This question asks for the probability of two independent events occurring in sequence: rolling a '1' on the first throw and rolling a '4' on the second throw of a fair six-sided dice.

Calculating Dice Roll Probability

Step 1: Probability of the First Event

The dice is fair and has six faces labelled '1' through '6'. The total number of possible outcomes for a single roll is 6.

• The desired outcome for the first roll is getting a '1'. There is only one way for this to happen.
• The probability of rolling a '1' on the first roll is calculated as: $ P(\text{1st roll is 1}) = \frac{\text{Number of ways to roll a 1}}{\text{Total number of outcomes}} = \frac{1}{6} $

Step 2: Probability of the Second Event

Similarly, for the second roll:

• The desired outcome is rolling a '4'. There is only one way for this to happen.
• The probability of rolling a '4' on the second roll is: $ P(\text{2nd roll is 4}) = \frac{\text{Number of ways to roll a 4}}{\text{Total number of outcomes}} = \frac{1}{6} $

Step 3: Combined Probability of Independent Events

Since the two dice rolls are independent events (the outcome of the first roll does not affect the outcome of the second roll), the probability of both events occurring in the specified sequence is the product of their individual probabilities.

• Combined Probability = $P(\text{1st roll is 1}) \times P(\text{2nd roll is 4})$
• $ P(\text{1st is 1 and 2nd is 4}) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} $

Therefore, the probability of getting a '1' on the first roll and a '4' on the second roll is $\frac{1}{36}$.