people}}\) = \(\frac{73}{400}\) = 0.1825.

    Injury No Injury Total
    Stretched 0.13 0.675 0.805
    Did Not Stretch 0.0525 0.1425 0.195
    Total 0.1825 0.8175 1

    Using the table, we can get the same answer very quickly by just taking the column total under Injury to get 0.1825. As we get more complicated probability questions, these contingency tables will help organize your data.

    b) Using the relative frequency contingency table, take the total of the row for all the members that did not stretch and we get the P(Did Not Stretch) = 0.195.

    Injury No Injury Total
    Stretched 0.13 0.675 0.805
    Did Not Stretch 0.0525 0.1425 0.195
    Total 0.1825 0.8175 1

    c) Using the relative frequency contingency table, take the intersection of the injury column with the did not stretch row and we get P(Injury and Did Not Stretch) = 0.0525.

    Injury No Injury Total
    Stretched 0.13 0.675 0.805
    Did Not Stretch 0.0525 0.1425 0.195
    Total 0.1825 0.8175 1

    3. Subjective Probability

    The probability of event A is estimated using previous knowledge and is someone’s opinion.

    Compute the probability of meeting Dolly Parton.

    Solution

    I estimate the probability of meeting Dolly Parton to be 1.2E-9 \(\neg\) 0.0000000012 (i.e. very, very small).

    What is the probability it will rain tomorrow?

    Solution

    A weather reporter looks at several forecasts, uses their expert knowledge of the region, and reports the probability that it will rain in Portland, OR, is 80%.

    5.2: Three Types of Probability (3)

    5.2: Three Types of Probability (2024)
    Top Articles
    Frauenkondome: Vor- und Nachteile interner Kondome
    Wie man Frauenkondome benutzt – wikiHow
    How to choose a blood pressure monitor
    Latest Posts
    Durex vs. Trojan: Pros and Cons
    Buy approved Blood Pressure Monitors | British Heart Foundation
    Recommended Articles
    Article information

    Author: Ms. Lucile Johns

    Last Updated:

    Views: 6369

    Rating: 4 / 5 (41 voted)

    Reviews: 88% of readers found this page helpful

    Author information

    Name: Ms. Lucile Johns

    Birthday: 1999-11-16

    Address: Suite 237 56046 Walsh Coves, West Enid, VT 46557

    Phone: +59115435987187

    Job: Education Supervisor

    Hobby: Genealogy, Stone skipping, Skydiving, Nordic skating, Couponing, Coloring, Gardening

    Introduction: My name is Ms. Lucile Johns, I am a successful, friendly, friendly, homely, adventurous, handsome, delightful person who loves writing and wants to share my knowledge and understanding with you.