Question

Two identical cubic dice are rolled simultaneously. Which of the following is the probability that at least one of the face values of the two dice is greater than 3?

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

Hint:

Use the complement rule: find the probability both dice show 3 or less, then subtract from 1.

Answer 1

The Correct Option is B

Step 1: Set up direct counting:
Let the two dice show values $x$ and $y$, each ranging from 1 to 6, giving a total of 36 equally likely outcomes when the dice are rolled together. We need to count the outcomes where $x > 3$ or $y > 3$, that is, at least one of the two values is 4, 5 or 6.
Step 2: Count outcomes where the first die is greater than 3:
If $x \in \{4, 5, 6\}$, there are 3 choices for $x$ and 6 choices for $y$, giving $3 \times 6 = 18$ outcomes.
Step 3: Count outcomes where the second die is greater than 3:
Similarly, if $y \in \{4, 5, 6\}$, there are 6 choices for $x$ and 3 choices for $y$, giving $6 \times 3 = 18$ outcomes.
Step 4: Remove the double counted outcomes:
Outcomes where both $x > 3$ and $y > 3$ are counted twice in the above two steps. There are $3 \times 3 = 9$ such outcomes. Using the inclusion exclusion principle, the number of favourable outcomes is $18 + 18 - 9 = 27$.
Step 5: Compute the probability:
The probability that at least one face value is greater than 3 is $\dfrac{27}{36} = \dfrac{3}{4}$, which matches the value obtained by the complement method and confirms option B as correct.
Final Answer:
\[ \boxed{P = \dfrac{3}{4}} \]