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Probability: Independent Events and Random Variables

Independent Events

Class Definition

Two events are defined as independent if the probability of their intersection is the simple product of their individual probabilities. Mathematically, for events A and B, they are independent if:P(A \cap B) = P(A) \cdot P(B)
This definition serves as a shortcut for messy computations, allowing complex intersection probabilities to be calculated by simply multiplying individual probabilities.

Conceptual Understanding

Conceptually, two events are independent if the occurrence of one has no effect on the probability of the other. This means if you found out event A happened, the probability of event B remains exactly what it was before, and vice versa.

Examples of Independent Events

Coin Flipping

  • Classical Example: Flipping a coin multiple times. Each flip is independent of previous flips. For instance, if a coin has landed on heads nine times in a row, the probability of the tenth flip being heads is still 1/2. The previous outcomes have no influence on the next one.

  • Computation Application: To find the probability of a sequence of independent events, like flipping a coin three times and getting heads each time (H1 \cap H2 \cap H3), you can use the independence property: P(H1 \cap H2 \cap H3) = P(H1) \cdot P(H2) \cdot P(H_3) = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \left(\frac{1}{2}\right)^3

  • More Complex Sequence: If you flip a coin seven times and want the probability of getting HTHTTTT (Heads, Tails, Heads, Tails, Tails, Tails, Tails), the calculation is:
    P(H \cap T \cap H \cap T \cap T \cap T \cap T) = \left(\frac{1}{2}\right)^7This is significantly simpler than calculating the total sample space and counting specific outcomes, especially for long sequences.

General Application

This principle applies whenever a sequence of measurable events occurs, and the previous stages in the sequence have no effect on the likelihood or probability of the next stage's outcomes.

Roulette Example

(Based on a version with 38 slices: 2 green, 18 red, 18 black).

  • Independence: Each spin of a roulette wheel is independent; the previous outcome does not affect the next.

  • Spin Probabilities:

    • Probability of Green (G): P(G) = \frac{2}{38}

    • Probability of Red (R): P(R) = \frac{18}{38}

    • Probability of Black (B): P(B) = \frac{18}{38}

  • Part A: Probability of Green, Green, Black, Black, Black (GGBBB)

    • We are asked to compute P(G1 \cap G2 \cap B3 \cap B4 \cap B_5).

    • Due to independence, this breaks down to:
      P(G) \cdot P(G) \cdot P(B) \cdot P(B) \cdot P(B) = \left(\frac{2}{38}\right)^2 \cdot \left(\frac{18}{38}\right)^3
      This method is much simpler than analyzing the entire five-spin sample space.

  • Part B: Probability of None of the Five Spins Being Red

    • This means not Red for all five spins: P(R1^c \cap R2^c \cap R3^c \cap R4^c \cap R_5^c).

    • The probability of a single spin not being red (P(R^c)) is the probability of it being green or black: P(R^c) = P(G) + P(B) = \frac{2}{38} + \frac{18}{38} = \frac{20}{38}.

    • Due to independence, the probability is:
      P(R^c)^5 = \left(\frac{20}{38}\right)^5

  • Part C: Probability that at Least One Spin is Red

    • The event