15.2 Independent Events and Repeated Bernoulli Trials

15.2 Independent Events and Repeated Bernoulli Trials

1. Independence of Events

In probability theory, two events are independent if the occurrence of one event does not affect the probability of occurrence of the other.

1.1 Definition of Independent Events

Two events $A$ and $B$ defined on a sample space $\Omega$ are said to be statistically independent (or simply independent) if and only if: $P(AB) = P(A) P(B)$

If $P(A) > 0$, this definition is logically equivalent to: $P(B|A) = P(B)$ Similarly, if $P(B) > 0$, it is equivalent to $P(A|B) = P(A)$.

1.2 Difference Between Independent and Mutually Exclusive Events

It is crucial in Gaokao examinations to distinguish between independence and mutual exclusivity:

• Mutually Exclusive ($A \cap B = \emptyset$): $A$ and $B$ cannot occur together ($P(AB) = 0$). If $P(A) > 0$ and $P(B) > 0$, mutually exclusive events cannot be independent, because $P(AB) = 0 \neq P(A)P(B)$.
• Independent ($P(AB) = P(A)P(B)$): The occurrence of $A$ provides zero information about whether $B$ occurs.

1.3 Properties of Independent Events

If events $A$ and $B$ are independent, then each of the following pairs is also independent:

1. $A$ and $\overline{B}$
2. $\overline{A}$ and $B$
3. $\overline{A}$ and $\overline{B}$

Proof for $A$ and $\overline{B}$: $P(A\overline{B}) = P(A - AB) = P(A) - P(AB) = P(A) - P(A)P(B) = P(A)(1 - P(B)) = P(A)P(\overline{B})$

1.4 Mutual Independence of $n$ Events

Events $A_1, A_2, \dots, A_n$ are mutually independent if for every subset of $k$ events ($2 \le k \le n$), the probability of their intersection equals the product of their individual probabilities: $P(A_{i_1} A_{i_2} \dots A_{i_k}) = P(A_{i_1}) P(A_{i_2}) \dots P(A_{i_k})$ Note: Pairwise independence ($P(A_i A_j) = P(A_i)P(A_j)$ for all $i \neq j$) does not imply mutual independence for $n \ge 3$.

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2. System Reliability Applications (Series and Parallel Systems)

In engineering and Gaokao probability problems, independence is widely used to evaluate system reliability.

2.1 Series System

A series system operates if and only if all component units operate independently. If components $C_1, C_2, \dots, C_n$ have reliability probabilities $p_1, p_2, \dots, p_n$: $P(\text{Series System Works}) = \prod_{i=1}^n p_i = p_1 \cdot p_2 \cdots p_n$

2.2 Parallel System

A parallel system operates if at least one component unit operates independently. $P(\text{Parallel System Works}) = 1 - P(\text{All Components Fail}) = 1 - \prod_{i=1}^n (1 - p_i)$

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3. Repeated Bernoulli Trials

A Bernoulli trial is a random experiment with exactly two possible outcomes: "Success" ($S$) with probability $p$, and "Failure" ($F$) with probability $q = 1 - p$, where $0 < p < 1$.

3.1 Definition of $n$-fold Bernoulli Trials

An experiment consisting of $n$ identical and independent repetitions of a Bernoulli trial is called an $n$-fold repeated Bernoulli trial (or $n$ independent Bernoulli trials). Key features:

1. The experiment is repeated $n$ times under identical conditions.
2. Each trial has only two mutually exclusive outcomes ($S$ and $F$).
3. The probability $p = P(S)$ remains constant across all $n$ trials.
4. The outcomes of different trials are mutually independent.

3.2 Binomial Probability Formula

Let $X$ denote the total number of successes in $n$ independent Bernoulli trials. The probability of obtaining exactly $k$ successes ($k = 0, 1, 2, \dots, n$) is given by the Binomial formula: $P(X = k) = C_n^k p^k (1-p)^{n-k} = C_n^k p^k q^{n-k}$

We write $X \sim B(n, p)$.

Derivation: Any specific sequence of $k$ successes and $n-k$ failures (e.g., $S S \dots S F F \dots F$) has probability $p^k (1-p)^{n-k}$ due to independence. The number of distinct sequences containing exactly $k$ successes out of $n$ trials is given by the combination $C_n^k$. Summing these mutually exclusive sequences yields $C_n^k p^k (1-p)^{n-k}$.

