Question

If a six-sided die is rolled three times, what is the probability that the die will land on an even number exactly twice and on an odd number exactly once?

• A. \(\frac{1}{8}\)
• B. \(\frac{1}{4}\)
• C. \(\frac{3}{8}\)
• D. \(\frac{1}{2}\)
• E. \(\frac{5}{8}\)

Hint:

This is a binomial probability problem. You can use the formula: \(P(k \text{ successes in } n \text{ trials}) = C(n,k) \cdot p^k \cdot (1-p)^{n-k}\). Here, \(n=3\) trials, \(k=2\) successes (even numbers), and \(p=1/2\) is the probability of success. The number of combinations is \(C(3,2) = \frac{3!}{2!(3-2)!} = 3\). So, the probability is \(3 \times (\frac{1}{2})^2 \times (\frac{1}{2})^1 = 3 \times \frac{1}{4} \times \frac{1}{2} = \frac{3}{8}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem involves calculating the probability of a specific combination of outcomes over multiple independent events. Each roll of the die is an independent event. We need to find the probability of getting two even numbers (E) and one odd number (O) in three rolls.
Step 2: Key Formula or Approach:
First, we determine the probability of a single event (rolling an even or an odd number). Then, we identify all possible sequences of three rolls that satisfy the condition (e.g., EEO, EOE, OEE). We calculate the probability of one such sequence and then multiply it by the total number of such sequences.
The probability of a sequence of independent events is the product of their individual probabilities.
Step 3: Detailed Explanation:
A standard six-sided die has the numbers {1, 2, 3, 4, 5, 6}.
The even numbers are {2, 4, 6}, so there are 3 even outcomes.
The probability of rolling an even number is \(P(E) = \frac{\text{Number of even outcomes}}{\text{Total outcomes}} = \frac{3}{6} = \frac{1}{2}\).
The odd numbers are {1, 3, 5}, so there are 3 odd outcomes.
The probability of rolling an odd number is \(P(O) = \frac{\text{Number of odd outcomes}}{\text{Total outcomes}} = \frac{3}{6} = \frac{1}{2}\).
We need the outcome to have exactly two evens and one odd. The possible arrangements for this are:

• Even, Even, Odd (EEO)
• Even, Odd, Even (EOE)
• Odd, Even, Even (OEE)

There are 3 such arrangements. Let's calculate the probability for one of these, for example, EEO. Since the rolls are independent:
\( P(\text{EEO}) = P(E) \times P(E) \times P(O) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \)
Each of the 3 arrangements (EEO, EOE, OEE) has the same probability of \(\frac{1}{8}\).
To find the total probability of the event (getting exactly two evens and one odd), we add the probabilities of these mutually exclusive arrangements:
\( P(\text{Total}) = P(\text{EEO}) + P(\text{EOE}) + P(\text{OEE}) = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8} \) Step 4: Final Answer:
The probability that the die will land on an even number exactly twice and on an odd number exactly once is \(\frac{3}{8}\).