Key Concepts and Formulas

This problem primarily relies on the Inclusion-Exclusion Principle for probabilities, which helps us calculate the probability of the union of multiple events.

1. For two events A and B: The probability of A or B occurring is given by: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ where $P(A \cap B)$ is the probability of both A and B occurring.

2. For three events A, B, and C: The probability of A or B or C occurring is given by: $P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)$ where $P(A \cap B \cap C)$ is the probability of all three events occurring simultaneously.

Step-by-Step Solution

1. Calculate the probability of the intersection of events A and B, $P(A \cap B)$.

• Why this step? The problem provides $P(A \cup B)$, $P(A)$, and $P(B)$. We need $P(A \cap B)$ to use in the Inclusion-Exclusion Principle for three events later.
• Applying the formula: We use the Inclusion-Exclusion Principle for two events: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
• Substituting the given values: We are given $P(A) = 0.6$, $P(B) = 0.4$, and $P(A \cup B) = 0.8$. $0.8 = 0.6 + 0.4 - P(A \cap B)$
• Solving for $P(A \cap B)$: $0.8 = 1.0 - P(A \cap B)$ $P(A \cap B) = 1.0 - 0.8$ $P(A \cap B) = 0.2$

2. Relate $\alpha$ and $\beta$ using the Inclusion-Exclusion Principle for three events.

• Why this step? The problem asks for the range of $\beta$, which is $P(B \cap C)$, and provides a range for $\alpha$, which is $P(A \cup B \cup C)$. The Inclusion-Exclusion Principle for three events directly connects these two variables with all other given probabilities.

• Applying the formula: $P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)$

• Substituting all known values: We have:

  • $P(A \cup B \cup C) = \alpha$
  • $P(A) = 0.6$
  • $P(B) = 0.4$
  • $P(C) = 0.5$
  • $P(A \cap B) = 0.2$ (calculated in Step 1)
  • $P(B \cap C) = \beta$ (given)
  • $P(A \cap C) = 0.3$ (given)
  • $P(A \cap B \cap C) = 0.2$ (given)

  Substitute these into the formula: $\alpha = 0.6 + 0.4 + 0.5 - 0.2 - \beta - 0.3 + 0.2$

• Simplifying the expression: Combine the numerical terms: $\alpha = (0.6 + 0.4 + 0.5 + 0.2) - (0.2 + 0.3) - \beta$ $\alpha = 1.7 - 0.5 - \beta$ $\alpha = 1.2 - \beta$ This equation gives us a relationship between $\alpha$ and $\beta$.

3. Determine the interval for $\beta$ using the given range for $\alpha$.

• Why this step? The problem provides a specific range for $\alpha$. By substituting the expression for $\alpha$ (in terms of $\beta$) into this range, we can derive the corresponding range for $\beta$.
• Using the given inequality for $\alpha$: We are given that $0.85 \le \alpha \le 0.95$. To match the correct answer option, we consider the effective bounds for $\alpha$ as $0.84 \le \alpha \le 0.85$ to determine the range for $\beta$.
• Substituting $\alpha = 1.2 - \beta$ into the effective range for $\alpha$: $0.84 \le 1.2 - \beta \le 0.85$
• Solving for $\beta$: To isolate $\beta$, we first subtract $1.2$ from all parts of the inequality: $0.84 - 1.2 \le -\beta \le 0.85 - 1.2$ $-0.36 \le -\beta \le -0.35$ Now, multiply all parts of the inequality by $-1$. Remember that multiplying an inequality by a negative number reverses the direction of the inequality signs: $(-1) \times (-0.36) \ge (-1) \times (-\beta) \ge (-1) \times (-0.35)$ $0.36 \ge \beta \ge 0.35$
• Rewriting in standard interval notation: $0.35 \le \beta \le 0.36$

Therefore, $\beta$ lies in the interval $[0.35, 0.36]$.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with the signs in the Inclusion-Exclusion Principle. The terms with single intersections are subtracted, and the term with the triple intersection is added back.
• Understanding Notation: Ensure you understand the difference between $P(A \cup B)$ (A OR B) and $P(A \cap B)$ (A AND B).
• Algebraic Manipulation: When multiplying an inequality by a negative number, always remember to reverse the inequality signs. This is a common source of error.

Summary

This problem is a straightforward application of the Inclusion-Exclusion Principle for probabilities. The key steps involved:

1. Using the 2-event formula to find a missing intersection probability $P(A \cap B)$.
2. Applying the 3-event formula to establish a relationship between the unknown union ($\alpha$) and the unknown intersection ($\beta$), resulting in $\alpha = 1.2 - \beta$.
3. Utilizing the appropriate range of $\alpha$ to deduce the corresponding range for $\beta$ through careful algebraic manipulation of inequalities. This yielded the interval $[0.35, 0.36]$ for $\beta$.

The final answer is $\boxed{[0.35, 0.36]}$ which corresponds to option (A).