8 pt 2

Probability Theory
- Independence
- Bernoulli Trials
- Random Variables
- Expected Values

Definition:
- The events E and F are independent if and only if the following equation holds:
  p(E \cap F) = p(E) \cdot p(F)
- Furthermore, the conditional probability of E given F is defined as:
  p(E | F) = \frac{p(E \cap F)}{p(F)} = p(E)
- Symmetrical property: If p(E \cap F) = p(E) \cdot p(F), then p(F | E) = p(F) as well.

Let E be the event of rolling an even number with an unbiased die, while F is the event that the resulting number is divisible by three.
Events defined as:
- E = {2, 4, 6}
- F = {3, 6}
- E \cap F = {6}
Calculate probabilities:
- p(E) = \frac{3}{6} = \frac{1}{2}
- p(F) = \frac{2}{6} = \frac{1}{3}
- p(E \cap F) = \frac{1}{6}
Verification:
- p(E \cap F) = p(E) \cdot p(F):
- \frac{1}{6} = \frac{1}{2} \cdot \frac{1}{3}
Conclusion: E and F are independent.

Let E be the event where a random bit string of length four begins with a 1. Let F be the event the bit string contains an even number of 1s.
Possible bit strings starting with 1:
- {1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111}
Possible bit strings with an even number of 1s:
- {0000, 0011, 0101, 0110, 1001, 1010, 1100, 1111}
Calculate probabilities:
- Total strings = 16
- p(E) = p(F) = \frac{8}{16} = \frac{1}{2}
- E \cap F = {1111, 1100, 1010, 1001}
- p(E \cap F) = \frac{4}{16} = \frac{1}{4}
Verification:
- p(E \cap F) = 1/4 = (1/2)(1/2) = p(E) \cdot p(F)
Conclusion: E and F are independent.

For a family with two children, let E be the event of having two boys (BB) and F be the event of having at least one boy (BB, BG, GB).
- E = {BB}, p(E) = 1/4
- F = {BB, BG, GB}, p(F) = 3/4
- E \cap F = {BB}, p(E \cap F) = 1/4
Independence check:
- p(E) \cdot p(F) = (1/4)(3/4) = 3/16
Conclusion: E and F are not independent since 1/4
eq 3/16.

A Bernoulli trial is defined as a random experiment with two possible outcomes, commonly referred to as a success (S) or failure (F).
Let p be the probability of success, and q = 1 - p be the probability of failure.
The relationship between probabilities dictates:
- p + q = 1
When conducting n independent Bernoulli trials, the probability of achieving exactly k successes is given by the formula:
- P(X = k) = C(n, k) p^k q^{n-k}
  where C(n, k) represents the binomial coefficient, the number of ways to choose k successes from n trials.

Consider the probability of obtaining two successes in five Bernoulli trials.
Let’s denote:
- Successful outcomes: S
- Failing outcomes: F
Example sequence: SSFFF
Probability calculation for this specific sequence:
- p^2 q^3
Another possible sequence: FSFSF with the same logic leads to the same probability which is p^2 q^3.
Total number of possible sequences with k successes in n trials is determined by the binomial coefficient C(n, k).

For a biased coin where the probability of heads (success) is 2/3, let's calculate the probability of exactly four heads in seven tosses.
Total ways to get four heads among seven tosses is given by the binomial coefficient:
- C(7, 4)
The probability of each combination is:
- (\frac{2}{3})^4 (\frac{1}{3})^3
Therefore, the total probability is given by the product:
- P(X = 4) = C(7, 4) \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^3 = \frac{560}{2187} \approx 0.2561

A random variable is defined as a function that maps outcomes from the sample space of an experiment to real numbers.
It quantifies the outcomes for mathematical analysis.

Definition: The distribution of a random variable X over sample space S consists of pairs denoted as {(r, P(X = r)) | r \in X(S)} where P(X = r) signifies the probability that X takes the value r.

Illustrative example: Flipping a coin three times, let X(s) represent the number of heads.
Outcomes defined:
- X(HHH) = 3,
- X(TTT) = 0,
- X(HHT) = X(HTH) = X(THH) = 2,
- X(TTH) = X(THT) = X(HTT) = 1.

The expected value of a random variable is a statistical measure that provides a weighted average of the possible outcomes when an experiment is repeated many times.
Each outcome contributes to the average based on its probability.

For a random variable with outcomes 1 and 2, where 1 occurs with P = 0.1 and 2 occurs with P = 0.9, the traditional arithmetic average is not sufficient as it neglects the probabilities.
True average value is computed as a weighted sum:
- E(X) = 0.1 imes 1 + 0.9 imes 2 = 0.1 + 1.8 = 1.9
The generalized formula for the expected value is:
- E(X) = \sum_{s \in S} P(s) X(s)

Flipping a fair coin three times:
Sample space S contains 8 outcomes. The expected value of the random variable X, which counts how many heads appear, is calculated as:
- E(X) = \frac{1}{8} [X(HHH) + X(HHT) + X(HTH) + X(THH) + X(TTH) + X(THT) + X(HTT) + X(TTT)]
- After computation:
- E(X) = 1.5

Random variable X defines the outcome on a fair die. Each face shows values 1-6 with equal probabilities of \frac{1}{6}.
Expected value calculated as:
- E(X) = \frac{1}{6} \cdot [1 + 2 + 3 + 4 + 5 + 6] = \frac{21}{6} = \frac{7}{2}

Random variable X denotes the sum of numbers when rolling two dice. The sample size yields 36 outcomes.
Distribution analysis shows that while outcomes are equally likely, the values of X do not share equal probabilities.
Probabilities of outcomes for specific X values:
- P(X = 2) = \frac{1}{36}
- P(X = 3) = \frac{2}{36} = \frac{1}{18}
- P(X = 4) = \frac{3}{36} = \frac{1}{12}
- P(X = 5) = \frac{4}{36} = \frac{1}{9}
- P(X = 6) = \frac{5}{36}
- P(X = 7) = \frac{6}{36} = \frac{1}{6}
- P(X = 8) = \frac{5}{36}
- P(X = 9) = \frac{4}{36} = \frac{1}{9}
- P(X = 10) = \frac{3}{36} = \frac{1}{12}
- P(X = 11) = \frac{2}{36} = \frac{1}{18}
- P(X = 12) = \frac{1}{36}
Expected value is calculated as:
- E(X) = 2 \cdot (\frac{1}{36}) + 3 \cdot (\frac{1}{18}) + 4 \cdot (\frac{1}{12}) + 5 \cdot (\frac{1}{9}) + 6 \cdot (\frac{5}{36}) + 7 \cdot (\frac{1}{6}) + 8 \cdot (\frac{5}{36}) + 9 \cdot (\frac{1}{9}) + 10 \cdot (\frac{1}{12}) + 11 \cdot (\frac{1}{18}) + 12 \cdot (\frac{1}{36}) = 7
Interpretation: Averaging repeated rolls yields an average sum of 7.

Theorem:
- If X and Y are random variables on a sample space S, and a and b are real constants:
- E(X + Y) = E(X) + E(Y)
- E(aX + b) = aE(X) + b
Following this theorem, the expected value of a sum of two dice can be simplified using their individual expected values:
- For two independent random variables X₁ and X₂ representing two dice, we find that:
- E(X₁) = E(X₂) = \frac{7}{2}
- Thus, E(X₁ + X₂) = E(X₁) + E(X₂) = 7

End of Notes