If y − 3 = 3x, which of the following sets represents possible inputs and outputs of the function, represented as ordered pairs?A. {(0, 3), (1, 6), (2, 9)}
B. {(3, 0), (6, 1), (9, 2)}
C. {(1, 3), (2, 6), (3, 9)}
D. {(3, 1), (6, 2), (9, 3)}

To find the probability that exactly n cards are dealt before the first ace appears, we can use the concept of a geometric distribution. In a geometric distribution, we're interested in the number of trials (in this case, card draws) required for a success to occur (in this case, drawing an ace) for the first time.

The probability of drawing an ace in a single draw from a well-shuffled pack of 52 cards is 4/52 because there are 4 aces out of 52 cards.

So, the probability of drawing a non-ace in a single draw is (52 - 4)/52 = 48/52.

Now, let X be the random variable representing the number of cards drawn before the first ace appears. X follows a geometric distribution with parameter p, where p is the probability of success on a single trial.

P(X = n) = (1 - p)^(n - 1) * p

In this case, p is the probability of drawing an ace on a single trial, which is 4/52, and n is the number of cards drawn before the first ace.

So, the probability that exactly n cards are dealt before the first ace appears is:

P(X = n) = (1 - 4/52)^(n - 1) * (4/52)

Now, to find the probability that exactly k cards are dealt in all before the second ace appears, we need to consider two scenarios:

1. The first ace appears on the nth card, and the second ace appears on the kth card after that. This is represented by P(X = n) * P(X = k).

2. The first ace appears on the kth card, and the second ace appears on the nth card after that. This is represented by P(X = k) * P(X = n).

So, the total probability that exactly k cards are dealt before the second ace appears is:

P(X = n) * P(X = k) + P(X = k) * P(X = n)

You can calculate this probability using the formula for the geometric distribution with p = 4/52 as mentioned earlier for both P(X = n) and P(X = k).