Probability Basics and Applications

Introduction to Basic Probability

• The video discusses the fundamental concepts of probability and its applications.

• It focuses on computing probability using a probability distribution.

• Mentioned types of probability distribution: uniform and non-uniform.

• Key method for computing probability: add all probabilities of individual outcomes that comprise a specific event.

Example Problem: Instructor Response Time

• A department tracks the number of days (denoted as d) it takes an instructor to respond to a student email.

• The data collected is represented by a probability distribution, which is partially summarized due to missing information at three days.

Finding the Missing Probability

• We need to calculate the missing probability corresponding to three days.

• To ensure a valid probability distribution, the sum of all probabilities must equal one: $17/100 + 28/100 + 32/100 + 8/100 + p = 1$

  • Rearranging gives:
    $p = 1 - (17/100 + 28/100 + 32/100 + 8/100)$

  • Calculating total of known probabilities:
    $17 + 28 + 32 + 8 = 85$

  • Therefore, $p = 1 - 85/100 = 15/100$

  • The missing probability for a student waiting exactly three days = 15 over 100.

Event A Probabilities

• Event A: A student must wait one day or less to receive an email response.

• Components of Event A:

  • Probability waiting 0 days: $17/100$

  • Probability waiting 1 day: $28/100$

• Thus, the probability of A:
  $P(A) = P(0 days) + P(1 day) = 17/100 + 28/100 = 45/100$

Event B Probabilities

• Event B: A student must wait 2 or 3 days to receive an email response.

• Components of Event B:

  • Probability waiting 2 days: $32/100$

  • Probability waiting 3 days: $15/100$

• Thus, the probability of B:
  $P(B) = P(2 days) + P(3 days) = 32/100 + 15/100 = 47/100$

Intersection and Union of Events

Intersection of Events A and B

• The intersection of events A and B involves finding common outcomes.

• A includes outcomes {0, 1}, and B includes outcomes {2, 3}.

• Overlap (intersection) is empty:

  • $P(A ext{ intersect } B) = 0/100$

Union of Events A and B Complement

• B Complement: Outcomes in sample space not in B: {0, 1, 4}.

• Union of A and B Complement: Combine outcomes of A (0, 1) with B Complement (4).

• Resulting set: {0, 1, 4}.

• Probability computation:

  • Total probabilities:
    $P(0) + P(1) + P(4) = 17/100 + 28/100 + 8/100 = 53/100$

• Thus, the probability that a student waits one day or less or does not wait 2 or 3 days = 53 over 100.

Probability Rules

Basic Properties of Probabilities

• If S is a sample space of an experiment, the following applies:

  • Probability of any event A is between 0 and 1 inclusively:
    $0 \ ext{ to } 1$

  • Probability of the empty set:
    $P( ext{empty set}) = 0$

  • Probability of the sample space:
    $P(S) = 1$

• Explanation:

  • Zero outcomes divided by total gives zero. Full outcomes divided by the same total yields one.

Union Rule

• To compute the probability of the union of two events A and B:
  $P(A ∪ B) = P(A) + P(B) - P(A ∩ B)$

• Justification: Avoids double counting outcomes common to both events.

Complement Rule

• To find the probability of event A's complement: $P(A') = 1 - P(A)$

  • Thus, the probability of A can also be rearranged:
    $P(A) = 1 - P(A')$

Special Considerations for Mutually Exclusive Events

• If events A and B are mutually exclusive,
  $P(A ∩ B) = 0$.

• Therefore, union rule can be simplified to:
  $P(A ∪ B) = P(A) + P(B)$

• Caution: This simplification only applies to mutually exclusive events.

Example Investigation with Dice

Experiment of Rolling Two Dice

• Description: One is green, and one is blue.

• Objective: Calculate specific probabilities for sums and outcomes of dice.

Part A: Probability of Sum Being 4 or 9

• Define desired probability:
  $P( ext{sum = 4}) + P( ext{sum = 9}) - P( ext{sum = 4 and 9})$

• Calculation steps:

  • Outcomes for a sum of 4: 3/36

  • Outcomes for a sum of 9: 4/36

  • Overlap (sum of both): 0/36

• Total probability for sums:
  $= 3/36 + 4/36 - 0/36 = 7/36$

Part B: Probability Green Die Not Showing 4

• Using complement rule:
  $P( ext{Green not 4}) = 1 - P( ext{Green = 4})$

• Outcomes where green die shows 4: 6/36.

• Thus:
  $= 36/36 - 6/36 = 30/36$

Part C: Probability Green Die Shows 2 or Blue Die Shows 5

• Combined event probability:
  $P( ext{Green = 2}) + P( ext{Blue = 5}) - P( ext{both})$

• Outcomes:

  • Green 2: 6/36

  • Blue 5: 6/36

  • Both conditions together: 1/36

• Calculation yields:
  $6/36 + 6/36 - 1/36 = 11/36$

Part D: Probability Sum Is 7 or Green Die Not 4

• Combined event probability calculation:
  $P( ext{Sum = 7}) + P( ext{Green not 4}) - P( ext{Sum = 7 and Green not 4})$

• Sum of 7: 6/36 (via specific rolls).

• Green not 4: previously calculated: 30/36.

• Subtract overlaps where sum is 7 with green die showing 4: 1 outcome.

• Final probability:
  $= 6/36 + 30/36 - 5/36 = 31/36$

Conclusion and Practical Applications

• The concepts and rules of probability discussed allow for a systematic approach to calculating probabilities in various scenarios, applicable in both theoretical and real-world contexts.

• Understanding how to use rules such as the union and complement is crucial, particularly in larger sample spaces where direct counting becomes impractical.