Oct. 8

Probability of At Least One Event
- When calculating the probability of at least one occurrence of an event, we often use the complement.
- Example Calculation:
- If the probability of an event occurring is P(A) = rac{1}{4}, then the probability of at least one occurrence is computed as:
  P( ext{at least one}) = 1 - P( ext{none}) = 1 - rac{1}{4} = rac{3}{4}.
- This aligns with the complement rule where the total probability adds up to 1.

Probability of Less than Two Heads
- "Less than two heads" refers to the occurrence of either one or zero heads; this can be confirmed with a probability table.
- Outcome Calculation:
- The probabilities for zero heads or one head sum to: P(0) + P(1) = rac{3}{4}.
Probability of Less than or Equal to Two Heads
- This scenario includes the cases for zero, one, or two heads.
- All possibilities can be assessed with total sample space, which is confirmed to be 7.

Sample Space
- The sample space size is crucial for determining distributions and total outcomes.

Mean Calculation:
- The mean (or expected value) of a probability distribution can be calculated using the formula: E(X) = ext{summation of }(x imes P(x))
  - This formula computes the average outcome over a series of trials.

Variance Calculation:
- Variance computes how much the outcomes vary around the mean. The formula involves squaring each value:
- ext{Variance} = ext{summation of }(x^2 imes P(x)) - E(X)^2
- Example: Suppose tossing a fair coin twice, the variance of observed head outcomes is calculated based on respective probabilities.
Standard Deviation:
- The square root of the variance provides the standard deviation, which indicates the spread of data points.

Expected Sum from Rolling Dice:
- When rolling two fair dice, the expected sum is typically 7, but the standard deviation indicates variability.
- Survey results show a concentration of expected outcomes around the average, backed by statistics indicating expected ranges of values (68% within one standard deviation, 95% within two).

Inspector's Item Analysis:
- An inspector evaluating three items with a known defect rate of 20% has probabilities for the number of defective items:
- Values of x could be 0, 1, 2, or 3 defective items. The probability for each case can be evaluated:
  - P(X=0) = P( ext{no defectives}) = 0.8^3
  - For one defective: count combinations (3 choose 1) times the probability of getting one defective and two good:
  - P(X=1) = inom{3}{1} imes 0.2^1 imes 0.8^2

Finding At Least One Defective:
- To find the probability of finding at least one defective item, calculate:
- P(X ext{ is at least } 1) = 1 - P(X=0)
- Expected calculations of defective items demonstrate distributions skewed towards lower defect rates.

Factorials:
- Defined as the product of all positive integers up to n, represented as n! = n imes (n - 1) imes … imes 1.
- Utilized to determine the number of arrangements for r items from a set of n items.
Combinations:
- When choosing r items from n without regard for order: C(n, r) = rac{n!}{r!(n-r)!}.
- Example: Choosing 5 items from a class of students to give or receive a bonus without caring about their arrangement.

All these concepts illustrate the importance of probability, statistics, and combinations in evaluating data in practical terms and forming reasoned conclusions based on expected values.