Question

Two dice are rolled together. The probability that the sum of the numbers obtained is divisible by 6, is :

• A. $\frac{1}{6}$
• B. $\frac{11}{36}$
• C. $\frac{1}{12}$
• D. $\frac{1}{4}$

Hint:

For questions involving two dice, you can remember the frequency of sums:
Sum of 2: 1 way
Sum of 3: 2 ways
Sum of 4: 3 ways
Sum of 5: 4 ways
Sum of 6: 5 ways
Sum of 7: 6 ways
Sum of 8: 5 ways
Sum of 9: 4 ways
Sum of 10: 3 ways
Sum of 11: 2 ways
Sum of 12: 1 way
Adding the frequencies for the multiples of 6 (Sums of 6 and 12) gives $5 + 1 = 6$ ways.
Dividing by 36 immediately gives $\frac{6}{36} = \frac{1}{6}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This question is from the topic of "Probability".
We are rolling two standard six-sided dice simultaneously.
We need to determine the probability that the sum of the numbers showing on top of both dice is divisible by 6.

Step 2: Key Formula or Approach: The classical probability of an event $E$ occurring is given by:
\[ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{n(E)}{n(S)} \] When two dice are rolled:

• The outcome of each die is an integer from the set $\{1, 2, 3, 4, 5, 6\}$.

• The total number of possible outcomes in the sample space $S$ is $6 \times 6 = 36$.

• The possible sum of numbers on the two dice ranges from a minimum of $1 + 1 = 2$ to a maximum of $6 + 6 = 12$.

Step 3: Detailed Explanation:

• Write down the total number of outcomes in the sample space $S$:
\[ n(S) = 36 \]

• The sum $S_d$ of the numbers on the two dice can be any integer such that $2 \le S_d \le 12$.

• We want the sum $S_d$ to be divisible by 6. The multiples of 6 in the range $[2, 12]$ are:
\[ \text{Possible sums} = \{6, 12\} \]

• Let us find all the individual ordered pairs $(x, y)$ that yield these sums, where $x$ is the outcome of the first die and $y$ is the outcome of the second die:

• Pairs that sum up to 6:
- $(1, 5)$ because $1 + 5 = 6$
- $(2, 4)$ because $2 + 4 = 6$
- $(3, 3)$ because $3 + 3 = 6$
- $(4, 2)$ because $4 + 2 = 6$
- $(5, 1)$ because $5 + 1 = 6$
This gives a total of $5$ favorable pairs.

• Pairs that sum up to 12:
- $(6, 6)$ because $6 + 6 = 12$
This gives a total of $1$ favorable pair.

• Sum the favorable outcomes to find the total count of favorable outcomes $n(E)$:
\[ n(E) = 5 + 1 = 6 \]

• Now, apply the probability formula:
\[ P(E) = \frac{n(E)}{n(S)} = \frac{6}{36} = \frac{1}{6} \]

Step 4: Final Answer:
The probability of getting a sum divisible by 6 is $\frac{1}{6}$, which corresponds to Option (A).