Probability Theory

Step 1: Understanding the Concept:
This is a conditional probability problem. We are asked to find the probability of an event A (even number on the first throw) happening, given that another event B (the sum of throws is 8) has already happened. The sample space is reduced to only the outcomes where the sum is 8.

Step 2: Key Formula or Approach:
The conditional probability of event A given event B is:
Step 3: Detailed Explanation:
Let's define the events:

Event B (the given condition): The sum of the two throws is 8.
Event A: An even number turns up on the first throw. \end{itemize} First, we list all possible outcomes where the sum of the two throws is 8. This is our reduced sample space (event B). The possible pairs are (first throw, second throw): The total number of outcomes in event B is .
Next, we identify which of these outcomes also satisfy event A (an even number on the first throw). These are the outcomes in the intersection of A and B ( ). Looking at the set B, the outcomes with an even number on the first throw are: The number of outcomes in is .
Now, we can calculate the conditional probability:
Step 4: Final Answer:
The probability that an even number turns up at the first throw, given that the sum is 8, is .

Step 1: Understanding the Concept:
This problem involves calculating the probability of the union of two events. Since the two events (drawing an ace and drawing a club) are not mutually exclusive (there is an ace of clubs), we must use the addition rule of probability.

Step 2: Key Formula or Approach:
The probability of event A or event B occurring is given by: where represents the event that both A and B occur.

Step 3: Detailed Explanation:
Let's define the events:

Event A: The card drawn is an ace.
Event B: The card drawn is a club. \end{itemize} A standard deck has 52 cards.

There are 4 aces in the deck. So, the probability of drawing an ace is .

There are 13 clubs in the deck. So, the probability of drawing a club is .
\end{itemize} The events are not mutually exclusive because there is one card that is both an ace and a club: the Ace of Clubs.

The event is drawing the Ace of Clubs. The probability of this is .
\end{itemize} Now, we apply the addition rule:
Step 4: Final Answer:
The probability of getting either an ace or a club is .