The probability of four events A, B, C and D are: P(A) = 0.1, P(B) = 0.2, P(C) = 0.3, P(D) = 0.4. Suppose that the events A and B are independent of the events C and D and that P(A... The probability of four events A, B, C and D are: P(A) = 0.1, P(B) = 0.2, P(C) = 0.3, P(D) = 0.4. Suppose that the events A and B are independent of the events C and D and that P(A ∩ B) = 0.1, P(C ∩ D) = 0.2. Calculate the following probabilities: (a) P(A ∪ B). (b) P(B ∪ D). (c) P(Aᶜ ∪ Bᶜ). (d) P((A ∪ C)ᶜ). (e) P(A | D). (f) P(C | D). (g) P(A ∪ Cᶜ). (h) Find the expression of P(A ∪ B ∪ C ∪ D) and indicate which of terms are unknown.

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Understand the Problem

The question presents a probability problem involving four events A, B, C, and D. It provides the probabilities of individual events and the intersections of some events. The task is to calculate various probabilities related to these events, including unions, complements, and conditional probabilities. Additionally, the question asks for the expression of the probability of the union of all four events and identification of unknown terms.

Answer

(a) $0.2$ (b) $0.52$ (c) $0.9$ (d) $0.63$ (e) $0.1$ (f) $0.5$ (g) $0.73$ (h) $0.6$
Answer for screen readers

(a) $P(A \cup B) = 0.2$ (b) $P(B \cup D) = 0.52$ (c) $P(A^c \cup B^c) = 0.9$ (d) $P((A \cup C)^c) = 0.63$ (e) $P(A | D) = 0.1$ (f) $P(C | D) = 0.5$ (g) $P(A \cup C^c) = 0.73$ (h) $P(A \cup B \cup C \cup D) = P(A) + P(B) + P(C) + P(D) - P(A \cap B) - P(A \cap C) - P(A \cap D) - P(B \cap C) - P(B \cap D) - P(C \cap D) + P(A \cap B \cap C) + P(A \cap B \cap D) + P(A \cap C \cap D) + P(B \cap C \cap D) - P(A \cap B \cap C \cap D) = 0.6$

Steps to Solve

  1. Calculate $P(A \cup B)$ Apply the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. $P(A \cup B) = 0.1 + 0.2 - 0.1 = 0.2$

  2. Calculate $P(B \cup D)$ Apply the formula $P(B \cup D) = P(B) + P(D) - P(B \cap D)$. Since A and B are independent of C and D, B and D are independent, which means $P(B \cap D) = P(B)P(D)$. $P(B \cap D) = 0.2 \times 0.4 = 0.08$ $P(B \cup D) = 0.2 + 0.4 - 0.08 = 0.52$

  3. Calculate $P(A^c \cup B^c)$ Apply De Morgan's Law: $P(A^c \cup B^c) = P((A \cap B)^c) = 1 - P(A \cap B)$. $P(A^c \cup B^c) = 1 - 0.1 = 0.9$

  4. Calculate $P((A \cup C)^c)$ $P((A \cup C)^c) = 1 - P(A \cup C)$ $P(A \cup C) = P(A) + P(C) - P(A \cap C)$. Since A and C are independent, $P(A \cap C) = P(A)P(C)$. $P(A \cap C) = (0.1)(0.3) = 0.03$ $P(A \cup C) = 0.1 + 0.3 - 0.03 = 0.37$ $P((A \cup C)^c) = 1 - 0.37 = 0.63$

  5. Calculate $P(A|D)$ $P(A|D) = \frac{P(A \cap D)}{P(D)}$. Since A and D are independent, $P(A \cap D) = P(A)P(D)$. $P(A \cap D) = (0.1)(0.4) = 0.04$ $P(A|D) = \frac{0.04}{0.4} = 0.1$

