Understanding the Problem: Probability with Two Die Throws

The question asks for the probability of getting at least one six when a standard six-sided die is thrown twice. This means we want to find the likelihood of either getting a six on the first throw, or on the second throw, or on both throws.

Determining the Sample Space

When a standard die is thrown once, there are 6 possible outcomes (1, 2, 3, 4, 5, 6). When it is thrown twice, the total number of possible outcomes is the product of the outcomes for each throw.

Total number of outcomes = Number of outcomes on 1st throw × Number of outcomes on 2nd throw

Total number of outcomes = $6 \times 6 = 36$.

We can list these outcomes as pairs $(a, b)$, where 'a' is the result of the first throw and 'b' is the result of the second throw. For example, (1, 1), (1, 2), ..., (6, 6).

Calculating Probability using the Complement Event

The event "at least one six" is the opposite (complement) of the event "no sixes in two throws". It is often easier to calculate the probability of the complement event and subtract it from 1 (the total probability).

Let A be the event "at least one six".

Let A' be the event "no sixes in two throws".

The probability of A is $P(A) = 1 - P(A')$.

Calculating the Probability of No Sixes

For a single throw, the outcomes that are not a six are {1, 2, 3, 4, 5}. There are 5 such outcomes.

The probability of not getting a six on a single throw is $\frac{\text{Number of outcomes that are not a six}}{\text{Total number of outcomes}} = \frac{5}{6}$.

Since the two throws are independent events, the probability of not getting a six on the first throw AND not getting a six on the second throw is the product of their individual probabilities.

$P(\text{no six on 1st throw and no six on 2nd throw}) = P(\text{no six on 1st throw}) \times P(\text{no six on 2nd throw})$

$P(A') = \frac{5}{6} \times \frac{5}{6} = \frac{25}{36}$.

Calculating the Probability of At Least One Six

Now we can find the probability of getting at least one six:

$P(A) = 1 - P(A')$

$P(A) = 1 - \frac{25}{36}$

To subtract, find a common denominator:

$P(A) = \frac{36}{36} - \frac{25}{36}$

$P(A) = \frac{36 - 25}{36}$

$P(A) = \frac{11}{36}$.

Alternative Method: Listing Favorable Outcomes

We can also list the outcomes where at least one six appears:

• Outcomes with a six on the first throw: (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) - 6 outcomes
• Outcomes with a six on the second throw (excluding (6,6) which is already counted): (1,6), (2,6), (3,6), (4,6), (5,6) - 5 outcomes

Total number of favorable outcomes (at least one six) = $6 + 5 = 11$.

The probability is $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{11}{36}$.

Both methods give the same result.

Summary of Probability Calculation

Total possible outcomes when rolling a die twice = 36.

Outcomes with no sixes = $5 \times 5 = 25$.

Outcomes with at least one six = Total outcomes - Outcomes with no sixes = $36 - 25 = 11$.

Probability of at least one six = $\frac{\text{Number of outcomes with at least one six}}{\text{Total number of outcomes}} = \frac{11}{36}$.

Revision Table: Key Probability Concepts

Additional Information: Understanding Probability of Multiple Events

When dealing with probability questions involving multiple events, like throwing a die multiple times, it's important to identify whether the events are dependent or independent. In this case, the outcome of the first die roll does not affect the outcome of the second die roll, so they are independent events.

The total number of outcomes in a sequence of independent events is found by multiplying the number of outcomes for each individual event. If you throw a die 'n' times, the total number of outcomes is $6^n$.

The phrase "at least one" in probability questions often suggests using the complement approach, as calculating the probability of "none" is frequently simpler than calculating the probability of "one OR two OR ... OR n".

Understanding how to calculate probabilities of simple events and how to combine them for multiple independent events is fundamental in probability theory.