Question:

The sample space for a single throw of a pair of unbiased normal dice consists of $6 \times 6 = 36$ equally likely outcomes. Step 1: Compute $P(E_1)$. The event $E_1$ is getting a prime numbered face on each die. The prime numbers on a standard die are $\{2, 3, 5\}$. There are 3 prime numbers. The number of favorable outcomes for $E_1$ is $3 \times 3 = 9$. $$P(E_1) = \frac{9}{36} = \frac{1}{4}$$ Step 2: Compute $P(E_2)$. The event $E_2$ is getting the same number on each die. The favorable outcomes are $\{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\}$. There are 6 such outcomes. $$P(E_2) = \frac{6}{36} = \frac{1}{6}$$ Step 3: Compute $P(E_3)$. The event $E_3$ is getting a total of 4 on the two dice. The favorable outcomes are $\{(1,3), (2,2), (3,1)\}$. There are 3 such outcomes. $$P(E_3) = \frac{3}{36} = \frac{1}{12}$$ Step 4: Evaluate the proposition that $P(E_1), P(E_2), P(E_3)$ are in A.P. For three probabilities to be in an arithmetic progression (A.P.), the condition $2P(E_2) = P(E_1) + P(E_3)$ must be satisfied. $$2P(E_2) = 2 \times \frac{1}{6} = \frac{1}{3}$$ $$P(E_1) + P(E_3) = \frac{1}{4} + \frac{1}{12} = \frac{3}{12} + \frac{1}{12} = \frac{4}{12} = \frac{1}{3}$$ Since $2P(E_2) = P(E_1) + P(E_3)$, the probabilities $P(E_1), P(E_2), P(E_3)$ are in A.P. This proposition is true. Step 5: Evaluate the proposition that the events $E_1$ and $E_2$ are independent. For two events to be independent, $P(E_1 \cap E_2) = P(E_1)P(E_2)$ must hold. The intersection $E_1 \cap E_2$ consists of outcomes where both dice show a prime number and the numbers are the same. These outcomes are $\{(2,2), (3,3), (5,5)\}$. There are 3 such outcomes. $$P(E_1 \cap E_2) = \frac{3}{36} = \frac{1}{12}$$ The product of the individual probabilities is: $$P(E_1)P(E_2) = \frac{1}{4} \times \frac{1}{6} = \frac{1}{24}$$ Since $P(E_1 \cap E_2) \neq P(E_1)P(E_2)$ (i.e., $\frac{1}{12} \neq \frac{1}{24}$), the events $E_1$ and $E_2$ are not independent. This proposition is false. Step 6: Evaluate the proposition that $P(E_3|E_1) = \frac{2}{9}$. The conditional probability $P(E_3|E_1)$ is given by $P(E_3|E_1) = \frac{P(E_3 \cap E_1)}{P(E_1)}$. The intersection $E_3 \cap E_1$ consists of outcomes where the total is 4 AND both faces are prime. The outcomes for $E_3$ are $\{(1,3), (2,2), (3,1)\}$. From these, only $(2,2)$ has both faces as prime numbers (1 is not a prime number). So, $E_3 \cap E_1 = \{(2,2)\}$. There is 1 such outcome. $$P(E_3 \cap E_1) = \frac{1}{36}$$ Using the conditional probability formula: $$P(E_3|E_1) = \frac{1/36}{9/36} = \frac{1}{9}$$ Since $\frac{1}{9} \neq \frac{2}{9}$, this proposition is false. Step 7: Evaluate the proposition that $P(E_1 \cup E_2) + P(E_2 \setminus E_3) = \frac{17}{36}$. First, calculate $P(E_1 \cup E_2)$: $$P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2)$$ Using the probabilities calculated in previous steps: $$P(E_1 \cup E_2) = \frac{1}{4} + \frac{1}{6} - \frac{1}{12} = \frac{3}{12} + \frac{2}{12} - \frac{1}{12} = \frac{4}{12} = \frac{1}{3}$$ Next, calculate $P(E_2 \setminus E_3)$: The event $E_2 \setminus E_3$ represents outcomes in $E_2$ but not in $E_3$. This can be calculated as $P(E_2) - P(E_2 \cap E_3)$. The intersection $E_2 \cap E_3$ consists of outcomes where the numbers are the same AND the total is 4. This is only the outcome $(2,2)$. $$P(E_2 \cap E_3) = \frac{1}{36}$$ $$P(E_2 \setminus E_3) = P(E_2) - P(E_2 \cap E_3) = \frac{6}{36} - \frac{1}{36} = \frac{5}{36}$$ Finally, sum these probabilities: $$P(E_1 \cup E_2) + P(E_2 \setminus E_3) = \frac{1}{3} + \frac{5}{36} = \frac{12}{36} + \frac{5}{36} = \frac{17}{36}$$ This proposition is true.