Consider on experiment of a single throw of a pair of unbiased normal dice. Let three events `epsilon_(1), epsilon_(2) and epsilon_(3)` are defined as follows `epsilon_(1)` : getting prime numbered face on each dice
`epsilon_(2):` getting the same number on each dice
`epsilon_(3):` getting the sum of 4 on two dice which of the following is not true?

Options

A. The probabilities `P(epsilon_(1)), P(epsilon_(2)), P(epsilon_(3))` are arithmetic progression.
B. The events `epsilong_(1) and epsilon_(2)` are dependent
C. `P((epsilon_(3))/(epsilon_(1)))=(2)/(9)`
D. `P((epsilon_(3))/(epsilon_(1)))=(1)/(9)`

Answer

To solve the problem, we need to analyze the three events defined in the context of rolling two unbiased dice. Let's break down each event and calculate the probabilities.

### Step 1: Define the Sample Space
When rolling two unbiased dice, the total number of outcomes is \(6 \times 6 = 36\).

### Step 2: Analyze Event \( \epsilon_1 \)
**Event \( \epsilon_1 \)**: Getting a prime numbered face on each die.

The prime numbers on a die are 2, 3, and 5. The possible outcomes for this event are:
- (2, 2)
- (2, 3)
- (2, 5)
- (3, 2)
- (3, 3)
- (3, 5)
- (5, 2)
- (5, 3)
- (5, 5)

Counting these outcomes, we find there are 9 favorable outcomes.

**Probability of \( \epsilon_1 \)**:
\[ P(\epsilon_1) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{9}{36} = \frac{1}{4} \]

### Step 3: Analyze Event \( \epsilon_2 \)
**Event \( \epsilon_2 \)**: Getting the same number on each die (doubles).

The possible outcomes are:
- (1, 1)
- (2, 2)
- (3, 3)
- (4, 4)
- (5, 5)
- (6, 6)

Counting these outcomes, we find there are 6 favorable outcomes.

**Probability of \( \epsilon_2 \)**:
\[ P(\epsilon_2) = \frac{6}{36} = \frac{1}{6} \]

### Step 4: Analyze Event \( \epsilon_3 \)
**Event \( \epsilon_3 \)**: Getting a sum of 4 on the two dice.

The possible outcomes are:
- (1, 3)
- (3, 1)
- (2, 2)

Counting these outcomes, we find there are 3 favorable outcomes.

**Probability of \( \epsilon_3 \)**:
\[ P(\epsilon_3) = \frac{3}{36} = \frac{1}{12} \]

### Step 5: Check if the Events are in Arithmetic Progression (AP)
For the probabilities to be in AP, the middle probability must equal the average of the other two:
\[ P(\epsilon_2) \text{ should equal } \frac{P(\epsilon_1) + P(\epsilon_3}}{2} \]
Calculating:
\[ \frac{P(\epsilon_1) + P(\epsilon_3)}{2} = \frac{\frac{1}{4} + \frac{1}{12}}{2} \]
Finding a common denominator (12):
\[ = \frac{\frac{3}{12} + \frac{1}{12}}{2} = \frac{\frac{4}{12}}{2} = \frac{2}{12} = \frac{1}{6} \]
Since \( P(\epsilon_2) = \frac{1}{6} \), the events are in AP.

### Step 6: Check if Events are Dependent
Events \( \epsilon_1 \) and \( \epsilon_2 \) are dependent if:
\[ P(\epsilon_1 \cap \epsilon_2) \neq P(\epsilon_1) \cdot P(\epsilon_2) \]
The intersection \( \epsilon_1 \cap \epsilon_2 \) includes outcomes (2, 2), (3, 3), and (5, 5) which are prime doubles:
- (2, 2)

So, \( P(\epsilon_1 \cap \epsilon_2) = \frac{1}{36} \).

Calculating \( P(\epsilon_1) \cdot P(\epsilon_2) \):
\[ P(\epsilon_1) \cdot P(\epsilon_2) = \frac{1}{4} \cdot \frac{1}{6} = \frac{1}{24} \]
Since \( \frac{1}{36} \neq \frac{1}{24} \), the events are dependent.

### Step 7: Calculate Conditional Probability \( P(\epsilon_3 | \epsilon_1) \)
Using the formula:
\[ P(\epsilon_3 | \epsilon_1) = \frac{P(\epsilon_3 \cap \epsilon_1)}{P(\epsilon_1)} \]
The intersection \( \epsilon_3 \cap \epsilon_1 \) includes (2, 2):
\[ P(\epsilon_3 \cap \epsilon_1) = \frac{1}{36} \]
Thus,
\[ P(\epsilon_3 | \epsilon_1) = \frac{\frac{1}{36}}{\frac{1}{4}} = \frac{1}{36} \cdot \frac{4}{1} = \frac{4}{36} = \frac{1}{9} \]

### Conclusion
- \( P(\epsilon_1) = \frac{1}{4} \)
- \( P(\epsilon_2) = \frac{1}{6} \)
- \( P(\epsilon_3) = \frac{1}{12} \)
- The events are in AP.
- The events are dependent.
- \( P(\epsilon_3 | \epsilon_1) = \frac{1}{9} \)

The statement that is **not true** is the one that claims the events are independent.