  6. Calculate $P(C|D)$ $P(C|D) = \frac{P(C \cap D)}{P(D)}$ $P(C|D) = \frac{0.2}{0.4} = 0.5$

  7. Calculate $P(A \cup C^c)$ $P(A \cup C^c) = P(A) + P(C^c) - P(A \cap C^c)$. $P(C^c) = 1 - P(C) = 1 - 0.3 = 0.7$. $P(A \cap C^c) = P(A)P(C^c)$ since A and $C^c$ are independent $P(A \cap C^c) = (0.1)(0.7) = 0.07$ $P(A \cup C^c) = 0.1 + 0.7 - 0.07 = 0.73$

  8. Find the expression for $P(A \cup B \cup C \cup D)$ Using the Principle of Inclusion-Exclusion: $P(A \cup B \cup C \cup D) = P(A) + P(B) + P(C) + P(D) - P(A \cap B) - P(A \cap C) - P(A \cap D) - P(B \cap C) - P(B \cap D) - P(C \cap D) + P(A \cap B \cap C) + P(A \cap B \cap D) + P(A \cap C \cap D) + P(B \cap C \cap D) - P(A \cap B \cap C \cap D)$

    Substitute known values: $P(A \cup B \cup C \cup D) = 0.1 + 0.2 + 0.3 + 0.4 - 0.1 - (0.1)(0.3) - (0.1)(0.4) - (0.2)(0.3) - (0.2)(0.4) - 0.2 + P(A \cap B \cap C) + P(A \cap B \cap D) + P(A \cap C \cap D) + P(B \cap C \cap D) - P(A \cap B \cap C \cap D)$

    $P(A \cup B \cup C \cup D) = 1 - 0.1 - 0.03 - 0.04 - 0.06 - 0.08 - 0.2 + P(A \cap B \cap C) + P(A \cap B \cap D) + P(A \cap C \cap D) + P(B \cap C \cap D) - P(A \cap B \cap C \cap D)$

    $P(A \cup B \cup C \cup D) = 0.49 + P(A \cap B \cap C) + P(A \cap B \cap D) + P(A \cap C \cap D) + P(B \cap C \cap D) - P(A \cap B \cap C \cap D)$

    Since A and B are independent of C and D: $P(A \cap B \cap C) = P(A \cap B)P(C) = (0.1)(0.3) = 0.03$ $P(A \cap B \cap D) = P(A \cap B)P(D) = (0.1)(0.4) = 0.04$

    Since C and D are independent of A and B: $P(A \cap C \cap D) = P(A)P(C \cap D) = (0.1)(0.2) = 0.02$ $P(B \cap C \cap D) = P(B)P(C \cap D) = (0.2)(0.2) = 0.04$

    $P(A \cap B \cap C \cap D) = P(A \cap B)P(C \cap D) = (0.1)(0.2) = 0.02$

    $P(A \cup B \cup C \cup D) = 0.49 + 0.03 + 0.04 + 0.02 + 0.04 - 0.02 = 0.6$

(a) $P(A \cup B) = 0.2$ (b) $P(B \cup D) = 0.52$ (c) $P(A^c \cup B^c) = 0.9$ (d) $P((A \cup C)^c) = 0.63$ (e) $P(A | D) = 0.1$ (f) $P(C | D) = 0.5$ (g) $P(A \cup C^c) = 0.73$ (h) $P(A \cup B \cup C \cup D) = P(A) + P(B) + P(C) + P(D) - P(A \cap B) - P(A \cap C) - P(A \cap D) - P(B \cap C) - P(B \cap D) - P(C \cap D) + P(A \cap B \cap C) + P(A \cap B \cap D) + P(A \cap C \cap D) + P(B \cap C \cap D) - P(A \cap B \cap C \cap D) = 0.6$

More Information

The principle of inclusion-exclusion allows us to calculate the probability of the union of multiple events. Independence simplifies the calculation of intersections.

Tips

A common mistake is forgetting to subtract the intersection when calculating the probability of the union of two events. Another common mistake involves incorrectly assuming independence when it is not explicitly stated or derivable from the given information. Also, De Morgan's laws are often used incorrectly.

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