Question

Let Ω={1,2,3,4,5}be the sample space with the events A={1,2,5},B={1,3,5} and c={2,3,5}.Let E^c denote the complement of an event E.Then P((A∩B)^c∪C^c) is

• A. \(\frac{1}{5}\)
• B. \(\frac{3}{5}\)
• C. \(\frac{2}{5}\)
• D. \(\frac{4}{5}\)
• E. 1

Answer 1

The Correct Option is D

Given:

• Sample space \( \Omega = \{1, 2, 3, 4, 5\} \)
• Events:
  • \( A = \{1, 2, 5\} \)
  • \( B = \{1, 3, 5\} \)
  • \( C = \{2, 3, 5\} \)

 

We need to find \( P((A \cap B)^c \cup C^c) \).

Step 1: Find \( A \cap B \): \[ A \cap B = \{1, 5\} \]

Step 2: Find its complement \( (A \cap B)^c \): \[ (A \cap B)^c = \Omega \setminus \{1, 5\} = \{2, 3, 4\} \]

Step 3: Find \( C^c \): \[ C^c = \Omega \setminus \{2, 3, 5\} = \{1, 4\} \]

Step 4: Compute the union \( (A \cap B)^c \cup C^c \): \[ \{2, 3, 4\} \cup \{1, 4\} = \{1, 2, 3, 4\} \]

Step 5: Calculate the probability: \[ P(\{1, 2, 3, 4\}) = \frac{4}{5} \]

The correct answer is (D) \( \frac{4}{5} \).

Answer 2

Given the sample space \( \Omega = \{1, 2, 3, 4, 5\} \), and the events:

\[ A = \{1, 2, 5\}, \quad B = \{1, 3, 5\}, \quad C = \{2, 3, 5\} \]

We want to find \( P((A \cap B)^c \cup C^c) \).

First, let's find \( A \cap B \):

\[ A \cap B = \{1, 5\} \]

Then, find the complement of \( (A \cap B) \):

\[ (A \cap B)^c = \Omega - (A \cap B) = \{2, 3, 4\} \]

Next, find the complement of \( C \):

\[ C^c = \Omega - C = \{1, 4\} \]

Now, let's find the union of \( (A \cap B)^c \) and \( C^c \):

\[ (A \cap B)^c \cup C^c = \{1, 2, 3, 4\} \]

Assuming each outcome in \( \Omega \) is equally likely, the probability of each outcome is \( \frac{1}{5} \). Therefore:

\[ P((A \cap B)^c \cup C^c) = \frac{\text{number of elements in } (A \cap B)^c \cup C^c}{\text{number of elements in } \Omega} = \frac{4}{5} \]