Question

Let \(E_1\) and \(E_2\) be two events such that the conditional probabilities \(P(E_1|E_2)=\frac 12\), \(P(E_2|E_1)=\frac 34\) and \(P(E_1∩E_2)=\frac 18\)⋅ Then:

• A. \(P(E_1∩E_2)=P(E_1)⋅P(E_2)\)
• B. \(P(E'_1∩E'_2)=P(E'_1)⋅P(E'_2)\)
• C. \(P(E_1∩E'_2)=P(E_1)⋅P(E_2)\)
• D. \(P(E'_1∩E_2)=P(E_1)⋅P(E_2)\)

Answer 1

The Correct Option is C

 We are given the following probabilities for events \(E_1\) and \(E_2\):

• \(P(E_1|E_2) = \frac{1}{2}\)
• \(P(E_2|E_1) = \frac{3}{4}\)
• \(P(E_1 \cap E_2) = \frac{1}{8}\)

We need to determine which of the given options is correct. Let's use the concept of conditional probabilities:

1. Calculating \( P(E_1 \cap E_2) \):

We already know \(P(E_1 \cap E_2) = \frac{1}{8}\). It is given.

1. Using the Conditional Probability Formula:

\(P(E_1|E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)}\) 
Substituting the known values, we have: 
\(\frac{1}{2} = \frac{\frac{1}{8}}{P(E_2)}\) 
Solving for \(P(E_2)\), we get: 
\(P(E_2) = \frac{1}{4}\)

1. Using the Conditional Probability Formula (again):

\(P(E_2|E_1) = \frac{P(E_1 \cap E_2)}{P(E_1)}\) 
Substituting the known values, we have: 
\(\frac{3}{4} = \frac{\frac{1}{8}}{P(E_1)}\) 
Solving for \(P(E_1)\), we get: 
\(P(E_1) = \frac{1}{6}\)

1. Finding \( P(E_1 \cap E'_2) \):

We need to check which statement about independence holds for the complements of the events. 
We know \(P(E_1 \cap E_2) = \frac{1}{8}\). 
Now consider: 
\(P(E_1 \cap E'_2) = P(E_1) - P(E_1 \cap E_2)\) 
\(= \frac{1}{6} - \frac{1}{8}\) 
\(= \frac{4}{24} - \frac{3}{24} = \frac{1}{24}\)

1. Verification of the Correct Option:

Compute the product of the probabilities: 
\(P(E_1) \cdot P(E_2) = \frac{1}{6} \times \frac{1}{4} = \frac{1}{24}\) 
As \(P(E_1 \cap E'_2) = P(E_1) \cdot P(E_2)\), the correct option is: 
 

\(P(E_1 \cap E'_2) = P(E_1) \cdot P(E_2)\)

 

Therefore, the correct answer is:

\(P(E_1 \cap E'_2) = P(E_1) \cdot P(E_2)\)