Understanding Probability with Dice Throws

This problem involves calculating the probability of two independent events occurring when a fair dice is thrown twice. The two events are: getting a multiple of 3 in the first throw, and getting an odd number in the second throw.

Sample Space for a Single Dice Throw

When a standard six-sided dice is thrown, the possible outcomes are {1, 2, 3, 4, 5, 6}. The total number of possible outcomes for a single throw is 6.

Probability of a Multiple of 3 in the First Throw

Let event A be getting a multiple of 3 in the first throw. The multiples of 3 in the sample space {1, 2, 3, 4, 5, 6} are {3, 6}.

• Number of favourable outcomes for event A = 2 (getting a 3 or a 6).
• Total number of possible outcomes = 6.

The probability of event A is calculated as:

\(P(A) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}\)

\(P(A) = \frac{2}{6} = \frac{1}{3}\)

Probability of an Odd Number in the Second Throw

Let event B be getting an odd number in the second throw. The odd numbers in the sample space {1, 2, 3, 4, 5, 6} are {1, 3, 5}.

• Number of favourable outcomes for event B = 3 (getting a 1, 3, or 5).
• Total number of possible outcomes = 6.

The probability of event B is calculated as:

\(P(B) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}\)

\(P(B) = \frac{3}{6} = \frac{1}{2}\)

Calculating the Combined Probability for Independent Events

Since the outcome of the second throw does not depend on the outcome of the first throw, the two events (getting a multiple of 3 in the first throw and getting an odd number in the second throw) are independent events.

The probability of two independent events A and B both occurring is given by the product of their individual probabilities:

\(P(A \text{ and } B) = P(A) \times P(B)\)

Substituting the probabilities we calculated:

\(P(\text{multiple of 3 in first throw and odd number in second throw}) = P(A) \times P(B)\)

\(P(\text{multiple of 3 and odd number}) = \frac{1}{3} \times \frac{1}{2}\)

\(P(\text{multiple of 3 and odd number}) = \frac{1 \times 1}{3 \times 2} = \frac{1}{6}\)

Thus, the probability of getting an odd number in the second throw and a multiple of 3 in the first throw is \(\frac{1}{6}\).

Revision Table: Probability Concepts

Additional Information: Understanding Independent vs. Dependent Events

It's important to distinguish between independent and dependent events when calculating probabilities. In this dice problem, the two throws are independent because the result of the first throw has no influence on the result of the second throw. The dice doesn't remember the previous result!

• Independent Events: \(P(A \text{ and } B) = P(A) \times P(B)\). Example: Flipping a coin twice, throwing a dice twice.
• Dependent Events: The occurrence of one event affects the probability of the other. For dependent events, \(P(A \text{ and } B) = P(A) \times P(B|A)\), where \(P(B|A)\) is the probability of event B occurring given that event A has already occurred. Example: Drawing cards from a deck without replacement. The probability of drawing a certain card on the second draw depends on what was drawn first.

In this specific problem, applying the rule for independent events is correct because the dice throws are separate, unconnected actions.