Question:
Probability of obtaining an even prime number on each die when a pair of dice is rolled is
Probability of obtaining an even prime number on each die when a pair of dice is rolled is
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"Even prime number" is a classic trick phrase in probability questions. Always remember that 2 is the only even prime number. 1 is neither prime nor composite.
Updated On: Apr 29, 2026
- 0
- $\frac{1}{6}$
- $\frac{1}{12}$
- $\frac{1}{36}$
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
This problem asks for the probability of a specific outcome when rolling two dice. The rolls of two dice are independent events, meaning the outcome of one die does not affect the outcome of the other.
Step 2: Key Formula or Approach:
1. Identify the set of "even prime numbers". 2. Find the probability of getting an even prime number on a single roll of a die. 3. Since the two rolls are independent, use the multiplication rule for independent events: $P(A \text{ and } B) = P(A) \times P(B)$.
Step 3: Detailed Explanation:
When a standard fair six-sided die is rolled, the sample space is $S = \{1, 2, 3, 4, 5, 6\}$. Let's analyze the properties of these numbers: - Prime numbers are numbers greater than 1 that have only two divisors: 1 and themselves. The prime numbers in our sample space are $\{2, 3, 5\}$. - Even numbers are divisible by 2. The even numbers in our sample space are $\{2, 4, 6\}$. The only number that is both even AND prime is $2$. So, "obtaining an even prime number" is the same as "rolling a 2". Let $E_1$ be the event of rolling a 2 on the first die. Number of favorable outcomes = 1 (just the number 2). Total number of possible outcomes = 6. \[ P(E_1) = \frac{1}{6} \] Let $E_2$ be the event of rolling a 2 on the second die. Similarly, $P(E_2) = \frac{1}{6}$. We want the probability of getting an even prime number on each die, which means $E_1$ occurs AND $E_2$ occurs. Because the dice rolls are independent: \[ P(E_1 \text{ and } E_2) = P(E_1) \times P(E_2) \] \[ P(\text{even prime on both}) = \left(\frac{1}{6}\right) \times \left(\frac{1}{6}\right) = \frac{1}{36} \]
Step 4: Final Answer:
The required probability is $\frac{1}{36}$.
This problem asks for the probability of a specific outcome when rolling two dice. The rolls of two dice are independent events, meaning the outcome of one die does not affect the outcome of the other.
Step 2: Key Formula or Approach:
1. Identify the set of "even prime numbers". 2. Find the probability of getting an even prime number on a single roll of a die. 3. Since the two rolls are independent, use the multiplication rule for independent events: $P(A \text{ and } B) = P(A) \times P(B)$.
Step 3: Detailed Explanation:
When a standard fair six-sided die is rolled, the sample space is $S = \{1, 2, 3, 4, 5, 6\}$. Let's analyze the properties of these numbers: - Prime numbers are numbers greater than 1 that have only two divisors: 1 and themselves. The prime numbers in our sample space are $\{2, 3, 5\}$. - Even numbers are divisible by 2. The even numbers in our sample space are $\{2, 4, 6\}$. The only number that is both even AND prime is $2$. So, "obtaining an even prime number" is the same as "rolling a 2". Let $E_1$ be the event of rolling a 2 on the first die. Number of favorable outcomes = 1 (just the number 2). Total number of possible outcomes = 6. \[ P(E_1) = \frac{1}{6} \] Let $E_2$ be the event of rolling a 2 on the second die. Similarly, $P(E_2) = \frac{1}{6}$. We want the probability of getting an even prime number on each die, which means $E_1$ occurs AND $E_2$ occurs. Because the dice rolls are independent: \[ P(E_1 \text{ and } E_2) = P(E_1) \times P(E_2) \] \[ P(\text{even prime on both}) = \left(\frac{1}{6}\right) \times \left(\frac{1}{6}\right) = \frac{1}{36} \]
Step 4: Final Answer:
The required probability is $\frac{1}{36}$.
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