3.3 Most Probable Number of Successes

The value $k_0 \in {0, 1, \dots, n}$ that maximizes $P(X = k)$ satisfies the inequality: $(n+1)p - 1 \le k_0 \le (n+1)p$

• If $(n+1)p$ is not an integer, $k_0 = \lfloor (n+1)p \rfloor$ is unique.
• If $(n+1)p$ is an integer, there are two most probable values: $k_0 = (n+1)p - 1$ and $k_0 = (n+1)p$.

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4. Worked Gaokao Exam Examples

Example 1: Best-of-Five Tournament Analysis

Problem: Two table tennis players, Player A and Player B, compete in a best-of-five match (the first player to win 3 games wins the match). The games are independent, and in each game, the probability that Player A wins is $p = \frac{2}{3}$.

1. Find the probability that Player A wins the match in exactly 4 games.
2. Find the overall probability that Player A wins the match.

Detailed Solution:

1. Winning in exactly 4 games: For Player A to win in exactly 4 games, Player A must win exactly 2 of the first 3 games, and then win the 4th game. The probability of winning 2 out of the first 3 games is given by the Binomial term $C_3^2 p^2 (1-p)^{3-2}$: $P(\text{A wins 2 of first 3}) = C_3^2 \left(\frac{2}{3}\right)^2 \left(\frac{1}{3}\right)^1 = 3 \times \frac{4}{9} \times \frac{1}{3} = \frac{4}{9}$ The 4th game is won by Player A with probability $p = \frac{2}{3}$. By independence of trials: $P(\text{Match ends at game 4 with A winning}) = \frac{4}{9} \times \frac{2}{3} = \frac{8}{27}$

2. Overall Probability of Player A winning the match: Player A can win the match in 3, 4, or 5 games. These scenarios are mutually exclusive:

   • Win in 3 games: A wins 3 straight games. $P_3 = p^3 = \left(\frac{2}{3}\right)^3 = \frac{8}{27}$
   • Win in 4 games: $P_4 = \frac{8}{27}$ (from part 1).
   • Win in 5 games: A wins 2 of the first 4 games, and then wins the 5th game. $P_5 = \left[ C_4^2 p^2 (1-p)^2 \right] \times p = \left[ 6 \times \left(\frac{2}{3}\right)^2 \left(\frac{1}{3}\right)^2 \right] \times \frac{2}{3} = \left[ 6 \times \frac{4}{9} \times \frac{1}{9} \right] \times \frac{2}{3} = \frac{24}{81} \times \frac{2}{3} = \frac{16}{81}$ Total probability that Player A wins: $P(A \text{ wins match}) = P_3 + P_4 + P_5 = \frac{8}{27} + \frac{8}{27} + \frac{16}{81} = \frac{24}{81} + \frac{24}{81} + \frac{16}{81} = \frac{64}{81}$

Example 2: Target Shooting System Reliability

Problem: Three shooters $S_1, S_2, S_3$ independently fire one shot at a target. Their hit probabilities are $p_1 = 0.8$, $p_2 = 0.7$, and $p_3 = 0.6$ respectively.

1. Find the probability that the target is hit at least once.
2. Find the probability that exactly two shooters hit the target.

Detailed Solution:

1. Target hit at least once: The target is hit at least once if not all shooters miss. The complement event is "all three shooters miss". $P(\overline{S_1}) = 1 - 0.8 = 0.2$, $P(\overline{S_2}) = 1 - 0.7 = 0.3$, $P(\overline{S_3}) = 1 - 0.6 = 0.4$. By independence: $P(\text{All miss}) = P(\overline{S_1}) P(\overline{S_2}) P(\overline{S_3}) = 0.2 \times 0.3 \times 0.4 = 0.024$ Thus, the probability of at least one hit is: $P(\text{At least 1 hit}) = 1 - P(\text{All miss}) = 1 - 0.024 = 0.976$

2. Exactly two shooters hit: "Exactly two hit" consists of three mutually exclusive events:

   • $E_1$: $S_1, S_2$ hit, $S_3$ misses $\implies P(E_1) = 0.8 \times 0.7 \times 0.4 = 0.224$
   • $E_2$: $S_1, S_3$ hit, $S_2$ misses $\implies P(E_2) = 0.8 \times 0.3 \times 0.6 = 0.144$
   • $E_3$: $S_2, S_3$ hit, $S_1$ misses $\implies P(E_3) = 0.2 \times 0.7 \times 0.6 = 0.084$ Summing these probabilities: $P(\text{Exactly 2 hit}) = 0.224 + 0.144 + 0.084 = 0.452$