Question

Two dice are thrown together. The probability that the sum of the numbers is divisible by 2 or 3 is

• A. \( \dfrac{1}{6} \)
• B. \( \dfrac{3}{4} \)
• C. \( \dfrac{1}{3} \)
• D. \( \dfrac{2}{3} \)

Hint:

For probability involving “or”, always use the inclusion–exclusion principle to avoid double counting.

Answer 1

The Correct Option is D

Step 1: Total number of outcomes.
When two dice are thrown, total possible outcomes are \[ 6 \times 6 = 36 \] Step 2: Sum divisible by 2.
A sum is divisible by 2 if it is even.
Number of outcomes giving even sum = 18.
Step 3: Sum divisible by 3.
Possible sums divisible by 3 are \(3,6,9,12\).
Number of outcomes giving these sums = 12.
Step 4: Subtract common cases.
Sums divisible by both 2 and 3 are divisible by 6.
Possible sums: \(6,12\).
Number of such outcomes = 6.
Step 5: Apply inclusion–exclusion principle.
\[ \text{Favourable outcomes} = 18 + 12 - 6 = 24 \] Step 6: Compute probability.
\[ P = \frac{24}{36} = \frac{2}{3} \]