Basic Probability and Applications

Basic Probability and Applications

Introduction to Probability and Venn Diagrams

  • Discussion about basic probability rules and applications.

  • Emphasis on utilizing Venn diagrams to visualize problems in scenarios where counting outcomes is not feasible.

Example 1: Events A and B in Sample Space S

  • Given information:

    • Probability of A: $P(A) = \frac{6}{10}$

    • Probability of B: $P(B) = \frac{3}{10}$

    • Probability of the union of A and B: $P(A \cup B) = \frac{7}{10}$

  • Objective: Compute the following probabilities:

    • Probability of A intersect B: $P(A \cap B)$

    • Probability of A complement intersect B: $P(A^c \cap B)$

    • Probability of A intersect B complement: $P(A \cap B^c)$

Step 1: Defining Regions in Venn Diagram

  • Identify mutually exclusive regions: w, x, y, z.

  • Using given probabilities to define regions:

    1. Probability of A ($P(A) = P(w) + P(x)$):

    • Conclude that regions w and x combined equal $ rac{6}{10}$.

    1. Probability of B ($P(B) = P(x) + P(y)$):

    • Conclude that regions x and y combined equal $ rac{3}{10}$.

    1. Probability of union of A and B ($P(A \cup B) = P(w) + P(x) + P(y)$):

    • Take union into account to find the probability of z: $P(z) = 1 - P(A \cup B) = 1 - \frac{7}{10} = \frac{3}{10}$.

Step 2: Utilizing Union Rule

  • Union rule formula:
    P(A)+P(B)−P(A∩B)=P(A∪B)P(A) + P(B) - P(A \cap B) = P(A \cup B)

  • Plugging in known values: 610+310−P(A∩B)=710\frac{6}{10} + \frac{3}{10} - P(A \cap B) = \frac{7}{10}

    • Solving gives:
      P(A∩B)=210P(A \cap B) = \frac{2}{10}

Step 3: Filling in the Venn Diagram

  • Knowing $P(A \cap B) = P(x) = \frac{2}{10}$, derive other probabilities:

    • For w: $P(A) = P(w) + P(x)$ leads to $\frac{6}{10} = P(w) + \frac{2}{10}$, thus $P(w) = \frac{4}{10}$.

    • For y: $P(B) = P(x) + P(y)$ leads to $\frac{3}{10} = \frac{2}{10} + P(y)$, thus $P(y) = \frac{1}{10}$.

Step 4: Answering Probabilities

  1. Probability of A intersect B:

    • $P(A \cap B) = P(x) = \frac{2}{10}$.

  2. Probability of A complement intersect B:

    • $A^c$ consists of regions w and z; thus:

    • $P(A^c \cap B) = P(y) = \frac{1}{10}$.

  3. Probability of A intersect B complement:

    • $A \cap B^c$: Regions w, y, z

    • $P(A \cap B^c) = P(w) + P(y) + P(z) = \frac{4}{10} + \frac{1}{10} + \frac{3}{10} = \frac{8}{10}$.

Example 2: Survey of Gardeners

  • A survey of 200 gardeners with the following statistics:

    • 50% use compost: $P(F) = \frac{100}{200}$.

    • 38% water daily: $P(W) = \frac{76}{200}$.

    • 29% neither compost nor water daily: $P(F^c \cap W^c) = \frac{58}{200}$.

Step 1: Creating Venn Diagram

  • Define events:

    • F: fertilize using compost

    • W: water daily.

Step 2: Utilizing Complementation Laws

  • Find probability of gardeners using compost or water daily:
    P(F∪W)=1−P(Fc∩Wc)=1−58200=142200P(F \cup W) = 1 - P(F^c \cap W^c) = 1 - \frac{58}{200} = \frac{142}{200}

Step 3: Using Union Rule for Intersection

  • Using union rule to find gardeners who compost and water daily:
    P(F∩W)=P(F)+P(W)−P(F∪W)P(F \cap W) = P(F) + P(W) - P(F \cup W)

  • Plugging in:
    142200=100200+76200−P(F∩W)\frac{142}{200} = \frac{100}{200} + \frac{76}{200} - P(F \cap W)

  • Solve to find:
    P(F∩W)=34200P(F \cap W) = \frac{34}{200}

Step 4: Last Probability Calculation

  • Determine probability of not composting but watering daily:

  • $F^c$ intersects with W;

  • Through Venn diagram definitions, find:
    P(Fc∩W)=P(d)=42200P(F^c \cap W) = P(d) = \frac{42}{200}

Conclusion

  • Reiterate importance of using Venn diagrams to visualize probabilistic problems.

  • The systematic use of formulas and definitions helps clarify complex relationships among events in probability.