A fair coin is tossed once. If the coin shows a tail, then a fair six faced die is tossed once. Given that the first toss of the coin results in a tail (so the die is definitely tossed), what is the probability that the die shows a number greater than 4?

Introduction / Context:
This question links a coin toss to a subsequent die roll. The die is tossed only if the first coin toss shows a tail. We are told that the first toss results in a tail, so the die is definitely rolled. The problem then reduces to a simple probability question about a fair die: what is the chance that the number obtained is greater than 4? This tests understanding of conditional probability in a context where the condition fixes which part of the experiment actually happens.

Given Data / Assumptions:

A fair coin is tossed once.

If the coin shows a tail, a fair six faced die is tossed once.

We are given that the first coin toss resulted in a tail, so the die is certainly rolled.

The die is fair, with faces numbered 1 to 6.

We want the probability that the die shows a number greater than 4.

Concept / Approach:
Once we know that the first toss was a tail, the event concerning the coin is already fixed and has probability 1 in the conditional world we are considering. The remaining random part of the experiment is the die roll. For a fair six faced die, the outcomes greater than 4 are 5 and 6. Each face has equal probability 1/6, so we can directly count favourable outcomes and divide by the total number of outcomes for the die.

Step-by-Step Solution:
Step 1: Possible outcomes of a fair die are the numbers 1, 2, 3, 4, 5 and 6.Step 2: The event of interest is that the die shows a number greater than 4.Step 3: Numbers greater than 4 on the die are 5 and 6.Step 4: Number of favourable outcomes for the die = 2 (namely 5 and 6).Step 5: Total possible outcomes when the die is rolled = 6.Step 6: Probability that the die shows a number greater than 4 = favourable / total = 2 / 6.Step 7: Simplify 2 / 6 by dividing numerator and denominator by 2 to get 1 / 3.

Verification / Alternative check:
If we were to consider the entire experiment including the coin, we would first note that the die is rolled only in half of all experimental runs (when the coin shows tail). However, the question conditions on the event that the first toss is tail, effectively restricting attention to exactly those runs. Within that restricted set, the die behaviour is unaffected and remains uniformly random over its six faces. Therefore, the unconditional structure does not change the conditional probability 1/3 for outcomes 5 or 6.

Why Other Options Are Wrong:
The value 2/3 would be correct if four outcomes of the die were considered favourable, which is not the case here. The probability 2/5 is associated with a five outcome sample space, which does not fit a standard die. The fraction 4/15 and 1/6 both arise from mixing coin and die stages incorrectly without recognising the conditioning; they no longer apply once we know that the die has indeed been rolled.

Common Pitfalls:
Some learners mistakenly include the coin toss probability after conditioning, for example, multiplying 1/2 for tail with 1/3 for the die and obtaining 1/6, which would be the joint probability of tail and a number greater than 4 without conditioning. However, the question already tells us that a tail occurred, so we must work within that conditional universe. Always read the phrase "given that" as a clear signal to condition on an event, which may simplify what remains.

Final Answer:
Given that the first coin toss is a tail, the probability that the die shows a number greater than 4 is 1/3.