Question

Two different dice are thrown together. Find the probability that the numbers obtained have : (i) even sum, (ii) even product.

Hint:

For product problems, it is usually easier to find the "odd" case first and subtract from total, since \( \text{odd} \times \text{odd} = \text{odd} \) is the only way to get an odd product.

Answer 1

Step 1: Understanding the Concept:
When two dice are thrown, there are \( 6 \times 6 = 36 \) total outcomes. We need to identify specific outcomes that satisfy the given conditions.
Step 2: Detailed Explanation:
Total possible outcomes \( = 36 \).
(i) For an even sum:
The sum is even if both dice show even numbers or both show odd numbers.
Odd numbers on one die: {1, 3, 5} (3 choices).
Even numbers on one die: {2, 4, 6} (3 choices).
Case 1: Both Odd \( \to 3 \times 3 = 9 \) outcomes.
Case 2: Both Even \( \to 3 \times 3 = 9 \) outcomes.
Total favorable outcomes \( = 9 + 9 = 18 \).
\[ P(\text{even sum}) = \frac{18}{36} = \frac{1}{2} \]
(ii) For an even product:
The product is even if at least one die shows an even number.
The product is odd ONLY if both dice show odd numbers.
Total odd product outcomes (both odd) \( = 3 \times 3 = 9 \).
Favorable outcomes (even product) \( = \text{Total} - \text{Odd product} = 36 - 9 = 27 \).
\[ P(\text{even product}) = \frac{27}{36} = \frac{3}{4} \]
Step 4: Final Answer:
(i) Probability of even sum is \(\frac{1}{2}\). (ii) Probability of even product is \(\frac{3}{4}\).