In a single throw with two fair dice, what is the probability that neither die shows an even number and neither die shows a multiple of 3?

Introduction / Context:
This probability question involves two fair six sided dice. We are asked to find the probability that on a single simultaneous throw, each die shows a number that is neither even nor a multiple of 3. In other words, both dice must land on numbers that are odd and not multiples of 3.

Given Data / Assumptions:

• Each die is fair and has faces numbered 1 to 6.
• The dice are thrown simultaneously, and each outcome pair is equally likely.
• A valid outcome must satisfy, for each die separately, that the number is not even and not a multiple of 3.

Concept / Approach:
We first identify which numbers on a single die are neither even nor multiples of 3. Then we count the number of favourable outcomes for the pair of dice and divide by the total number of possible outcomes, which is 6 * 6 = 36. This uses basic counting and independence of the two dice.

Step-by-Step Solution:
Numbers on a die: 1, 2, 3, 4, 5, 6.Even numbers are 2, 4, 6. Multiples of 3 are 3 and 6.Numbers that are not even and not multiples of 3 must be odd and not divisible by 3.Check: 1 is odd and not a multiple of 3, 3 is a multiple of 3, 5 is odd and not a multiple of 3.Therefore, the allowed values on each die are 1 and 5, so there are 2 favourable numbers per die.Since the dice are independent, the number of favourable ordered outcomes for the pair is 2 * 2 = 4.Total possible ordered outcomes when throwing two dice = 6 * 6 = 36.So the required probability = favourable / total = 4 / 36 = 1 / 9 after simplification.

Verification / Alternative check:
We can explicitly list the favourable outcomes: (1,1), (1,5), (5,1) and (5,5). There are exactly four. Since there are 36 total ordered outcomes, the fraction is 4 / 36 which again reduces to 1 / 9, confirming the calculation.

Why Other Options Are Wrong:
The probabilities 11/36 and 25/36 are much larger than 1/9 and correspond to including outcomes where one or both dice show even numbers or multiples of 3, which violates the condition. The value 5/6 is far too large and would suggest that almost all outcomes are favourable, which is incorrect because only 4 out of 36 outcomes meet the strict condition. Only 1/9 matches the correct counting.

Common Pitfalls:
Some learners misinterpret the wording and only require that one die avoids even numbers and the other avoids multiples of 3, instead of imposing the condition on each die separately. Others forget that 3 and 6 are both multiples of 3, leading to too many allowed values. Carefully checking each number against both conditions prevents these misunderstandings.

Final Answer:
The probability that neither die shows an even number and neither shows a multiple of 3 is 1/9.