Let A₁, A₂, A₃ be events satisfying 0 < P(A_i) < 1 for i = 1, 2, 3. Which of the following statements is true? 

Concept:

Conditional Probability:

For two events A and B, the conditional probability of A occurring given that B has occurred is denoted as

\(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\), provided that P(B) > 0.

$P(A_3 | A_2)$ Explanation:

Option 1:

Counter example: 

Let, \( P(A_1) = 0.8\)

 \(P(A_2) = 0.4\) ,  \(P(A_3) = 0.3\) 

\(P(A_1 \cap A_2) = 0.25\) ,  \(P(A_2 \cap A_3) = 0.2\) and \(P(A_1 \cap A_3) = 0.05\) 

\( P(A_1 | A_2) = \frac{0.25}{0.4} = 0.625 \) , \(P(A_2 | A_3) = \frac{0.2}{0.3} = 0.6667\) and
 
\(P(A_1 | A_3) = \frac{0.05}{0.3} = 0.1667\)

Left-hand side:  \(0.625 \times 0.6667 = 0.4167\) 

Right-hand side:  0.1667 

Here, the left-hand side 0.4167 is greater than the right-hand side 0.1667, which violates the inequality. 

Thus, this provides a counter example where

\(P(A_1 | A_2) P(A_2 | A_3) > P(A_1 | A_3)\)

So, option 1) is false.

Option 2:

Let us define three events \(A_1, A_2, A_3\) in a probability space with the following probabilities:

 \(P(A_1) = 0.5 \), \(P(A_2) = 0.6\)

 \(P(A_3) = 0.4\)

\(P(A_1 \cap A_2) = 0.3\) , \(P(A_3 \cap A_2) = 0.25 \)

\(P(A_1 \cap A_3 \cap A_2) = 0.2\)

\(P(A_1 | A_2) = \frac{P(A_1 \cap A_2)}{P(A_2)} = \frac{0.3}{0.6} = 0.5\)

\(P(A_3 | A_2) = \frac{P(A_3 \cap A_2)}{P(A_2)} = \frac{0.25}{0.6} \approx 0.4167\)

\(P(A_1 \cap A_3 | A_2) = \frac{P(A_1 \cap A_3 \cap A_2)}{P(A_2)} = \frac{0.2}{0.6} \approx 0.3333 \)

Left-hand side: \(P(A_1 | A_2) P(A_3 | A_2) = 0.5 \times 0.4167 = 0.2083\)

Right-hand side: \( P(A_1 \cap A_3 | A_2) = 0.3333\)

Hence, \(P(A_1 | A_2) P(A_3 | A_2) < P(A_1 \cap A_3 | A_2)\)

So option 2) is false.

Option 3:
   
This statement looks plausible because the probability of the union of two events is generally less than

or equal to the sum of the individual probabilities. This would make the product of the conditional probabilities

potentially greater than or equal to the union probability. Hence, this option is true.

Option 4:
 

\(P(A_1) = 0.6\)

\(P(A_2) = 0.5\)

\(P(A_3) = 0.4\)

\(P(A_1 \cap A_2) = 0.3\)

 \(P(A_2 \cap A_3) = 0.2\)

\(P(A_1 \cap A_3) = 0.1\)

\(P(A_1 | A_2) = \frac{P(A_1 \cap A_2)}{P(A_2)} = \frac{0.3}{0.5}= 0.6\) 

 \(P(A_2 | A_3) = \frac{P(A_2 \cap A_3)}{P(A_3)} = \frac{0.2}{0.4} = 0.5\)

\(P(A_1 | A_3) = \frac{P(A_1 \cap A_3)}{P(A_3)} = \frac{0.1}{0.4} = 0.25\)

Left-hand side: \(P(A_1 | A_2) + P(A_2 | A_3) = 0.6 + 0.5 = 1.1\)

Right-hand side: \(P(A_1 | A_3) = 0.25\)

The correct answer is Option 3).