A fair six-sided die is rolled twice. What is the probability of getting a 4 on the first roll and not getting a 6 on the second roll?

Introduction / Context:
This question involves two successive rolls of a fair six-sided die and asks for the probability of a specific ordered event: the first roll shows 4 and the second roll shows any number except 6. It is a typical example of a probability problem with independent events.

Given Data / Assumptions:

• A fair six-sided die with faces 1, 2, 3, 4, 5, 6 is rolled twice.
• Each roll is independent of the other.
• We are interested in the event: first roll = 4 and second roll ≠ 6.
• Each face has probability 1/6 on any single roll.

Concept / Approach:
For independent events A and B, the probability of both A and B occurring is P(A) * P(B). Here, event A is getting 4 on the first roll, and event B is not getting 6 on the second roll. We find each probability separately and multiply them. Alternatively, we can think in terms of the 36 ordered pairs of outcomes and count the favourable ones.

Step-by-Step Solution:
Event A: first roll is 4. P(A) = 1/6. Event B: second roll is not 6. There are 5 favourable faces (1, 2, 3, 4, 5). So P(B) = 5/6. Because the rolls are independent, P(A and B) = P(A) * P(B). Compute P(A and B) = (1/6) * (5/6) = 5/36. Therefore, the required probability is 5/36.

Verification / Alternative check:
We can also view this as an ordered pair problem. Total outcomes = 36 pairs (first, second). The first must be 4, giving 6 possibilities for the second roll, but we exclude the second roll equal to 6. So there are 5 favourable pairs: (4,1), (4,2), (4,3), (4,4), (4,5). Thus, favourable outcomes = 5, total outcomes = 36, giving probability 5/36, consistent with the multiplication approach.

Why Other Options Are Wrong:
1/36: This would be the probability of a single specific ordered pair, such as (4,6), not of 5 different favourable pairs.
1/12 and 1/9: These values do not equal 5/36 when simplified and correspond to different counts of favourable outcomes.
None of these: This is incorrect because 5/36 is provided as one of the answer choices and is correct.

Common Pitfalls:
A typical mistake is to confuse "not 6" with "greater than 4" or some other condition, which changes the number of favourable outcomes. Another error is to forget independence and try to treat the probabilities additively. Always remember that for independent events, the joint probability is the product of the individual probabilities.

Final Answer:
Hence, the probability of getting a 4 on the first roll and not getting a 6 on the second roll is 5/36.