Given a fair six-faced dice where the faces are labelled ‘1’, ‘2’, ‘3’, ‘4’, ‘5’, and ‘6’, what is the probability of getting a ‘1’ on the first roll of the dice and a ‘4’ on the second roll?
This question involves calculating the probability of two separate, independent events happening in sequence when rolling a fair six-faced dice.
Probability is a measure of how likely an event is to occur. It is calculated as:
$$ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} $$
We are dealing with a fair six-faced dice, meaning each face (1, 2, 3, 4, 5, 6) has an equal chance of appearing on any given roll. The total number of possible outcomes for a single roll is 6.
The event we are interested in for the first roll is getting a '1'.
$$ P(\text{Roll is 1}) = \frac{1}{6} $$
The event we are interested in for the second roll is getting a '4'.
$$ P(\text{Roll is 4}) = \frac{1}{6} $$
Since the two dice rolls are independent events (the outcome of the first roll does not affect the outcome of the second roll), we can find the probability of both events occurring by multiplying their individual probabilities.
$$ P(\text{1 on first roll AND 4 on second roll}) = P(\text{Roll is 1}) \times P(\text{Roll is 4}) $$
Substituting the probabilities we found:
$$ \text{Probability} = \frac{1}{6} \times \frac{1}{6} $$
$$ \text{Probability} = \frac{1}{36} $$
The probability of getting a '1' on the first roll and a '4' on the second roll of a fair six-faced dice is $\frac{1}{36}$.
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